Transcript
Fundamentals of Electrical Engineering / Electronics
Alternating Current Technology
4th Edition – V 0102
Fundamentals of Electrical Engineering / Electronics
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Code: 0.3.6
Alternating Current Technology Preface
Preface This manual
Alternating Current Technology is designed for use in conjunction with hps training equipment, to provide information through experimentation on the properties of the most important standard circuits. The individual topics are divided into the following sections: •
General
•
Experiments, consisting of the set tasks (experiments) and the procedures to be used.
The section General contains a short introduction to the respective experiments. Because of the intensive subject matter involved, a detailed theoretical description has been purposely dispensed with. We recommend you to read the commercially available textbooks for more profound theoretical knowledge and as background reading to the experiments. All tables and diagrams required in the experiments are provided, in order to simplify procedures. The Appendix to this manual contains a detailed section of solutions to the tasks and questions set in the experiments and enables you to check your own results. hps SystemTechnik offers several training systems for conducting the experiments. These systems are geared to individual requirements.
Alternating Current Technology
I
Contents
Contents 1
Establishing and Displaying Characteristics in AC Technology ....................................................5
1.1
General...................................................................................................................................................5
1.2
Characteristics of the Sine-Wave Voltage .............................................................................................5
1.3
Active Power with Sine-Wave Voltage ...................................................................................................9
1.4
Characteristics of Square-Wave AC Voltage .......................................................................................13
2
Three-Phase Alternating Current..................................................................................................... 13
2.1
General................................................................................................................................................ 13
2.2
Potential Gradient in Three-Phase Current Systems.......................................................................... 15
2.3
Loads in Star Circuit ............................................................................................................................ 19
2.4
Loads in Delta Circuit .......................................................................................................................... 23
2.5
Measurements on a Defective Star Circuit ......................................................................................... 27
2.6
Measurements on a Defective Delta Circuit........................................................................................ 31
3
Capacitor in the AC Circuit .............................................................................................................. 35
3.1 3.2
General................................................................................................................................................ 35 Charging and Discharging Process of a Capacitor ............................................................................. 37
3.3
Phase Shift between Current and Voltage on the Capacitor .............................................................. 41
3.4
Capacitive Reactance of a Capacitor.................................................................................................. 43
3.5
Series Circuiting of Capacitors............................................................................................................ 45
3.6
Parallel Circuiting of Capacitors .......................................................................................................... 47
3.7
Reactive Power of a Capacitor............................................................................................................ 49
4
Coil in the AC Circuit ........................................................................................................................ 53
4.1
General................................................................................................................................................ 53
4.2
Switch-On and Switch-Off Process of a Coil....................................................................................... 55
4.3
Phase Shift between Current and Voltage on a Coil........................................................................... 59
4.4
Inductive Reactance of a Coil ............................................................................................................. 61
4.5
Series Circuiting of Coils ......................................................................................................................63
4.6
Parallel Circuiting of Coils ....................................................................................................................65
4.7
Reactive Power of a Coil ......................................................................................................................67
II
Alternating Current Technology Contents
5
Interconnecting Resistor, Capacitor and Coil .................................................................................71
5.1
General .................................................................................................................................................71
5.2
Series Circuiting of Resistor and Capacitor..........................................................................................71
5.3
Parallel Circuiting of Resistor and Capacitor........................................................................................75
5.4
Series Circuiting of Resistor and Coil...................................................................................................79
5.5
Parallel Circuiting of Resistor and Coil .................................................................................................83
5.6
Series Circuiting of Capacitor and Coil.................................................................................................87
5.7
Parallel Circuiting of Capacitor and Coil...............................................................................................91
5.8
Series Circuiting of Resistor, Capacitor and Coil .................................................................................95
5.9
Parallel Circuiting of Resistor, Capacitor and Coil ...............................................................................99
5.10
Active, Reactive and Apparent Power................................................................................................103
6
Transformers ....................................................................................................................................107
6.1
General ...............................................................................................................................................107
6.2
Coupling Factor ..................................................................................................................................107
6.3
Transformation Ratio ..........................................................................................................................109
6.4
Resistance Transformation ................................................................................................................111
Alternating Current Technology
7
Establishing and Displaying Characteristics in AC Technology
1
Establishing and Displaying Characteristics in AC Technology
1.1
Fundamentals
sinusoidal voltage
+U 0
t
Alternating current changes its direction of flow continuously, in contrast to direct current which always flows in the same directi-
-U
square-wave voltage
+U 0
t
on. The wave (function) of the current or required voltage respectively may adopt different shapes. Figure 1.1.1 shows some of the
-U
delta voltage
+U
common functions in electrical and electronics engineering. In AC technology a distinc-
0
t
-U
sawtooth voltage
+U
tion is made additionally between singlephase and multiphase AC voltages and currents. The public power supply, for example,
0
t
-U
trapezoidal voltage
+U
supplies a three-phase alternating current as a rule. The following experiments are restricted to
0
t -U
Figure 1.1.1
the sine-wave and square-wave voltages which are the most frequently occuring forms in electrical and electronic engineering.
They deal with the characteristics frequency, root mean square value, peak value, peak-to-peak value, phase shift and power. Experiments in three-phase alternating current are conducted separately in the next chapter.
1.2
Characteristics of the Sine-Wave Voltage
1.2.1 General If a sine-wave AC voltage is applied to an ohmic resistor, it generates an AC current in it with the same function.
Explanation of variables: ip = peak value of current (positive or negative) in A ipp = peak-to-peak value of current in A Irms = root mean square value of current in A
8
Alternating Current Technology Establishing and Displaying Characteristics in AC Technology
T
+U
= period duration in s up = peak value of voltage in V, amplitude, peak value (positive or negative)
+ up
u pp
upp = peak-to-peak value of voltage in V
U rms
Urms = root mean square value of voltage in V
0
t -up
T
-U Figure 1.2.1.1
Figure 1.2.1.1 shows the function in time of a sinewave AC voltage. During one period the sine-wave voltage reaches its positive maximum value after every quarter period, crosses the zero point, reaches its negative maximum value and then returns to zero. Other characteristic values and the formulae for calculating them are specified below.
Frequency f in Hz The frequency f gives the number of periods per second.
f =
1 T
1 Hz =
Radian frequency
1 1s in 1/s
The radian frequency ω is the product of 2 . π and the frequency. ω=
2.π.f
Momentary values u, i
u = up . sin (ω t) Wavelength
i = ip . sin (ω t) in m (km)
The wavelength results from the propagation speed (v in m/s) and the frequency f. λ=
v f
Root mean square values U rms, Irms This is the root mean square value of a sinusoidal shaped voltage or current.
U rms = u p ⋅
1 2
I rms = i p ⋅
1 2
Alternating Current Technology
9
Establishing and Displaying Characteristics in AC Technology
1.2.2 Experiments Experiment Display the sine-wave voltage on an ohmic resistor with an oscilloscope and establish the following values from the screen image.
A
Y1
up = peak voltage ip = peak current
G
U
R 1 kΩ
upp = peak-to-peak voltage Urms= root mean square voltage
0V
B
⊥
oscilloscope Figure 1.2.2.1
Irms = root mean square current T = period duration f
= frequency
ω = radian frequency λ = wavelength
u = momentary value (after a third of a period)
Experiment procedure •
Set up the experiment according to circuit (figure 1.2.2.1), connect the function generator and connect the oscilloscope to test points A and B.
•
Set the oscilloscope according to the specifications made in the grid (figure 1.2.2.2).
•
Then the sine-wave AC voltage shown in the grid should be set on the function generator and the values specified under experiment determined.
Settings: X = 0.1 ms / division Y1 = 1 V / division
- 0 (Y 1 )
Figure 1.2.2.2
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Alternating Current Technology Establishing and Displaying Characteristics in AC Technology
Peak voltage u p
Period duration T
up =
T=
Peak current ip
Frequency f
ip =
up
=
f =
R
1 = T
Peak-to-peak voltage upp
Radian frequency
u pp = 2 ⋅ u p
ω = 2⋅π ⋅ f
Root mean square voltage U rms
Wavelength (at a propagation speed of 300 · 10 m/s) 6
1 U rms = u p ⋅
2
Root mean square current I rms
I rms =
U rms R
=
λ=
v = f
Momentary voltage u after a third of a period Note of solution: ωt is the angle with the unit „rad“
u = up . sin (ω t) =
Alternating Current Technology
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Establishing and Displaying Characteristics in AC Technology
1.3
Active Power with Sine-Wave Voltage
1.3.1 General
U I
u i
0
t
Figure 1.3.1.1
If a sine-wave voltage is applied to a purely ohmic load a sinusoidal current flows through the load. AC current and AC voltage are in phase, i. e. they both achieve their positive and negative maximum value at the same time. The power which is converted in the purely ohmic load is known as the active power. It results from the product of current and voltage. The power curve can be displayed by multiplying the momentary values of voltage and current in a line diagram.
P =I U .
p=u
P = I2 ⋅R
P=
U2 R
(root mean square values)
P; p
= active power in W
U; u
= voltage in V
I; i
= current in A
R
= ohmic resistance in Ω
. i (momentary values)
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Alternating Current Technology Establishing and Displaying Characteristics in AC Technology
1.3.2 Experiments Experiment Scan the voltage and current on an ohmic resistor with an oscilloscope, transcribe the curves onto the given diagram and construct the power curve by multiplying the momentary values. Experiment procedure •
Set up the experiment according to circuit (figure 1.3.2.1), connect the function generator and set the following voltage with the oscilloscope: up = 7 V (sinusoidal); f = 1 kHz A U
G
Y1
•
test point B to channel 2 (Y2), inverse - test point C to ground
R 1 kΩ •
up = 7 V
C
(sinusoidal)
UM
⊥
RM 100 Ω
Settings on the oscilloscope: - Sweep X: 0,1 ms / division - channel 1 (Y1): 2 V / division (voltage U) - channel 2 (Y2): (measuring voltage UM
B
0V
Connect the oscilloscope: - test point A to channel 1 (Y 1)
Y2
0.5 V / division (invert) ⇒
current I A)
- Center the zero line of channels 1 and 2. - Triggering: Y1
oscilloscope
Figure 1.3.2.1
•
Draw the scanned voltages U and UM in the given diagram (figure 1.3.2.2).
•
Determine the momentary values u, u M, i and p at the times specified in table 1.3.2.2 and then construct the power curve in the diagram (figure 1.3.2.2).
Attention: Make sure that the test point C is not connected to test points B or A through the ground of the devices used (function generator, oscilloscope). Use an isolating transformer if necessary.The 100 Ω resistor in the circuit serves as a measuring resistor. The voltage U M at this resistor is proportional to the current flowing through both resistors. The current is calculated with the following formula:
I=
UM UM = RM 100 Ω
(Ohm’s Law)
The reference point of the two voltages has been put between the resistors 1 k Ω and 100 Ω for simultaneous representation of the voltage U and the measuring voltage U M. It must be taken into account that the two voltages are displayed with a 180° phase shift. The correct representation of the voltage curves is achieved by inverting one of the two voltages with the oscilloscope (in the experiment voltage UM; channel 2; Y2).
Alternating Current Technology
13
Establishing and Displaying Characteristics in AC Technology
Time t [ms]
Voltage u [V]
Voltage uM [V]
Current i [mA]
Active power p [mW]
0 0.1 0.15 0.25 0.35 0.4 0.5 0.6 0.65 0.75 0.85 0.9 1.0 Table 1.3.2.1
P [ mW ]
U [V]
I [ mA]
40
20
30 20
6 10
10 0
8
4 2
0
0 0.1
-10 -20
-2 -10
-30 -40
-6 -20
-50 -60
Figure 1.3.2.2
-4
-8 -10
-30
-12
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
t [ ms]
14
Alternating Current Technology Establishing and Displaying Characteristics in AC Technology
Question 1:
What does the power curve tell us?
Answer:
Question 2:
What is the root mean square value of the power?
Answer:
P=U.I
P= or
P=
up ⋅ ip
=
2
U = up ⋅
1
=
2
I = ip ⋅ 1 = 2
Alternating Current Technology
15
Establishing and Displaying Characteristics in AC Technology
1.4
Characteristics of the Square-Wave AC Voltage
1.4.1 General Figure 1.4.1.1 shows the function in time of a square-wave AC voltage. If such a voltage is applied to an ohmic load a DC current flows through it which changes its direction at certain intervals. T +U
= period duration in s
up = peak voltage in V (positive or negative) upp = peak-to-peak voltagein V u pp
+u p
Correspondingly for the current: ip = peak currentin A (positive or negative)
0 t
ipp = peak-to-peak current in A
-u p
-U
T
Figure 1.4.1.1
If the time intervals of the change in direction are equal and if the voltages acting in both directions are also equal one refers to a symmetrical square-wave voltage. Other characteristic values and the formulas for their calculation are specified below.
Root mean square values U rms und Irms In a symmetrical square-wave voltage or current respectively the root mean square value for the voltage and current is equal to the peak values.
Urms = up
Irms = ip
Power P in W (only on ohmic loads) This is calculated using the rms values of the voltage and current. 2 P = I rms ⋅R
P = U rms ⋅ I rms
P=
2 U rms
R
Frequency f in Hz The frequency f gives the number of periods per second.
f =
1 T
1 Hz =
1 s
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Alternating Current Technology Establishing and Displaying Characteristics in AC Technology
1.4.2 Experiments Experiment Display the square-wave AC voltage on an ohmic resistor with the oscilloscope and establish the following values from the screen image. A
G
U (square wave)
Y1
up
= peak voltage
ip = peak current Urms, Irms = root mean square voltage and current
R 1 kΩ
B
⊥
P
= power
T
= period duration
f
= frequency
0V
oscilloscope
Figure 1.4.2.1
Experiment procedure •
Set up the experiment according to circuit (figure 1.4.2.1), connect the function generator and connect the oscilloscope to test points A and B.
•
Set the oscilloscope according to the specifications made in the grid (figure 1.4.2.2).
•
Then the square-wave AC voltage shown in the grid (figure 1.4.2.2) should be set on the function generator in connection with the oscilloscope and the values of the characteristics specified under experiment determined.
Settings: X = 0.1 ms / division Y1 = 1 V / division
- 0 (Y 1 )
Figure 1.4.2.2
Alternating Current Technology
17
Establishing and Displaying Characteristics in AC Technology
Peak voltage up
Peak-to-peak voltage upp
up =
upp = 2 . up =
Peak current ip
ip =
up
=
R
Power P
P = Urms . Irms = p = up . ip =
Root mean square voltage U rms
Period duration T
Urms = up =
T=
Root mean square current Irms
Frequency f
Irms = ip =
f = 1 = T
18
Alternating Current Technology Establishing and Displaying Characteristics in AC Technology
Notes:
Alternating Current Technology
19
Three-Phase Alternating Current
2
Three-Phase Alternating Current
2.1
Fundamentals U1
U2
U1
U3
U
0
t
° 0 2 1
1
2
0 °
120°
120°
Figure 2.1.1
U3
U2
Figure 2.1.2
Three-phase current systems result when several phase-shifted voltages are connected together. Generation of the phase-shifted, sinusoidal AC voltages is achieved by an appropriate arrangement of coils in the three-phase current generator. The most common three-phase generator which is usually used in the public power supply network supplies three sinusoidal AC voltages which are linked to each other and phase-shifted by 120°. Figure 2.1.1 shows the three voltages as a line diagram and figure 2.1.2 shows them as a pointer diagram.
20
Alternating Current Technology Three-Phase Alternating Current
Two basic circuit types are common in the coils of the three-phase current generator and on the load side when connecting motors, for example; the star circuit and the delta circuit (see also figure 2.1.3 and figure 2.1.4).
L1
L1 phase star point
3 L , 1 L
2 L , 1 L
U
U
L2 rotation direction N , 1 L
U
L
L
L
L
3 L , 2 L
N , 2 L
U
U N , 3 L
L2 L3
L L
U
L3
N
Figure 2.1.3 Star circuit of the generator coils
Figure 2.1.4 Delta circuit of the generator coils
L1, L2, L3
= conductor (phase)
N
= zero conductor (neutral wire)
UL1,L2 / UL2,L3 / UL3,L1
= conductor voltages between the conductors specified in the index (in V)
UL1,N / UL2,N / UL3,N
= phase voltages (star voltages) between the lines specified in the index (in V)
The conductor voltages in the delta circuit are also known as the delta voltage.
In the star circuit three conductor voltages (400 V) are available between the conductors L1, L2 and L3 as well as three phase voltages (230 V) between the conductors L1, L2 and L3 and the zero conductor (N). In the delta circuit three voltages are available between the conductors L1, L2 and L3 with 400 V each. The specified voltages are rms values. The frequency of the sinusoidal AC voltages is 50 Hz. In the following experiments the behaviour of voltages, currents and powers in star and delta circuits is investigated. The three necessary sinusoidal AC voltages are not taken directly from the mains, they are generated electronically by phase shifters. For riskless experimentation the voltage 230 V is limited to 7 V and the voltage 400 V is limited to 12 V.
Alternating Current Technology
21
Three-Phase Alternating Current
2.2
Potential Gradient in Three-Phase Current Systems
2.2.1
General L1
U
L
, 3
L
U
L
L
U
1
1
U L3 ,
N
° 0 2
1
2 0
L
2
°
U
N
1 ,
120°
L2 ,N
U L2, L3
L3
L2
Figure 12.2.1.1 Voltage pointer diagram for the three-phase current system
The three sinusoidal, 120° phase-shifted AC voltages generated by a three-phase current generator are interlinked in such a way that three conductor voltages and three phase voltages can be taken off at conductors L1, L2, L3 and N in starcircuited coils. The voltage pointer diagram (figure 2.2.1.1) shows the relationships between the conductor voltages and the phase voltages with respect to the phase relation and the voltage ratios. The conductor voltage UL and the phase voltage Uphase are linked with factor, also called the linkage factor. In the public power supply network therefore, at a phase voltage of 230 V there is a conductor voltage of:
UL
=
3 ⋅ U phase
=
1.73 ⋅ 230 V
=
400 V
22
Alternating Current Technology Three-Phase Alternating Current
2.2.2
Experiments
Experiment Display the phase voltages of a three-phase current system on the oscilloscope, draw the displayed voltage curves in a diagram (figure 2.2.2.2) and determine the angle of phase shift between the individual voltages. Then measure the phase and conductor voltages with a multimeter and establish the linkage factor. What is the peak value of the phase and conductor voltages? L1
Experiment procedure
channel 1 (Y1)
Settings on the oscilloscope:
L2
- Center zero conductor of channel 1 and channel 2
channel 2 (Y2)
G L3
3
N ground
⊥
- Sweep X:
2 ms / division
- Channel 1 (Y1):
5 V / division
- Channel 2 (Y2):
5 V / division
- Triggering:
Channel 1
oscilloscope 7 V / 12 V Figure 2.2.2.1
•
Apply voltage UL1,N to channel 1 and voltage UL2,N to channel 2 of the oscilloscope. Draw the displayed voltage curves in the provided (figure 2.2.2.2). Apply voltage UL3,N to channel 2 and draw this curve in the diagram, too.
12 U [V] 10 8 6 4 2 0 5
10
15
20
t [ms]
-2 -4 -6 -8 -10 -12
Figure 12.2.2.2
Alternating Current Technology
23
Three-Phase Alternating Current
Angle of phase shift
Linkage factor
Period duration T = 20 ms =ˆ 360°
The linkage factor results from the ratio of the con-
Time shift of individual voltages according to the dia-
ductor voltage UL to the phase voltage Uphase.
UL
gram (figure 2.2.2.2)
t
=
ϕ
=
U phase
=
Frequency f
f
=
1 T
=
Phase voltages Uphase
Peak value of the phase voltage u s
UL1,N =
up
= U phase ⋅
2
=
UL2,N =
UL3,N =
Conductor voltages UL
Peak value of the conductor voltage u s
UL1,L2 =
up
UL2,L3 =
UL3,L1 =
=UL ⋅
2
=
24
Alternating Current Technology Three-Phase Alternating Current
Notes:
Alternating Current Technology
25
Three-Phase Alternating Current
2.3
Loads in Star Circuit
2.3.1
General
If loads are connected in star circuit to a three-phase current network as shown in figure 2.3.1.1, the same current flows in lines L1, L2 and L3 at symmetrical load (R 1 = R2 = R3), the zero line is currentless.
L1
I L1
L1, L2, L3
R1 N
IN
L2
I L2
L3
I L3
= conductor (phase)
N
= zero conductor
IL1, IL2, IL3
= conductor currents in A
IN
= zero conductor current in A
R1, R2, R3 = loads R3
R2
Figure 2.3.1.1
Conductor current IL The individual conductor currents IL which are also referred to as phase currents (I phase) are calculated according to Ohm's Law with the formula:
IL
=
UR R
I phase
=
U phase R
On an unsymmetrical load a compensation current I N flows through the zero conductor.
Power Ptot The power consumption Ptot on a load in star circuit consists of the individual phase powers. In the experiment the individual phase powers are designated PR1, PR2 and PR3 corresponding to the ohmic resistors.
Ptot = PR1 + PR2 + PR3 A symmetrical load gives the following relationships:
Ptot = 3 . Pphase
Pphase = Uphase . Iphase
26
Alternating Current Technology Three-Phase Alternating Current
On inductive and capacitive load:
Active power Ptot
Ptot
=
Ptot
=
3 ⋅ U phase ⋅ I phase ⋅ cos ϕ 3 ⋅UL ⋅ IL
⋅
cos ϕ
Apparent power Stot
S tot
=
S tot
=
3 ⋅ U phase ⋅ I phase 3 ⋅UL ⋅ IL
Reactive power Qtot
Qtot
=
Qtot
=
3 ⋅ U phase ⋅ I phase ⋅ sin ϕ 3 ⋅UL ⋅ IL
⋅
sin ϕ
Alternating Current Technology
27
Three-Phase Alternating Current
2.3.2
Experiments
Experiment Measure the currents IL and IN and the voltages UL and Uphase (rms values) with a multimeter in a three-phase AC current network on a star circuited, symmetrical and unsymmetrical ohmic load and then calculate the powers Pphase and Ptot. L1
G
L2
3
L3
Ω
R1 = R2 = R3 = 1 k (symmetrical load) R1 = 1 k Ω
N I L1
I L3
IN
I L2
7 V / 12 V
R2 = 680
Ω
R3 = 330
Ω
(unsymmetrical load) R1
R3
R2
Figure 2.3.2.1
Experiment procedure •
Connect symmetrical load according to the circuit in figure 2.3.2.1.
•
Measure voltages and currents according to the following table 2.3.2.1 with the multimeter and calculate the powers.
•
Then repeat these measurements and calculations on an unsymmetrical load.
•
Enter all the measured values in the table. Star circuit
Symmetrical load IL1
Conductor currents IL, IN (Iphase)
IL2 IL3 IN UL1, L2
Conductor voltages UL
UL2, L3 UL3, L1
Phase voltages Uphase
UL1, N UL2, N UL3, N PR1
Phase powers Pphase
PR2
Total power Ptot
Ptot
PR3
Table 2.3.2.1
Unsymmetrical load
28
Alternating Current Technology Three-Phase Alternating Current
Calculating the phase power and total power: The phase powers correspond to the powers P R 1, PR 2 and PR 3.
Symmetrical load
PR1 = UL1,N . IL1 = PR2 = UL2,N . IL2 = PR3 = UL3,N . IL3 = Ptot = 3 . Pphase =
Unsymmetrical load
PR1 = UL1,N . IL1 = PR2 = UL2,N . IL2 = PR3 = UL3,N . IL3 = Ptot = PR1 + PR2 + PR3 =
Alternating Current Technology
29
Three-Phase Alternating Current
2.4
Loads in Delta Circuit
2.4.1
General
If loads are connected in delta circuit to three-phase AC mains (figure 2.4.1.1) every load is applied directly to the conductor voltage. The conductor voltage is equal to the phase voltage:
UL = Uphase The conductor currents I L1, IL2 and IL3 are divided in the branching points into the phase currents I R1; IR2; IR3.
L1
I L1
L1, L2, L3 I R1 R3
N
I R3
generator
L2
I L2
L3
I L3
R1
= conductor (phase)
N
= zero conductor
IL1, IL2, IL3
= conductor currents in A
R1, R2, R3 = loads IR1, IR2, IR3
= phase currents I phase in A
R2 I R2
loads
Figure 2.4.1.1
Phase current Iphase On symmetrical load (R 1 = R2 = R3) the phase currents are
I phase
=
3 smaller than the conductor currents:
IL 3
Power Ptot The power consumption of a load in delta circuit consists of the individual phase powers P phase. In the experiment the individual phase powers are designated PR1, PR2 and PR3 corresponding to the ohmic resistors:
Ptot = PR1 + PR2 + PR3 A symmetrical load gives the following relationships:
Ptot = 3 . Pphase Phase power Pphase The phase power Pphase results respectively from the phase current I phase and the phase voltage Uphase:
Pphase = Iphase . Uphase
30
Alternating Current Technology Three-Phase Alternating Current
Below are some more formulae which are used for calculating the powers on an inductive and capacitive symmetrical load.
Active power Ptot
Ptot = 3 . Uphase . Iphase . cos ϕ
Ptot =
3 . UL . IL . cos ϕ
Stot =
3 . UL . IL
Qtot =
3 . UL . IL . sin ϕ
Apparent power Stot
Stot = 3 . Uphase . Iphase Reactive power Qtot
Qtot = 3 . Uphase . Iphase . sin ϕ
2.4.2
Experiments
Experiment Measure the phase currents and conductor currents as well as the conductor voltages in a three-phase AC mains on a delta circuited symmetrical and unsymmetrical load and then calculate the phase power and total power. L1
G 3
R1 = R2 = R3 = 1 kΩ
L2
(symmetrical load)
L3
R1 = 1 kΩ
N I L3
I L1
I L2
7 V / 12 V
R2 = 680
Ω
R3 = 330
Ω
(unsymmetrical load) I R1 R3
R1
I R3 R2
I R2
Figure 2.4.2.1
Experiment procedure •
Connect symmetrical load according to the circuit in figure 2.4.2.1.
•
Measure voltages and currents according to the following table 2.4.2.1 with the multimeter and calculate the powers. Then repeat these measurements and calculations on an unsymmetrical load.
•
Enter all the measured values in the table.
Alternating Current Technology
31
Three-Phase Alternating Current
Delta circuit
Symmetrical load IL1
Conductor currents IL
IL2 IL3 IR1
Phase currents Iphase
IR2 IR3 UL1, L2
Conductor voltages UL (Uphase)
UL2, L3 UL3, L1 PR1
Phase powers Pphase
PR2
Total power Ptot
Ptot
PR3
Table 2.4.2.1
Calculating the phase power and total power:
Symmetrical load
PR1 = IR1 . UL1,L2 = PR2 = IR2 . UL2,L3 = PR3 = IR3 . UL3,L1 = Ptot = 3 . Pphase =
Unsymmetrical load
PR1 = IR1 . UL1,L2 = PR2 = IR2 . UL2,L3 = PR3 = IR3 . UL3,L1 = Ptot = PR1 + PR2 + PR3 =
Unsymmetrical load
32
Alternating Current Technology Three-Phase Alternating Current
Notes:
Alternating Current Technology
33
Three-Phase Alternating Current
2.5
Measurements on a Defective Star Circuit
2.5.1
General
If an outer conductor (L1, L2 or L3) fails in a star circuit connected to the three-phase AC mains, the srcinally consumed power on symmetrical load is reduced by a third. If the zero conductor also fails, the power consumption is reduced by half as a result of two loads in series being applied to the remaining conductor voltage. If two outer conductors are broken in a star circuit with symmetrical load, the power consumed is only a third of the srcinal power consumption.
2.5.2
Experiments
Experiment Measure the conductor currents, the conductor voltages and the phase voltages on a defective star circuit and then calculate the phase and total powers.
G
L1
I L1
L2
I L2
L3
I L3
N
IN
R1 = R2 = R3 = 1 kΩ
3
A
7 V / 12 V
R1 R3
D
R2
C
B
Figure 2.5.2.1
Experiment procedure •
Connect load according to the circuit (figure 2.5.2.1).
•
Make the respective break in the conductor at the generator output in accordance with table 2.5.2.1 and make the necessary measurements. Measure the phase and conductor voltages directly at test points A ... D.
•
Enter all the measured values in the table 2.5.2.1.
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Alternating Current Technology Three-Phase Alternating Current
Use the following formulae to calculate the powers:
Phase powers Pphase
Pphase = Uphase . Iphase
(e.g. PR 1 = UL1,N . IL1)
Total power Ptot
Ptot = PR 1 + PR 2 + PR 3
Conductor breaks Star circuit none
Conductor currents IL (Iphase) in mA
Conductor voltages UL in V
Phase voltages Uphase in V
Phase powers Pphase in mW
Total power Pin totmW Table 2.5.2.1
P
IL1
7.0
IL2
7.0
IL3
7.0
IN
0
UL1, L2
12.1
UL2, L3
12.1
UL3, L1
12.1
UL1, N
7.0
UL2, N
7.0
UL3, N
7.0
PR 1
49.0
PR 2
49.0
PR 3
49.0
tot
147.0
L1
L1 / L2
L1 / N
Alternating Current Technology Three-Phase Alternating Current
Calculation of the phase and total powers:
Conductor break L1
PR1 = UL1,N . IL1 = PR2 = UL2,N . IL2 = PR3 = UL3,N . IL3 =
Ptot = PR1 + PR2 + PR3 =
Conductor breaks L1 and L2
PR1 = UL1,N . IL1 = PR2 = UL2,N . IL2 = . PR3 = UL3,N IL3 = Ptot = PR1 + PR2 + PR3 =
Conductor breaks L1 and N
PR1 = UL1,N . IL1 = PR2 = UL2,N . IL2 = PR3 = UL3,N . IL3 =
Ptot = PR1 + PR2 + PR3 =
35
36
Alternating Current Technology Three-Phase Alternating Current
Notes:
Alternating Current Technology
37
Three-Phase Alternating Current
2.6
Measurements on a Defective Delta Circuit
2.6.1
General
If outer conductors or phases are broken in the case of a delta circuited load connected to the three-phase AC mains, the consumed power can be at the worst only a sixth of the maximum power. The following table 2.6.1.1 lists the power consumptions in different defective delta circuits. Break one phase
P=2
3
⋅ Pmax
two phases
P=1
3
one outer conductor
P=1
⋅ Pmax
2
one phase and one outer conductor
⋅ Pmax
P= 1
3
⋅ Pmax
P= 1
6
⋅ Pmax
Table 2.6.1.1
2.6.2
Experiments
Experiment Measure the conductor currents, the phase currents and the conductor voltages (phase voltages) on a defective delta circuit and then calculate the phase and total powers. Experiment procedure
L1
•
Connect load according to the circuit (figure
G
L2
3
L3
conductor and phase in accordance with
N
table 2.6.2.1 and make the necessary mea-
2.6.2.1). Make the respective break in the
surements. 7 V / 12 V •
A I R1 R3
R1
re the conductor voltages (phase voltages) at test points A ... C.
I R3 C R2
Figure 2.6.2.1
Make the breaks in the phase by pulling out the appropriate resistors (R1, R2, R3). Measu-
I R2
B
•
Enter all the measured values in the table 2.6.2.1.
38
Alternating Current Technology Three-Phase Alternating Current
Use the following formulas to calculate the powers:
Phase powers Pphase
Pphase = Uphase . Iphase
(e.g. PR 1 = UL1,L2 . IR 1)
Total power Ptot
Ptot = PR 1 + PR 2 + PR 3
Conductor and phase breaks Delta circuit
none
Conductor currents IL in mA
Phase currents Iphase in mA
IL1
21.3
IL2
21.3
IL3
21.3
IR 1
12.3
IR 2
12.3
IR 3
12.3
UL1, L2
12.1
Conductor voltages U L (Uphase) in V UL2, L3
12.1
UL3, L1
12.1
PR 1
148.8
PR 2
148.8
PR 3
148.8
tot
446.4
Phase powers Pphase in mW
Total power Pin tot mW Table 2.6.2.1
P
R2
L1
R1 / R3
L2 / R
2
L1 / R
2
Alternating Current Technology
39
Three-Phase Alternating Current
Calculation of the phase and total powers:
Phase breaks R2
Phase breaks R1 and R3
PR1 = UL1,L2 . IR1 =
PR1 = UL1,L2 . IR1 =
PR2 = UL2,L3 . IR2 =
PR2 = UL2,L3 . IR2 =
PR3 = UL3,L1 . IR3 =
PR3 = UL3,L1 . IR3 =
Ptot = PR1 + PR2 + PR3 =
Ptot = PR1 + PR2 + PR3 =
Conductor break L1
Conductor and phase breaks L2 and R 2
PR1 = UL1,L2 . IR1 =
PR1 = UL1,L2 . IR1 =
PR2 = UL2,L3 . IR2 =
PR2 = UL2,L3 . IR2 =
PR3 = UL3,L1 . IR3 =
PR3 = UL3,L1 . IR3 =
Ptot = PR1 + PR2 + PR3 =
Ptot = PR1 + PR2 + PR3 =
Conductor and phase breaks L1 and R 2
PR1 = UL1,L2 . IR1 = PR2 = UL2,L3 . IR2 = PR3 = UL3,L1 . IR3 =
Ptot = PR1 + PR2 + PR3 =
40
Alternating Current Technology Three-Phase Alternating Current
Notes:
Alternating Current Technology
41
Capacitor in the AC Circuit
3
Capacitor in the AC Circuit
3.1
Fundamentals
Next to the resistors capacitors are the most frequently used components in electronic circuits. Their applications are very versatile:
Symbol
• isolation of DC and AC current • phase shifting of current and voltage Figure 3.1.1
• short-circuiting of AC voltages • as reactive resistor, timing delay element and • energy store • smoothing of DC voltages • setting up filters and resonant circuits etc.
Capacitors have a variety of different designs, the major ones are: - wirewound capacitors - ceramic capacitors - electrolytic capacitors - variable capacitors
The most important characteristics of these components are: Capacitance C This is the capacity for the load factor Q and is given by the dielectric constant ( ε0 . εr) and the effective plate surface divided by the distance apart of the plates.
ε ⋅ε ⋅ A C= 0 r d C = capacitance, unit: farad (F) or As / V
ε = electric field constant (8.85 . 10
-12
As / Vm)
0
εr = dielectric constant A = plate surface in m
2
d = distance between plates in m
Rated voltage The rated voltage is the highest permissible continuous voltage which may be applied to the capacitor.
42
Alternating Current Technology Capacitor in the AC Circuit
Peak voltage The peak voltage is the briefly permissible maximum voltage (peak value or peak-to-peak value) Insulating resistance Rp The insulating resistance is the specific resistance of the insulator used. The insulating resistance should be as large as possible ( > 1 G Ω ) so that the residual current flowing through the charged capacitor remains as small as possible. Capacitive reactance XC The capacitive reactance is determined by the value of capacitance and the frequency of the applied AC voltage.
XC =
1 ω ⋅C
XC = capacitive reactancein Ω
ω = radian frequency in Hz ( 2 . π . f )
Dielectric loss factor tan The dielectric loss factor gives the ratio between the capacitive reactance X C and the effective resistance (insulating resistance) RP of a capacitor.
tan δ =
XC RP
tan δ = dielectric loss factor Rp = insulating resistance in Ω
Charge Q The charge Q stored in a capacitor is dependent on the charging current and the charging time.
Q=I.t Q = charge in As = C (Coulomb) t
= time in s
I
= charging current in A (constant)
In a loaded capacitor:
Q=C.U C = capacitance in F U = maximum charging voltage in V
Alternating Current Technology
43
Capacitor in the AC Circuit
3.2
Charging and Discharging Process of a Capacitor
3.2.1
General
When charging and discharging a capacitor on DC voltage, current and voltage run according to an e-function whereby the voltage has risen to 63 % of the final voltage after 1 τ during charging and has sunk to 37 % of the initial voltage during discharging. τ is the time constant given by resistance x capacitance. During charging or discharging, the current sinks to 37 % of its initial value after 1 τ . After about 5 τ the charging or discharging process is ended.
τ=R.C
τ = time constant in s C = capacitance in F R = resistance to current limiting in Ω
The momentary current and the momentary voltage when charging and discharging a capacitor are calculated with the following formulas:
Momentary voltage value uC when charging
(
uC = U ⋅ 1 − e − t / τ
)
U = charging voltage e = base number 2.718 t
= charging time in s
Momentary voltage value uC when discharging
u C = U ⋅ e −t / τ U = voltage on the charged capacitor t
= charging time in s
Momentary current value iC when charging U iC = ⋅ e − t / τ R R = resistance to the current limiting in Ω
Momentary current value iC when discharging U iC = − ⋅ e − t / τ R
44
Alternating Current Technology Capacitor in the AC Circuit
3.2.2
Experiments
Experiment Display the charge and discharge curve of voltage and current of a capacitor on the oscilloscope and determine the following values from the curve: - time constant τ - capacitance C - momentary voltage u C after a charging time of 2 ms - momentary current iC at a discharging time of 2.5 ms - Then check the values determined from the curves by calculation. - How great is the charge Q after a charging time of 5 ms?
Experiment procedure • Set up the experiment according to circuit (figure 3.2.2.1) and connect the function generator with a
positive square-wave voltage:
U = 6 V;
f = 100 Hz;
A
Y1
B
Y
V=2
4.7 k Ω
G
up = 6 V
2
f = 100 Hz
0.22 µF
0V
C
⊥
oscilloscope Figure 3.2.2.1
• Connect the oscilloscope:
- Test point A to channel 1 (Y 1), for recording the input voltage - Test point C to ground - Test point B to channel 2 (Y 2), for recording the capacitor voltage or displaying the capacitor currents • To display the capacitor current switch over the resistor • 4.7 kΩ and the capacitor 0.22 µF in the circuit; the voltage applied to the resistor, which is proportional to
the capacitor current, is then displayed on the oscilloscope. Make other settings on the oscilloscope in accordance with the specifications below the grids (figure 3.2.2.2 and figure 3.2.2.3). • Draw the displayed voltage curves into the grids (figure 3.2.2.2 and figure 3.2.2.3) respectively. • Determine the values specified in the experiment from the recorded voltage curves.
Alternating Current Technology
45
Capacitor in the AC Circuit
Settings: X = 1 ms / division Y1 = 2 V / division Y2 = 2 V / division - 0 (Y 1 )
Triggering: Y1
Remarks: Y1: Input voltage Y2: Capacitor voltage - 0 (Y 2 ) Figure 3.2.2.2
Settings: X = 1 ms / division Y1 = 2 V / division Y2 = 2 V / division - 0 (Y1 )
Triggering: Y1
- 0 (Y 2 )
Remarks: Y1: Input voltage Y2: Voltage on the resistor (proportional to the capacitor current)
Figure 3.2.2.3
Time constant
From screen image:
Calculated:
46
Alternating Current Technology Capacitor in the AC Circuit
Capacitance C
From screen image:
Calculated:
Momentary voltage value u C after a charging time of 2 ms
From screen image:
Calculated:
Momentary current value iC after a discharging time of 2.5 ms
From screen image:
Calculated:
Charge Q after a charging time of 5 ms
Only calculated:
Alternating Current Technology
47
Capacitor in the AC Circuit
3.3
Phase Shift between Current and Voltage on the Capacitor
3.3.1
General If a sinusoidal AC voltage is applied to a capacitor, UC
UC
it is periodically charged and discharged.
IC
IC
Because of the AC voltage the polarity of the capacitor’s charge also changes periodically.
0
The current IC reaches its maximum value every
t
time the voltage UC crosses zero. 90°
The current IC in a capacitor and the voltage U C have a 90° phase shift. Figure 3.3.1.1
3.3.2
Experiments
Experiment Display the current and voltage curves of a capacitor on the oscilloscope and determine the phase shift between the current IC and the voltage UC from the screen image. A
Y1
Experiment procedure • Set up the experiment according to the circuit (figure 3.3.2.1) and connect the functi-
1 kΩ
on generator:
G
u pp = 3 V
C
(sinusoidal)
⊥
upp = 3 V (sinusoidal);
f = 1 kHz
Attention: 0.22 µ F
Make sure that test point C is not connected with test points A or B through ground of the
0V
B
Y2
oscilloscope
Figure 3.3.2.1
equipment used (function generator, oscilloscope). Use an isolating transformer if necessary.
The resistor (1 k Ω) in the circuit serves as a measuring resistor. The voltage U R applied to it is proportional to the capacitor current IC. For simultaneously displaying the capacitor voltage U C and the capacitor current I C (UR) the reference point of the voltages to be measured has been placed between the capacitor and the measuring resistor (1 k Ω) (test point C). It must be taken into account that the two voltages are displayed with a 180° shift to each other. The actual display of the voltage curves is made by inverting one of the two voltages with the oscilloscope (in the experiment UC, channel 2, Y2).
48
Alternating Current Technology Capacitor in the AC Circuit
• Connect the oscilloscope:
- Test point A to channel 1 (Y 1) - Test point B to channel (Y 2), inverse - Test point C to ground Make other settings on the oscilloscope in accordance with the specifications below the grid (figure 3.3.2.2). • Draw the displayed voltage curves into the grid (figure 3.3.2.2) and determine the phase shift between the
capacitor voltage UC and the capacitor current I C (UR).
Settings: X = 0.1 ms / division - 0 (Y 1 )
Y1 = 1 V / division Y2 = 1 V / division (inverse) Triggering: Y1 Remarks: Y1: Voltage UR (capacitor current IC)
- 0 (Y 2 )
Y2: Capacitor voltage UC
Figure 3.3.2.2
Phase shift between the capacitor current and the capacitor voltage:
Period duration T
T=
Phase shift
ϕ=
Alternating Current Technology
49
Capacitor in the AC Circuit
3.4
Capacitive Reactance of a Capacitor
3.4.1
General
A capacitor has a current limiting effect in the AC circuit which results from the countervoltage during charge reversal. This current limiting effect is referred to as the capacitive reactance X C. The value of the capacitive reactance depends on the capacitance of a capacitor and the frequency of the applied AC voltage. The capacitive reactance is calculated on a sinusoidal AC voltage with the following formula:
XC =
XC =
3.4.2
1 2 ⋅π ⋅ f ⋅ C
UC IC
XC
= capacitive reactance of the capacitor in Ω
2 ⋅π ⋅ f
= radian frequency ωin 1/s
C = capacitance in F
If capacitor current and capacitor voltage are known the capacitive reactance can be c alculated with Ohm's Law:
Experiments
Experiment Display the current and voltage curve of different capacitors at different frequencies on the oscilloscope. The respective capacitive reactance XC is to be determined from the screen image by means of the peak-to-peak values and then checked by calculation. Experiment procedure A
Y1
• Set up the experiment according to the cir-
cuit (figure 3.4.2.1), set a voltage u pp = 8 V 1 kΩ
G
(sinusoidal); f = 0.1 kHz on the function generator and connect it with the circuit.
u pp = 8 V
C
(sinusoidal)
⊥
• Connect the oscilloscope:
- Test point A to channel 1 (Y 1) C
- Test point B to channel 2 (Y 2) - Test point C to ground
0V
B
Y2
oscilloscope
•
Read off the peak-to-peak values of UR and UC for the frequencies and capacitors
Figure 3.4.2.1 C = 0.22 / 0.47 / 1 µF
specified in table 3.4.2.1 and enter them in the table. • Calculate values IC and X C and enter these in table 3.4.2.1 also. • Transfer the X C values to diagram (figure 3.4.2.2) for constructing the characteristic curve X C = f (f).
Attention: Make sure that test point C is not connected with test points A or B through ground of the equipment used (function generator, oscilloscope). Use an isolating transformer if necessary. The resistor (1Ωk) in the circuit serves as a measuring resistor. The voltage U R applied to it is proportional to the capacitor currentCI. The respective capacitor current is calculated with the formula I C = UR / R.
50
Alternating Current Technology Capacitor in the AC Circuit
f [kHz]
0.1
0.2
0.3
0.4
0.5
0.6
1.0 F UC [V]
0.47 F 0.22 F 1.0 F
UR [V]
0.47 F 0.22 F 1.0 F
IC [mA]
0.47 F 0.22 F 1.0 F
XC [k ]
0.47 F 0.22 F
Table 3.4.2.1
Figure 3.4.2.2
Question:
What do the curves tell us?
Answer:
Check the capacitive reactance X C of the capacitor C = 0.47 µF at f = 600 Hz by calculation. 1 = XC = 2 ⋅π ⋅ f ⋅ C
0.7
Alternating Current Technology
51
Capacitor in the AC Circuit
3.5
Series Circuiting of Capacitors
3.5.1
General C1
I
C2
When capacitors are circuited in series the total
C3
capacitance is less than the smallest individual
+
capacitance. It is calculated with the following UC 1
UC 2
UC 3
formula:
Ctot =
U C tot
U
1 1 1 1 + + ⋅⋅⋅ C1 C 2 C 3
-
When only two capacitors are circuited in series the following results:
Figure 3.5.1.1
Ctot =
C1 ⋅ C 2 C1 + C 2
The partial voltages on the individual capacitors behave inversely proportional to the respective capacitance, their sum gives the total voltage UC tot. The capacitor current in a series circuit of capacitors is the same at all points. The same applies to the electrical charge shift.
3.5.2
Experiments
Experiment Prove, by measuring the current and voltage and on the assumption that the capacitive reactance of a capacitor is XC = 1/ω . C, that in a series circuit of capacitors the total capacitance is less than the smallest individual capacitance.
A
R1
1 kΩ
B
C1
0.22 µ F
C
C2
D
0.47 µF
C3
1µ F
Experiment procedure E
• Set up the experiment according to the circuit
(figure 3.5.2.1) and connect the function generator:
G
Urms = 3 V (sinusoidal) f = 1kHz
• Urms = 3 V (sinusoidal);
f = 1 kHz
• Measure the voltages at test points A ... E
with a multimeter and enter the values in the 0V
Figure 3.5.2.1
table 3.5.2.1. • The voltage at the measuring resistor R 1
(1 kΩ) is proportional to the capacitor current IC and is used for its calculation.
52
Alternating Current Technology Capacitor in the AC Circuit
Partial and total voltage [V] Test points A-B (UR 1)
B-C (UC 1)
C-D (UC 2)
D-E (UC 3)
Table 3.5.2.1
• Calculate the capacitor current, capacitive reactances and capacitances. • Check the total capacitance Ctot by calculation.
Calculation of the capacitor current across the measuring resistor 1 k Ω:
IC =
U R1 = RM
Calculation of the capacitive reactances
U C1
Calculation of capacitances
=
C1 =
1 = ω ⋅ X C1
X C2 = UC2 = IC
C2 =
1 = ω ⋅ X C2
C3 =
1 = ω ⋅ X C3
X C1 =
X C3 =
IC
U C3 IC
X C tot =
=
U C tot IC
=
C tot =
Checking the total capacitance by calculation
C tot =
1
1 1
1
C1 + C 2 + C 3
=
1 = ω ⋅ X C tot
B-E (UC tot)
Alternating Current Technology
53
Capacitor in the AC Circuit
3.6
Parallel Circuiting of Capacitors
3.6.1
General
I C tot
In a parallel circuit of capacitors the total capacitance is
+ IC 1
IC 2
IC 3
C1
C2
C3
equal to the sum of the individual capacitances.
Ctot = C1 + C2 + C3 ...
U
The partial currents behave proportionally to the respective capacitance, their sum gives the total current
-
IC tot.
Figure 3.6.2.1
The voltage UC is the same in a parallel circuit of all capacitors.
3.6.2
Experiments
Experiment Prove, by measuring the current and voltage and on the assumption that the capacitive reactance of a capacitor is XC = 1 / ω . C, that the total capacitance in a parallel circuit of capacitors is equal to the sum of the individual capacitances. A
G
I C tot
U rms = 3 V (sinusoidal) f = 1 kHz
B
Experiment procedure • Set up the experiment according to the circuit
C
E
G
IC 1
IC 2
IC 3
(figure 3.6.2.1), connect the function generator
D
F
H
and set the following voltage:
C1 0.22 µ F
C2 0.47 µ F
C3 1µF
Urms = 3 V (sinusoidal);
f = 1 kHz
I 0V
Figure 3.6.2.1
• Measure the total current I C tot and the partial
currents IC 1, IC 2, IC 3 with a multimeter and enter the values in the table 3.6.2.1.
• Calculate the capacitive reactances X C tot, XC 1, XC 2, XC 3 with the formula X C = UC / IC. • Establish the individual capacitances and total capacitance with the formula C = 1 / ω . XC. • Check the total capacitance C tot established in the experim ent by calculation.
54
Alternating Current Technology Capacitor in the AC Circuit
Partial and total voltage [mA]
Capacitor voltage [V]
Test points A-B (IC tot)
C-D (IC 1)
E-F (IC 2)
Test points G-H (IC 3)
D-I (UC 1)
F-I (UC 2)
Table 3.6.2.1
Calculation of the capacitive reactances
U C1
Calculation of capacitances
=
C1 =
1 = ω ⋅ X C1
=
C2 =
1 = ω ⋅ X C2
X C3 = U C3 = IC3
C3 =
1 = ω ⋅ X C3
X C1 =
X C2 =
IC1
UC2
X C tot =
IC2
U C tot I C tot
=
Checking the total capacitance by calculation
Ctot = C1 + C2 + C3 =
C tot =
1 = ω ⋅ X C tot
H-I (UC 3)
Alternating Current Technology
55
Capacitor in the AC Circuit
3.7
Reactive Power of a Capacitor
3.7.1
General
If a capacitor is applied to a sinusoidal AC voltage a phase shift results between the current and the voltage caused by the capacitive reactance. The reactive power Q which occurs in the capacitor can be represented in the line diagram (figure 3.7.1.1) by multiplication of the current and voltage values. Energy is fed to the capacitor during the positive half-wave. During the negative half-wave the capacitor radiates this stored energy.
Q U U
I
I
Q
0 t
Figure 3.7.1.1
No active power is converted in the ideal capacitor.
The following definitions are valid: QC
= UC . IC (rms values)
qC
= uC . iC (momentary values)
QC; qC
= capacitive reactancein var
UC; uC
= capacitor voltagein V
IC; iC
= capacitor currentin A
56
Alternating Current Technology Capacitor in the AC Circuit
3.7.2
Experiments
Experiment Display the current and voltage curves of a capacitor on the oscilloscope, transfer the curves to a diagram and construct the power curve by multiplication of the momentary values of current and voltage. A UR
G
Y1
Experiment procedure • Set up the experiment according to the circuit (fi-
gure 3.7.2.1), connect the function generator and
1 kΩ
set the following voltage:
up=4 V (sinusoidal)
C
⊥
f = 1 kHz UC
up = 4 V (sinusoidal);
f = 1 kHz
• Connect the oscilloscope:
0.22 µ F
- Test point A to channel 1 (Y1) B
0V
Y2
oscilloscope
- Test point B to channel 2 (Y 2), inverse - Test point C to ground
Figure 3.7.2.1
• Settings on the oscilloscope:
- Sweep:
X = 0.1 ms / division
- Channel 1:
Y1 = 1 V / division (voltage UR ~ capacitor current IC)
- Channel 2: Y2 = 1 V / division (capacitor voltage UC) - Center zero line of Y1 and Y2 - Triggering:
Y1
• Draw the screen image on the diagram (figure 3.7.2.2) and establish the values for construction of the po-
wer curve by multiplication of the voltage and current values (e. g. every 0.1 ms).
Attention: Make sure that test point C is not connected with test points A or B through ground of the equipment used (function generator, oscilloscope). Use an isolating transformer if necessary. The resistor (1 k Ω) in the circuit serves as a measuring resistor. The voltage U R applied to it is proportional to the capacitor current IC (in mA). For simultaneously displaying the capacitor voltage UC and the capacitor current I C (UR) the reference point of the voltages to be measured has been placed between the capacitor and the measuring resistor (1 k Ω) (test point C). It must be taken into account that the two voltages are displayed with a 180° shift to each other. The actual display of the voltage curves is made by inverting one of the two voltages with the oscilloscope (in the experiment UC, channel 2, Y2).
Alternating Current Technology
57
Capacitor in the AC Circuit
Time t [ms]
Current [mA] iC
Voltage u
[V]C
Reactive power q
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Table 3.7.2.1
5 I [mA] C U C [V] QC [mvar]
4 3 2 1 0 0.2 -1 -2 -3 -4 -5
Figiure 3.7.2.2
0.4
0.6
0.8
1.0
t [ms]
C
[mvar]
58
Alternating Current Technology Capacitor in the AC Circuit
Notes:
Alternating Current Technology
59
Coil in the AC Circuit
4
Coil in the AC Circuit
4.1
Fundamentals
Coils are generally wirewound (usually copper) whereby the number of windings and the gauge of the wire can be extremely varied. Coils are widely used in electrical en-
Symbol
gineering and electronics, e. g.: - generating induction voltages (transformer)
Figure 4.1.1
- generating magnetic forces (electric motor, relay, contactors) - filtering out AC currents - storing magnetic energy - function as reactance
Coils are manufactured in a variety of designs and sizes depending on their applications. The two major categories are air coils and coils with a magnetic core. The choice of the magnetic core material and the design of the core are greatly influential in determining the characteristics of a coil.
The most important characteristics of a coil are: Inductance L The inductance specifies how great the generated countervoltage is at a predefined change in current. The inductance depends on the number of windings and the coil constant.
L = N 2 ⋅ AL L = inductance, unit:
Henry (H)
N = number of windings AL = coil constant in Henry (H)
Coil constant AL
AL =
µ0 ⋅ µ r ⋅ A d
-6 µ0 = magnetic field constant (1.257 . 10 Vs / Am)
µr = relative permeability A = core cross section in m d = length of coil in m
2
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Alternating Current Technology Coil in the AC Circuit
Inductive reactance XL The inductive reactance is determined by the value of inductance and the frequency of the applied AC voltage.
XL = ω . L XL = inductive reactance in Ω
ω = radian frequency in Hz ( 2 . π . f )
Loss factor tan The loss factor tanδ gives the ratio between the equivalent resistance of a coil and the inductive reactanceL.X
tan δ =
δ
RV XL
= loss angle
tan δ = loss factor RV
= equivalent resistance in Ω
The equivalent resistance consists mainly of the ohmic resistance of the coil, hysteresis and capacitance losses as well as losses due to eddy currents.
Coil quality Q The coil quality, as the name implies, specifies the quality of a coil and is equal to the reciprocal of the loss factor tan δ.
Q=
XL RV
Q = coil quality
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Coil in the AC Circuit
4.2
Switch-On and Switch-Off Process of a Coil
4.2.1 General Current and voltage run according to an e-function when switching the DC voltage of a coil on and off. If a DC voltage is applied, the current increases to 63 % of its final value within 1 τ and sinks to 37 % of its initial value when switching off. After 5 τ the current reaches approximately its final value or zero.
τ is the time constant. It results from the inductance of the coil and the ohmic resistance effective in the circuit. τ=
L R
τ = time constant in s L = inductance in H R = ohmic resistance in Ω (coil, line, internal resistance of the source)
The voltage on the coil is reduced to 37 % of its maximum value (on applying the C voltage) after 1 τ and reaches its smallest value, which is dependent on the ohmic coil resistance, after about 5 τ. If the DC voltage on the coil is switched off the self-induction voltage is active in the circuit. This has reverse polarity to the srcinally applied DC voltage and reaches zero after about 5 τ. The momentary current value iL and the momentary voltage value u L when switching on and off a DC voltage are calculated for the coil according to the following formulas:
Momentary value of current iL in the switch-on process
iL =
(
U 1 − e−t / τ R
)
(
i L = I max 1 − e −t / τ
Momentary value of voltage uL in the switch-on process
)
U = applied DC voltage in V
u L = U ⋅ e −t / τ U = maximum coil voltage
R = ohmic resistance active in the circuit in Ω t
= switch-on duration in s
e = base number 2.718 Momentary value of current iL in the
Momentary value of voltage uL in the
switch-off process
switch-off process
iL =
U
(e −t / τ )
i L = I max e −t / τ
u L = − U e −t / τ
R U = max. self-induction voltage of the coil
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Alternating Current Technology Coil in the AC Circuit
4.2.2 Experiments Experiment Display the current and voltage curve on the oscilloscope when switching a DC voltage on and off on the coil and establish the following values from the curves: - time constant τ - inductance L - momentary current iL at a switch-on duration of 0.2 ms.
Experiment procedure Set up the experiment according to the circuit (figure 4.2.2.1) and connect the function generator with a positive square-wave voltage: U = 6 V;
f = 1 kHz;
V=2 A
Y1
B
Y2
1 kΩ
G
up = 6 V f = 1kHz
100 mH
0V
C
⊥
oscilloscope Figure 4.2.2.1
•
Connect the oscilloscope: - Test point A to channel 1 (Y 1), for oscilloscoping the input voltage - Test point B to channel 2 (Y 2), for oscilloscoping the coil voltage or displaying the coil current - Test point C to ground
• To display the coil current, switch over the resistor 1 k Ω and the coil 100 mH in the circuit; the voltage at
the resistor is then displayed on the oscilloscope. This voltage is proportional to the coil current. • Make other settings on the oscilloscope in accordance with the specifications below the grids (figure
4.2.2.2 and figure 4.2.2.3).
• Establish the values specified in the experiment from the recorded voltage curves. • The values established from the curves should then be checked by calculation.
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63
Coil in the AC Circuit
Settings: X = 0.1 ms / division Y1 = 2 V / division Y2 = 2 V / division - 0 (Y1 ) - 0 (Y 2 )
Triggering: Y1 Remarks: Y1: Input voltage Y2: Coil voltage
Figure 4.2.2.2
Settings: X = 0.1 ms / division Y1 = 2 V / division Y2 = 2 V / division - 0 (Y 1 )
Triggering: Y1 Remarks: :I
Y1 nput voltage Y2: Voltage on the resistor (proportional to the coil current) - 0 (Y 2 ) Figure 4.2.2.3
Time constant
Momentary value iL
From screen image:
at a switch-on duration of 0.2 ms From screen image:
Calculated: Calculated: Inductance L From screen image:
Calculated:
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Alternating Current Technology Coil in the AC Circuit
Notes:
Alternating Current Technology
65
Coil in the AC Circuit
4.3
Phase Shift between Current and Voltage on a Coil
4.3.1 General If an AC voltage is applied to a coil, the coil current IL
UL
cannot follow the rapid voltage changes on account
UL
IL
of the inductance. The current IL in a coil follows the voltage UL by 90°. In the case of a sinusoidal AC voltage, for example,
0 t
the current IL reaches the negative peak value a quarter of a period (90°) later than the voltage U L
90°
(figure 4.3.1.1). Figure 4.3.1.1
The equivalent resistance of the coil is ignored in this consideration.
4.3.2 Experiments Experiment Display the current and voltage curves of a coil on the oscilloscope and determine the phase shift between the current IL and the voltage UL from the screen image. A
Y1
Experiment procedure • Set up the experiment according to the circuit
(figure 4.3.2.1), connect the function generator
1 kΩ
and set the following voltage:
G
u pp = 3 V
C
(sinusoidal)
⊥
100 mH
upp = 3 V (sinusoidal);
f = 1 kHz
Attention: Make sure that test point C is not connected with
0V
B
Y2
oscilloscope Figure 4.3.2.1
test points A or B through ground of the equipment used (function generator, oscilloscope). Use an isolating transformer if necessary.
The resistor (1 k Ω) in the circuit serves as a measuring resistor. The voltage U R applied to it is proportional to the capacitor current IL. For simultaneously displaying the coil voltage UL and the coil current I L (UR) the reference point of the voltages to be measured has been placed between the coil and the measuring resistor (1 k Ω) (test point C).
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Alternating Current Technology Coil in the AC Circuit
It must be taken into account that the two voltages are displayed with a 180° shift to each other.The actual display of the voltage curves is made by inverting one of the two voltages with the oscilloscope (in the experiment UL, channel 2, Y2). Settings: X = 0.1 ms / division - 0 (Y1 )
Y1 = 1 V / division Y2 = 1 V / division (inverse) Triggering: Y1 Remarks: Y1: Voltage UR (coil current I L) Y2: Coil voltage UL
- 0 (Y 2 )
Figure 4.3.2.2
• Connect the oscilloscope:
- Test point A to channel 1 (Y 1) - Test point B to channel 2 (Y 2), inverse - Test point C to ground • Make other settings on the oscilloscope in accordance with the specifications below the grid (figure
4.3.2.2). • Draw the displayed voltage curves into the grid (figure 4.3.2.2) and determine the phase shift between the
coil voltage UL and the capacitor current I L (UR).
Phase shift between the coil current I L and the coil voltage UL:
Period duration T
T=
Phase shift
ϕ=
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Coil in the AC Circuit
4.4
Inductive Reactance of a Coil
4.4.1 General A coil has a current limiting effect in the AC circuit which results from the counter voltage generated in it. This current limiting effect is referred to as the inductive reactance X L. The value of the inductive reactance depends on the inductance of a coil and the frequency of the applied AC voltage. The inductive reactance is calculated on a sinusoidal AC voltage with the following formula: If coil current and coil voltage are known the inductive
XL = 2 . π . f . L
reactance can be calculated with the formula:: XL
= inductive reactance of the capacitor in Ω
2 . π .f
= radian frequency ω in 1/s
L
XL =
UL IL
= inductance in H
4.4.2 Experiments Experiment Display the current and voltage curves of different coils at different frequencies on the oscilloscope and record the curve X L = f (f). The respective inductive reactance is determined from the screen image by means of the peak-to-peak values. A
Y1
Experiment procedure • Set up the experiment according to the cir-
cuit (figure 4.4.2.1), set a voltage u pp = 8 V
470 Ω
(sinusoidal); f = 1 kHz on the function ge-
u pp = 8 V
G
nerator and connect it with the circuit.
1…6 kHz
C
⊥
(sinusoidal)
• Connect the oscilloscope:
- Test point A to channel 1 (Y 1)
L
- Test point B to channel 2 (Y 2), inverse - Test point C to ground B
0V
Y2
• Read off the peak-to-peak values of UR and
oscilloscope Figure 4.4.2.1
L = 8 mH / 100 mH 8 mH = transformer coil N = 900
UL for the frequencies and capacitors specified in table 4.4.2.1 and enter them in the table.
• Calculate values IL and X L and enter these in table 4.4.2.1 also. • Transfer the X L values to diagram 4.4.2.2 for constructing the characteristic curve X L = f (f).
Attention: Make sure that test point C is not connected with test points A or B through ground of the equipment used (function generator, oscilloscope). Use an isolating transformer if necessary. The resistor (470 Ω) in the circuit serves as a measuring resistor. The voltage U R applied to it is proportional to the coil current I L. The respective coil current I L is calculated with the formula: IL = UR / R
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Alternating Current Technology Coil in the AC Circuit
f [kHz] UL [V]
1
2
3
4
5
6
8 mH 100 mH
UR [V]
8 mH 100 mH
IL [mA]
8 mH 100 mH
XL [k ]
8 mH 100 mH
Table 4.4.2.1
5
4
3
XL / k 2
1
0 01234567
f / kHz Figure 4.4.2.2
Question:
What do the curves tell us?
Answer: Calculate the inductive reactance X L of the coil 100 mH with the formula XL = ω . L (at f = 2kHz) and compare it with the value established in the experiment.
XL = ω . L = Value established in the experiment:
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69
Coil in the AC Circuit
4.5
Series Circuiting of Coils
4.5.1 General L1
IL
L2
When coils are circuited in series the total
L3
inductance is equal to the sum of the individual
+
inductances. UL1
UL 2
UL 3
Ltot = L1 + L2 + L3 ... The partial voltages on the individual coils
U L tot
U
behave proportionally to the respective inductance, their sum gives the total voltage
-
UL tot.
Figure 4.5.1.1
The coil current in a series circuit of coils is the same at all points.
4.5.2 Experiments Experiment Prove, by measuring the current and voltage, that in a series circuit of coils the total inductance is equal to the sum of the individual inductances. Between the inductive reactance and the inductance of a coil there is a relationship: XL = ω . L A
I1
B
L1
C
L2
D
L3
E
Experiment procedure • Set up the experiment according to the circuit (fi-
100 mH G
8 mH
8 mH
U rms = 5 V f = 1kHz (sinusoidal)
gure 4.5.2.1) and set a voltage of Urms = 5 V (sinusoidal); f = 1 kHz on the function generator and connect it with the circuit.
0V
Figiure 4.5.2.1
L = 8 mH / 100 mH 8 mH = transformer coil N = 900
• Measure the current IL between test points A and B as well as the voltages at the test points B ... E with the
multimeter and enter the values in table 4.5.2.1. • Calculate the individual inductive reactances, inductances and the total inductance in that order from the
measured values.
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Alternating Current Technology Coil in the AC Circuit
Coil current [mA]
Partial and total voltage [V]
Test points A-B (IL)
Test points B-C 1) (UL
C-D (U 2)
D-E (U 3) L
L
Table 4.5.2.1
Calculation of the inductive reactances
Calculation of inductances
X L1 =
U L1 = IL
L1 =
X L1 = ω
X L2 =
U L2 = IL
L2 =
X L2 = ω
X L3 =
U L3 = IL
L3 =
X L3 = ω
X L tot =
U L tot I
L
=
Ltot =
X L tot
ω
=
B-E (U
L tot)
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Coil in the AC Circuit
4.6
Parallel Circuiting of Coils
4.6.1 General In a parallel circuit of coils the total inductance is less
I L tot +
than the smallest individual inductance. It is calculated IL 1
IL 2
L1
IL 3
L2
with the following formula:
Ltot =
L3
U
1 1 1 1 + + ⋅⋅⋅ L1 L2 L3
-
With only two parallel-circuited coils the following
Figure 4.6.1.1
formula is used:
Ltot =
L1 ⋅ L2 L1 + L2
The current through the individual coils behaves inversely proportional to the respective inductance. The voltage UL is the same in a parallel circuit on all coils.
4.6.2 Experiments Experiment Prove, by measuring the current and voltage, that the total inductance in a parallel circuit of coils is less than the smallest individual inductance. Between the measured inductive reactances and the inductance there is a relationship: XL = ω . L A
B I L tot
G
U rms = 5 V f = 5 kHz (sinusoidal)
C
E
G
IL 1
IL 2
IL 3
D
F
H
L1
L2
L3
100 mH
8 mH
8 mH
I 0V
Figure 4.6.2.1
8 mH = transformer coil N = 900
Experiment procedure • Set up the experiment according to the circuit (figure 4.6.2.1), connect the function generator and set the
following voltage: Urms = 5 V (sinusoidal); f = 5 kHz • Measure the total current I L tot and the partial currents I L 1, IL 2, IL 3 with a multimeter and enter the values in
the table 4.6.2.1.
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Alternating Current Technology Coil in the AC Circuit
Partial and total current [mA]
Partial voltage UL [V]
Test points A-B (IL tot)
C-D (IL 1)
E-F (IL 2)
Test points G-H (IL 3)
D-I (UL 1)
F-I (UL 2)
Table 4.6.2.1
• Calculate the inductive reactances X L tot, XL 1, XL 2 and X L 3 with the formula X L = UL / IL. • Calculate the individual inductances and the total inductance with the formula L = XL / ω. • Check the value of total inductances L tot established in the experiment by calculation.
Calculation of the inductive reactances
X L1 =
XL2 =
X L3 =
UL1 I L1
UL2 IL2
UL3 IL3
X L tot =
=
L1 =
=
L2 =
=
L3 =
U L tot I L tot
=
Checking the total inductance by calculation
Ltot =
Calculation of inductances
1 = 1 1 1 + + L1 L2 L3
X L1
ω
X L2
ω
X L3
Ltot =
ω
=
=
=
X L tot
ω
=
H-I (UL 3)
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73
Coil in the AC Circuit
4.7
Reactive Power of a Coil
4.7.1 General If a coil is applied to a sinusoidal AC voltage a phase shift of 90° results between the current and the voltage caused by the inductive reactance. The reactive power Q which occurs in the coil can be represented in the line diagram (figure 4.7.1.1) by multiplication of the current and voltage values. Energy is fed to the coil during the positive half-wave. During the negative half-wave the coil radiates this stored energy. No active power is converted in the ideal coil (no ohmic resistance).
Q U
U
I
I Q
0
t
Figure 4.7.1.1
The following definitions are valid:
qL
= UL . IL (rms values) = uL . iL (momentary values)
QL; qL
= inductive reactance in var
UL; uL
= coil voltage in V
IL; iL
= coil current in A
QL
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Alternating Current Technology Coil in the AC Circuit
4.7.2 Experiments Experiment Display the current and voltage curves of a coil on the oscilloscope, transfer the curves to a diagram and construct the power curve by multiplication of the momentary values of current and voltage.
A
Y1
Experiment procedure • Set up the experiment according to the cir-
cuit (figure 4.7.2.1), connect the function
UR
G
1 kΩ
up = 4 V f = 1 kHz (sinusoidal)
generator and set the following voltage: C
⊥
up = 4 V (sinusoidal);
f = 1 kHz
• Connect the oscilloscope: UL
- Test point A to channel 1 (Y 1)
100 mH B
Y2
- Test point B to channel 2 (Y 2), inverse - Test point C to ground
0V
oscilloscope
Figure 4.7.2.1
• Settings on the oscilloscope:
- Sweep X:
0,1 ms / division
- Channel 1, Y1:
1 V / division ((voltage UR ~ coil current I L)
- Channel 2, Y2:
1 V / division (coil voltage UL)
- Center zero line of Y1 and Y2. - Triggering:
Y1
• Draw the screen image on the diagram (figure 4.7.2.2) and establish the values for construction of the po-
wer curve by multiplication of the voltage and current values (e. g. every 0.1 ms). Enter the values into the diagram (figure 4.7.2.2). Attention: Make sure that test point C is not connected with test points A or B through ground of the equipment used (function generator, oscilloscope). Use an isolating transformer if necessary. The resistor (1 k Ω) in the circuit serves as a measuring resistor. The voltage U R applied to it is proportional to the coil current I L (in mA). The actual display of the voltage curves is made by inverting one of the two voltages with the oscilloscope (in the experiment UL, channel 2, Y2). For simultaneously displaying the coil voltage UL and the coil current I L (UR) the reference point of the voltages to be measured has been placed between the coil and the measuring resistor (1 k Ω) (test point C). It must be taken into account that the two voltages are displayed with a 180° shift to each other.
Alternating Current Technology
75
Coil in the AC Circuit
Time t [ms]
Current [mA] iL
Voltage u
[V] L
Reactive power q
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Table 4.7.2.1
5 I L [mA] U [V] L Q L [mvar]
4 3
2
1
0 0.2 -1
-2
-3 -4
-5 Figure 4.7.2.2
0.4
0.6
0.8
1.0
t [ ms]
L
[mvar]
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Alternating Current Technology Coil in the AC Circuit
Notes:
Alternating Current Technology
77
Interconnecting Resistor, Capacitor and Coil
5
Interconnecting Resistor, Capacitor and Coil
5.1
Fundamentals
The following experiments deal with the interaction of resistor, capacitor and coil on a sinusoidal AC voltage. The aim is to measure and calculate the phase relation of current and voltage as well as the resulting values in parallel and series circuits of resistor, capacitor and coil. The values in accordance with phase and amount (rms values) can be shown either by a pointer diagram or with the oscilloscope. The experiments should be carried out with a constant frequency of 1 kHz. Experiments dealing with the frequency behaviour of resistor, capacitor and coil (e. g. high-pass, low-pass and band-pass) have not been taken into consideration here.
15.2
Series Circuiting of Resistor and Capacitor
5.2.1 General If a sinusoidal AC voltage is applied to a series circuit of resistor and capacitor the same current flows through both components. R
I
UR
C
UC
U
Figure 5.2.1.1
Phase shifts occur between the voltages UR, UC and U due to the capacitive reactance X C of the capacitor. These can be shown by a line diagram or pointer diagram (see figure 5.2.1.2).
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Alternating Current Technology
Interconnecting Resistor, Capacitor and Coil
The phase shift between the active voltage UR and the reactive voltage UC is constantly 90°, the same applies to the phase shift between the current and the reactive voltage U C. The phase shift between the apparent voltage U and the voltages U R and UC is determined by the voltage ratio UC to UR or the resistor ratio XC to R.
UR = active voltage in V I = current in A U = apparent voltage, total voltage in V UC = reactive voltage, capacitor voltage in V ϕ = phase angle in ° (degrees)
Figure 5.2.1.2 Pointer diagram for voltages
The resistances can be represented by a pointer diagram as follows: R = active resistance in Ω XC = capacitive reactance in Ω Z = apparent resistance in Ω
Figure 5.2.1.3 Pointer diagram for resistors
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79
Interconnecting Resistor, Capacitor and Coil
Due to the phase shift between current and voltage, a simple numeric addition of the partial voltages and partial resistances in the series circuit of resistor and capacitor is not possible. They are calculated with the following formulas: Apparent voltage U
U = U R2 + U C2
Capacitive reactance XC
U=Z
XC = Z . sin ϕ
.I
Apparent resistance Z
Z=
Tangent of the phase angle
U Z= I
R 2 + X C2
tan ϕ =
UC UR
=
XC R
Active resistance R
R = Z . cos ϕ
5.2.2 Experiments Experiment Measure and calculate the active voltage U R, the reactive voltage UC, the current I, the phase angle ϕ, the apparent resistance Z and the capacitive reactance X C in a series circuit of resistor and capacitor. Then draw the pointer diagrams for voltage and resistance from the measured values.
A
B C
1 kΩ U rms = 5 V
G
(sinusoidal)
~
D
f = 1 kHz 0.22 µF
0V
E
Figure 5.2.2.1
Experiment procedure • Set up the experiment according to the circuit (figure 5.2.2.1), connect the function generator and set the
following voltage in connection with the multimeter:: Urms = 5 V (sinusoidal);
f = 1 kHz
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Alternating Current Technology
Interconnecting Resistor, Capacitor and Coil
Current I
Active voltage UR
(test points A - B)
(test points C - D)
I=
UR =
Reactive voltage UC (test points D - E)
UC =
• Calculate the phase angle ϕ, apparent resistance Z and capacitive reactance X C.
Phase angle
tan δ =
UC UR
Apparent resistance Z
Z=
=
U = I
Capacitive reactance XC
XC = Z . sin ϕ
(sin ϕ =
UC U
=
)
XC =
• Draw pointer diagrams from the measured values.
0.5 V / division
100 Ω / division
Figure 5.2.2.2 Pointer diagram for voltages
Figure 5.2.2.3 Pointer diagram for resistances
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Interconnecting Resistor, Capacitor and Coil
5.3
Parallel Circuiting of Resistor and Capacitor
5.3.1 General If a sinusoidal AC voltage is connected to the parallel circuit of resistor and capacitor, the same voltage is applied to both components. The current I is divided into the capacitor current I C and the active current IR. Phase shifts occur between the currents I, I C and IR due to the capacitive reactance XC of the capacitor. These can be represented in a line diagram or a pointer diagram (see figure 5.3.1.2).
Figure 5.3.1.1
The voltage U is in phase with the active current I R whereas the reactive current IC of the capacitor, the voltage U and the active current are constantly phase shifted by 90°. The phase shift between the apparent current I and the currents I R and IC is determined by the current ratio I C to IR or the conductance ratio B C to G. IR IC I U ϕ
Figure 5.3.1.2 Pointer diagram for currents
= active current in A = reactive current, capacitor current in A = apparent current, total current in A = voltage in V = phase angle in ° (degrees)
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Alternating Current Technology
Interconnecting Resistor, Capacitor and Coil
The conductances can also be shown on a pointer diagram.
Y = apparent conductance in S (Siemens) G = active conductance in S BC = reactive conductance (capacitive) in S
Figure 5.3.1.3 Pointer diagram for conductances
Direct addition of the partial currents I C and IR and the conductances G and B C for determining the apparent conductance Y is not possible on account of the phase shifts between current and voltage on the capacitor; the following formulas must be used.
Apparent current I
Cosine, sine and tan of the angle
I = I R2 + I C2
I G cos ϕ = IR = Y
I=
U Z
I B sin ϕ = C = C I Y I B tan ϕ = C = C IR G
Apparent conductance Y
Y = G 2 + BC2 Y=
1 Z
Z = apparent resistance
Alternating Current Technology
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Interconnecting Resistor, Capacitor and Coil
5.3.2 Experiments Experiment Measure the apparent current I, the reactive current I C, the active current I R in a parallel circuit of resistor and capacitor and calculate the reactive conductance Y, the apparent resistance Z and the phase angle ϕ from the measured values. Then draw the pointer diagrams for current and conductance from the measured values.
Experiment procedure • Set up the experiment according to the circuit (figure 5.3.2.1), connect the function generator and set the
following voltage in connection with the multimeter: Urms = 5 V (sinusoidal); I
A
IC
G
~
f = 1 kHz
B
IR
U rms = 5 V (sinusoidal)
C
E
f = 1 kHz
D
F
0.22 µF
1 kΩ
0V
Figure 5.3.2.1
Apparent current I
Reactive current IC
Active current IR
(test points A - B)
(test points C - D)
(test points E - F)
I=
IC =
IR =
• Calculate the reactive conductance Y, the apparent resistance Z and the phase angle ϕ.
Reactive conductance Y
Apparent resistance Z
Y = G 2 + BC2 =
Z=
G=
1 = R
1 BC = = XC
1 = Y
Phase angle
1 =ω ⋅C = 1 ω ⋅C
I tan ϕ = C = IR
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• Construct the pointer diagrams from the measured values.
1 mA / division
0.2 mS / division
Figure 5.3.2.2 Pointer diagram for currents
Figure 5.3.2.3 Pointer diagram for conductances
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Interconnecting Resistor, Capacitor and Coil
5.4
Series Circuiting of Resistor and Coil
5.4.1 General
I
R
L
UR
UL
If a sinusoidal AC voltage is applied to a series circuit of resistor and coil the same current flows through both components.
U
Figure 5.4.1.1
Phase shifts occur between the voltages UR, UL and U due to the inductive reactance X L of the coil. These can be shown by a line diagram or pointer diagram (figure 5.4.1.2). The resistances can also be represented by a pointer diagram (figure 5.4.1.3) as follows:
Figure 5.4.1.2 Pointer diagram for voltages
Figure 5.4.1.3 Pointer diagram for resistances
UR = active voltage in V
R = active resistance in Ω
I
XL = inductive reactance in Ω
= current in A
U = apparent voltage, total voltage in V UL = reactive voltage, capacitor voltage in V ϕ = phase angle in ° (degrees)
Z
= apparent resistance in Ω
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Due to the phase shift between current and voltage, a simple numeric addition of the partial voltages and partial in the series circuit of resistor and coil is not possible. They are calculated with the following formulas:
Apparent voltage U
U = U R2 + U L2
Inductive reactance XL
XL = Z . sin ϕ
U=Z.I Apparent resistance Z
Tan and cosine of the phase angle
Z = R 2 + X L2
tan ϕ =
Z=
U I
Active resistance R
R = Z . cos ϕ
cos ϕ =
UL XL = UR R UL U
=
XL Z
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Interconnecting Resistor, Capacitor and Coil
5.4.2 Experiments Experiment Measure the coil voltage UL, the active voltage UR, the current I and the phase shift between the apparent voltage U and the active voltage UR with the oscilloscope on a series circuit of resistor and coil. Calculate the inductive reactance X L of the coil, the apparent resistance Z and the phase shift between the apparent voltage U and the coil voltage UL. The equivalent resistance of the coil can be ignored in the calculations on account of its low value (13 Ω).
Experiment procedure • Set up the experiment according to the circuit (figure 5.4.2.1), connect the function generator and set the
following voltage in connection with the multimeter: upp = 6 V (sinusoidal); I A
f = 1 kHz
• Measure the peak-to-peak voltage on the coil (test
points A - B) and the resistor (test points B - C) with the oscilloscope.
100 mH
G
u pp = 6 V (sinusoidal) f = 1 kHz
B
Attention: When measuring, make sure that the oscilloscope and generator grounds are not connected through mains;
1 kΩ
0V
use an isolating transformer if necessary. C
Figure 5.4.2.1
Coil voltage u L test points A - B
uL = Active voltage u R test points B - C
uR = The current i can be calculated by way of the voltage drop across the 1 k Ω resistor. Current i
u i= R = R
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Interconnecting Resistor, Capacitor and Coil
• Connect the oscilloscope as follows to determine the phase angle between the apparent voltage U and the
active voltage UR: Test point A to channel 1 (Y 1) Test point B to channel 2 (Y 2) Test point C to ground The other settings on the oscilloscope should be made according to the specifications below grid (figure 5.4.2.2). Settings: X = 0.1 ms / division Y1 = 1 V / division Y2 = 1 V / division Triggering: Y1 - 0 (Y 1 ,Y2 )
Remarks: Y1: Apparent voltage U Y2: Active voltage UR
Figure 5.4.2.2 • Draw the displayed curves in the grid (figure 5.4.2.2) and determine the phase angle ϕ.
Period duration T
Apparent resistance Z
T=
Z = R 2 + X L2 =
Phase angle
Phase angle cos
ϕ=
between the apparent voltage U and the coil voltage UL
cos ϕ = Inductive reactance XL
XL = ω . L =
UL U
=
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Interconnecting Resistor, Capacitor and Coil
5.5
Parallel Circuiting of Resistor and Coil
5.5.1 General I
If a sinusoidal AC voltage is applied to a parallel circuit of resistor and coil the same voltage is applied to both components. The current I is divided into the coil current IL and the active current I R. IL
IR
Phase shifts result between the currents I, I L and IR due to the inductive reactance XL of the coil. These can be represented by a line diagram
L
R
or a pointer diagram (figure 5.5.1.2).
U
Figure 5.5.1.1
IR = active current in A U = voltage in V I
= apparent current, total current in A
IL = reactive current, coil current in A ϕ = phase angle in ° (degrees)
Figure 5.5.1.2 Pointer diagram for currents
The voltage U is in phase with the active current U R whereas the reactive current I L is constantly 90° later than the voltage U and the active current I R. The phase shift between the apparent current I and the currents I R and IC is determined by the current ratio I L to IR or the conductance ratio B L to G.
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The conductances can also be represented in a pointer diagram which has the following form:
Y = apparent conductance in S (Siemens) G = active conductance in S BL = reactive inductance in S
Figure 5.5.1.3 Pointer diagram for conductances
Direct addition of the partial currents I L and IR and the conductances G and B L for determining the apparent current would lead to an incorrect result due to the phase shift between current and voltage. They are calculated by way of a geometric addition with the following formulas:
Apparent current I
I = I R2 + I L2
I=
U Z
Apparent conductance Y
Y = G 2 + BL2
Y=
1 Z
(Z = apparent resistance)
Cosine, sine and tan of the angle
I G cos ϕ = R = I Y
sin ϕ = I L = BL I Y I B tan ϕ = L = L IR G
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Interconnecting Resistor, Capacitor and Coil
5.5.2 Experiments Experiment Measure the apparent current I, the reactive current I L, the active current I R on a parallel circuit of resistor and coil and calculate the active conductance G, the reactive conductance BL, the apparent conductance Y and the phase angle ϕ. Then construct the pointer diagrams for current and conductance.
I
A
Experiment procedure
B
• Set up the experiment according to circuit
IL U rms = 5 V (sinusoidal)
G
f = 1 kHz
~
E
D
F
100 mH
1 kΩ
Figure 5.5.2.1
• Measuring with the multimeter:
(test points A - B)
I= Reactive current IL (test points C - D)
IL = Active current IR (test points E - F)
IR =
tor and set the following voltage in connecti-
C
0V
Apparent current I
(figure 5.5.2.1), connect the function genera-
IR
on with the multimeter: Urms = 5 V (sinusoidal);
f = 1 kHz
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Alternating Current Technology
Interconnecting Resistor, Capacitor and Coil
• Calculate the reactive conductance B L, apparent conductance Y, phase angle ϕ and active conductance G
and construct the pointer diagrams from the measured values. Active conductance G
G=
1 = R
Apparent conductance Y
Y = G 2 + BC2 =
Reactive conductance B L
Phase angle
1 BL = = ω ⋅L
I sin ϕ = L = I
1 mA / division
0.2 mS / division
Figure 5.5.2.2 Pointer diagram for currents
Fiure 5.5.2.3 Pointer diagram for conductances
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Interconnecting Resistor, Capacitor and Coil
5.6
Series Circuiting of Capacitor and Coil
5.6.1 General C
I
L
If a sinusoidal AC voltage is applied to a series circuit of coil and capacitor the partial voltage UC follows the current I by 90°
UC
UL
and the partial voltage UL precedes the current I by 90°, i. e. both partial voltages (also called reactive voltages) are in opposite phase (180°).
U
Figure 5.6.1.1
If both voltages have the same value, they equalize each other (resonance case). If one of the partial voltages is greater, the
UL
total voltage U is either inductive or capacitive, i. e. the current I 90 °
is 90° later. The losses of the coil and the capacitor are ignored in this
I 90 °
consideration.
UC
Figure 5.6.1.2 Pointer diagram for voltages (UL = UC)
UL
UL
UC
I UL - UC
UC - U L I UL
UC
UC
Figure 5.6.1.3 Pointer diagram for voltages (UL > UC)
Figure 5.6.1.4 Pointer diagram for voltages (UL < UC)
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Alternating Current Technology
Interconnecting Resistor, Capacitor and Coil
5.6.2 Experiments Experiment Measure the partial voltages UC and UL in the series circuits (coil / capacitor) and calculate whether the total voltage U precedes or follows the current I. Then construct the pointer diagrams.
Experiment procedure A
• Set up the experiment according to the circuit (fi-
gure 5.6.2.1), connect the function generator and 1 µF
set the following voltage:
U rms = 3 V
G
(sinusoidal)
~
B
Urms = 3 V (sinusoidal);
f = 1 kHz
f = 1 kHz 100 mH (8 mH) 0V
Figure 5.6.2.1
C
L = 8 mH / 100 mH 8 mH = transformer coil N = 900
• Measure voltages UC and U L. Then construct the pointer diagrams.
Series circuit of C = 1 µF; L = 100 mH:
Series circuit of C = 1 µF; L = 8 mH:
Capacitor voltage UC
Capacitor voltage UC
(test points A - B)
(test points A - B)
UC =
UC =
Coil voltage UL
Coil voltage UL
(test points B - C)
(test points B - C)
UL =
UL =
Total voltage U
Total voltage U
U = UL - UC =
U = UC - UL =
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Interconnecting Resistor, Capacitor and Coil
1 V / division
1 V / division
I
I
Figure 5.6.2.2 Pointer diagram: C = 1 µF; L = 100 mH;
f = 1 kHz
Figure 5.6.2.3 Pointer diagram: C = 1 µF; L = 8 mH;
f = 1 kHz
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Alternating Current Technology
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Notes:
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Interconnecting Resistor, Capacitor and Coil
5.7
Parallel Circuiting of Capacitor and Coil
5.7.1 General If capacitor and coil are circuited in parallel and supplied with a sinusoidal
I
AC voltage, the same voltage is applied to both components. The total current I is divided into the coil current I L and the capacitor current I C, whereby IL
IC
L
C
IL precedes the voltage U by 90°. The currents IL and IC are at opposite phase (180°) and, depending on their size, equalize each other wholly or partly (see pointer diagrams).
U
Figure 5.7.1.1
When I C = IL the total current I is zero (resonance case). IC
90°
U 90°
IL
Figure 5.7.1.2 Pointer diagram for currents (IL = IC)
IC
IC
IL
U IC - IL
IL - IC U
IC
IL
IL
Figure 5.7.1.3 Pointer diagram for currents (IC > IL)
Figure 5.7.1.4 Pointer diagram for currents (IC < IL)
When I C > IL a capacitive residual current is left over, i. e. the total current I is capacitive and precedes voltage U by 90°. When I C < IL an inductive residual current is left over, i.e. the total current I is inductive and follows the voltage U by 90°. Coil and capacitor losses have been ignored in these considerations.
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Alternating Current Technology
Interconnecting Resistor, Capacitor and Coil
5.7.2 Experiments Experiment Measure the total current I, the coil current I L and the capacitor current IC on two parallel circuits (coil / capacitor). Using the measured values, determine whether the total current I precedes or follows the voltage U. Then draw the corresponding pointer diagrams.
I
A
B
Experiment procedure IL
U rms = 3 V (sinusoidal)
G
f = 1 kHz
~
• Set up the experiment according to the circuit
IC
(figure 5.7.2.1), connect the function generator
C
E
D
F
and set the following voltage: Urms = 3 V (sinusoidal);
f = 1 kHz
C
L
0V
Figure 5.7.2.1 L / C-Kombination: 100 mH / 0.47 µF;
100 mH / 0.1 µF
• Measure the currents I, I L and IC Then construct the pointer diagrams.
Parallel of L = 100 mH; C = 0.47 µF
Parallel of L = 100 mH; C = 0.1 µF
Capacitor current IC
Capacitor current IC
(test points E - F)
(test points E - F)
IC =
IC =
Coil current IL
Coil current IL
(test points C - D)
(test points C - D)
IL =
IL =
Total current I
Total current I
(test points A - B)
(test points A - B)
I=
I=
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Interconnecting Resistor, Capacitor and Coil
2 mA / division
1 mA / division
U
U
Figure 5.7.2.2 Pointer diagram: L = 100 mH; C = 0.47 µF;
f = 1 kHz
Figure 5.7.2.3 Pointer diagram: L = 100 mH; C = 0.1 µF;
f = 1 kHz
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Interconnecting Resistor, Capacitor and Coil
Notes:
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101
Interconnecting Resistor, Capacitor and Coil
5.8
Series Circuiting of Resistor, Capacitor and Coil
5.8.1 General R
I
C
When connecting a sinusoidal AC voltage to a
L
series circuit of resistor, capacitor and coil, the same current flows through all three components. UR
UC
UL
The voltage UR is phase equal with the current I. The voltages UR, UC, UL and U are phase-shifted. The resistances behave in accordance with the voltages (see also pointer diagrams figure 5.8.1.2
U
Figure 15.8.1.1
and figure 5.8.1.3).
XL
UL UR
R I
ϕ
ϕ
UC - U L
XC - X L
Z
U
UC
UL
XC
XL
Figure 5.8.1.2 Pointer diagram voltages (UC > UL)
Figure 5.8.1.3 Pointer diagram resistances (X C > XL)
UR = active voltage in V
XL = inductive reactance, coil resistance in Ω
UL = reactive voltage (inductive), coil voltage in V
XC =
UC =
reactive voltage (capacitive), capacitor voltage in V
capacitive reactance, capacitor resistance in Ω
R =
active resistance in Ω
U =
phase voltage, total voltage in V
Z
apparent resistance in Ω
I
current in A
=
ϕ =
=
phase angle in ° (degrees)
The pointer diagrams show the case in which the voltage U C or the resistance XC is greater than the voltage UL or the resistance X L respectively, i. e. the capacitive part is superior. The voltage U follows the active voltage UR. If the inductive part is superior (U L > UC) the relationships are accordingly reversed. If the inductive and capacitive parts are equal they equalize each other due to the phase shift of 180°; in this case the voltage U is equal to the active voltage UR and the apparent resistance Z is equal to the active resistance R.
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Below are some formulas for calculating values in a series circuit of resistor, coil and capacitor.
Apparent voltage U
U = U R2 + (U L − U C ) 2
.I
U=Z
Apparent resistance Z
Z = R 2 + (X L − X C ) 2
Tan of the phase angle
tan ϕ =
U L − UC X − XC = L UR R
Z=
U I
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Interconnecting Resistor, Capacitor and Coil
5.8.2 Experiments Experiment Measure the voltages UR, UC and UL in a series circuit of resistor, capacitor and coil, determine whether the total voltage U precedes or follows the voltage UR and measure the phase angle ϕ with the oscilloscope. A L
Experiment procedure • Set up the experiment according to the
100 mH
circuit (figure 5.8.2.1), connect the function U rms = 3 V
G
(sinusoidal)
~
B C
0.47 µ F
f = 1 kHz
C R
generator and set the following voltage: Urms = 3 V (sinusoidal);
f = 1 kHz
470 Ω
0V
D
Figure 5.8.2.1
• Measuring with the multimeter:
Coil voltage UL (test points A - B)
UL = Capacitor voltage UC (test points B - C)
UC = Active voltage UR (test points C - D)
UR =
• Determination of phase relation of U to U R:
............................................................................... ...............................................................................
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• Connect the oscilloscope as follows to determine the phase angle ϕ between the total voltage U and the
active voltage UR: Test point C to channel 1 (Y 1) Test point A to channel 2 (Y 2) Test point D to ground • Make the other settings on the oscilloscope according to the specifications below the grid (figure 5.8.2.2). • Draw the displayed voltage curves in the grid (figure 5.8.2.2) and determine the phase angle ϕ.
Settings: X = 0.1 ms / division Y1 = 2 V / division Y2 = 2 V / division - 0 (Y 1)
Triggering: Y1 Remarks:
- 0 (Y 2)
Y1: Active voltage UR Y2: Total voltage, apparent voltage U
Figure 5.8.2.2
Period duration T
T=
Phase angle
ϕ=
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Interconnecting Resistor, Capacitor and Coil
5.9
Parallel Circuiting of Resistor, Capacitor and Coil
5.9.1 General If a sinusoidal AC voltage is connected to a parallel circuit of I
resistor, capacitor and coil the voltage on all components is the same.
IC
IL
The total current I is divided into active current I R, capacitor current IC and coil current I L. A phase shift occurs between the
IR
currents IL, IC, IR and I due to the reactances X L of the coil and XC of the capacitor (see pointer diagram figure 5.9.1.2). U
C
L
R
Figure 5.9.1.1
I
IC
= total current, apparent current in A
IC = reactive current (capacitive), capacitor current in A
IL
IL = reactive current (inductive), coil current in A I
IC
ϕ
IR = active current in A
- IL U
U = voltage in V ϕ = phase angle in ° (degrees)
IR IL
Figure 5.9.1.2 Pointer diagram for currents (IC > IL)
The current IC precedes current IR constantly by 90°, whilst the current I L follows the active current I R constantly by 90°. Currents IC and IL are therefore in opposite phase (180°) and equalize each other wholly or partly depending on their size. When I C = IL the two currents equalize each other, the total current I is equal to the active current in phase and value (resonance). When I C > IL a capacitive residual current is left over, the total current I precedes the active current I R. When I C < IL an inductive residual current is left over, the total current I follows the active current I R.
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Interconnecting Resistor, Capacitor and Coil
The pointer diagram figure 5.9.1.3 shows how the conductances B L, BC, G and Y behave in a parallel circuit of resistor, capacitor and coil.
BC = reactive conductance (capacitive) in S BC
BL
BL = reactive conductance (inductive) in S Y = apparent conductance in S
Y BC B L
G = active conductance in S
ϕ
G BL
Figure 5.9.1.3 Pointer diagram for conductances (BC > BL)
The pointer diagrams are prepared for the case I C > IL or BC > BL.
Formulas for calculating the values in a parallel circuit of R, L and C:
Apparent current I
I = I R2 + (I C − I L ) 2
Apparent conductance Y
Y = G 2 + (BC − BL ) 2
Tan of the phase angle
I − IL tan ϕ = C IR B −B tan ϕ = C G L
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Interconnecting Resistor, Capacitor and Coil
5.9.2 Experiments Experiment ϕ. Measure the current I, LI , IC and IR in a parallel circuit of resistor, capacitor and and calculate the phase angle
Then construct the pointer diagrams for currents and conductances.
I
A
Experiment procedure • Set up the experiment according to the
B
circuit (figure 5.9.2.1), connect the funcIR
IL
tion generator and set the following vol-
IC E
G
D
F
H
Urms = 3 V
G
(sinusoidal)
~
f = 1 kHz
tage:
C
R 470 Ω
C 0.47 µF
Urms = 3 V (sinusoidal);
L
100 mH
0V
Figure 5.9.2.1
• Measuring with the multimeter:
Total current I
Coil current IL
(test points A - B)
(test points G - H)
I=
IL =
Active current IR
Phase angle
(test points C - D)
I − IL tan ϕ = C = IR
IR =
Capacitor current IC (test points E - F)
IC =
f = 1 kHz
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Alternating Current Technology
Interconnecting Resistor, Capacitor and Coil
Calculations for constructing the pointer diagrams:
Reactive conductance (capacitive) BC
BC = ω . C =
Active conductance G
G=
1 = R
Reactive conductance (inductive) BL
Apparent conductance Y
1 BL = = ω⋅L
Y = G 2 + (BC − BL ) 2 =
2 mA / division
1 mS / division
Figure 5.9.2.2 Pointer diagram for currents
Figure 5.9.2.3 Pointer diagram for conductances
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Interconnecting Resistor, Capacitor and Coil
5.10 Active, Reactive and Apparent Power 5.10.1 General The previous experiments treated exclusively the behaviour I
of voltage, current and resistance connecting capacitors, coils and ohmic resistors. The object of the experiment are IC
U (sinusoidal)
UC
IL UL
IR UR
the resulting powers. Like the voltages and currents the powers are also mutually phase-shifted due to the reactances.
Figure 5.10.1.1
In figures 5.10.1.2 to 5.10.1.4 the power ratios in a parallel circuit of resistor, capacitor and coil (figure 5.10.1.1) are shown as pointer diagrams.
QC
QC
QL
P S QC - QL ϕ
ϕ
Q L- Q C
S P QL
Figure 5.10.1.2 Pointer diagram for powers (Q C > QL) S
= apparent power in VA
P
= active power in W
QL
QC
Figure 5.10.1.3 Pointer diagram for powers Q L > QC
QL = inductive reactive power in var QC = capacitive reactive power in var ϕ
= phase angle in ° (degrees)
If the capacitive reactive power QC is greater than the inductive reactive power QL (QC > QL) the apparent power S precedes the active power P (figure 5.10.1.2). In the reverse case (Q L > Q) the apparent power S follows the active power P (figure 5.10.1.3).
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Alternating Current Technology
Interconnecting Resistor, Capacitor and Coil
If both reactive powers are the same (QL = QC) they neutralize each other and the apparent power S is equal to the active power P (figure 5.10.1.4).
QC
The representation of the power ratios in a series circuit of resistor, capacitor and coil are similar to those in a parallel circuit except that the
P (S)
power pointers QC and QL are reversed. QL
Figure 5.10.1.4 Pointer diagram for powers QL = QC
The individual powers are calculated with the following formulas:
Apparent power S
S = P 2 + (QL − QC ) 2
S=U
.I
Active power P
P = U . I . cos ϕ
P = S . cos ϕ
Reactive power QL, QC
Q = U . I . sin ϕ
sin ϕ =
Q S
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Interconnecting Resistor, Capacitor and Coil
5.10.2 Experiments Experiment Measure the apparent power S, the active power P, the reactive powers Q C and Q L and calculate the phase angle ϕ in a parallel circuit of resistor, capacitor and coil and then construct the corresponding pointer diagram.
Experiment procedure I
A
B
• Set up the experiment according to the
IR
circuit (figure 5.10.2.1), connect the func-
IL
tion generator and set the following vol-
IC C
E
G
D
F
H
Urms = 3 V
G
(sinusoidal)
~
f = 1 kHz
R 470 Ω
C 0.47 µF
tage: Urms = 3 V (sinusoidal);
L
100 mH
0V
Figure 5.10.2.1
• Measure the individual currents with the multimeter in order to calculate the powers.
Total current I (test points A - B)
I= Active current IR (test points C - D)
IR = Capacitor current IC (test points E - F)
IC = Coil current IL (test points G - H)
IL =
f = 1 kHz
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Interconnecting Resistor, Capacitor and Coil
Calculate the powers and the phase angle with the given formulae and then draw the pointer diagram.
Active power P
Apparent power S
P =U . IR =
S = U.I=
Reactive power (capacitive) QC
Phase angle
QC =U . IC =
Reactive power (inductive) Q L
QL =U . IL =
6 mW (mvar, mVA) / division
Figure 5.10.2.2 Pointer diagram for power
cos ϕ =
P = S
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113
Transformers
6
Transformers
6.1
Fundamentals
Transformers consist of two or more windings (coils) which are magnetically coupled by a core. They are used for voltage, current and resistance matching and transformation as well as for isolation of electrical circuits (galvanic isolation). In the ideal, i. e. lossless, transformer, the power consumed is equal to the power produced. In practice losses occur in the winding and in the core of the transformer so that only a part of the consumed power is passed on to the load resistor R L. The losses consist mainly of the copper losses (ohmic resistance of the coil windings), the core losses and losses caused by introduction of an air gap into the cross section of the core to improve the transformation properties.
softmagnetic core
I1
input side
G
I2
U1
primary coil
U2
RL
output side
secondary coil
Figure 6.1.1
6.2
Coupling Factor
6.2.1 General To get a fixed magnetic coupling between the primary and secondary coils of a transformer, they are connected through a core. Depending on the coupling factor the magnetic flux in the core is greater or smaller and an accordingly higher or lower voltage is induced in the secondary coil. To avoid distortion of the signals to be transformed and premature magnetic saturation of the core material by DC current, the coupling factor is reduced by a break in the core (air gap).
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Alternating Current Technology Transformers
6.2.2 Experiments Experiment Urms = 4 V f = 1 kHz
Determine the size of the magnetic coupling betU1
G ~
N1 900
ween two coils by doing voltage measurements for:
N2 900
U2
- coils with core - coils with core and air gap - coils without core
0V
Experiment procedure Figure 6.2.2.1
•
Put together a primary and secondary coil with 900 windings each with the split tape core as
Primary coil
Secondary coil
shown in the diagram (figure 6.2.2.2). •
Connect the function generator to the primary coil as shown in the circuit diagram (figure 6.2.2.1) and set a sinusoidal voltage Urms = 4 V ; f = 1 kHz.
•
Measure the secondary voltage U 2 with the multimeter and enter the value in the table 6.2.2.1 under „Coils with core“.
N2 900
N1 900
•
To measure the secondary voltage on coils with core and air gap, place a piece of paper between
Figure 6.2.2.2
the upper and lower halves of the split tape core to simulate the air gap. •
N1
N2
900
900
To determine the secondary voltage in „Coils without core“ arrange the coils as shown in figure 6.2.2.3.
Figure 6.2.2.3
Question: Why do the secondary voltages differ?
Coupling factor U[rms] 1 Coils with core
4V
Coils with core and air gap
4V
Coils without core
4V
Table 6.2.2.1
U
2
[rms]
Answer:
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115
Transformers
6.3
Transformation Ratio
6.3.1 General In a transformer the ratio of the number of windings on the primary side to the number of windings on the secondary side is referred to as the transformation ratio.
I1
I2
U1, I1
= primary voltage and current
U2, I2
= secondary voltage and current
N1, N2 = primary and secondary windings N1
N2
U1
U2
Figure 6.3.1.1
This ratio corresponds to the ratio of primary voltage U 1 to secondary voltage U2 when the transformer is on no-load. The transformation ratio of the current is inverse to the voltage or winding ratio. Transformation ratio r:
r =
N1 N2
=
U1 U2
=
I2 I1
6.3.2 Experiments Experiment Determine the transformation ratio on the primary and secondary sides of a transformer with different numbers of windings by measuring voltage and current. A Urms = 6 V f = 1 kHz U 1
G ~
N1 = 900
N2 = 300 900
U2
0V
Figure 6.3.2.1
V
G ~
Urms = 6 V f = 1 kHz
I1
N1 = 900
I2 N2 = 300 900
0V
Figure 6.3.2.2
Experiment procedure •
Put together a primary and secondary coil with 900 and 300 windings each with the split tape core as shown in the diagram (figure 6.3.2.3).
•
Connect the function generator to the secondary coil as shown in the circuit diagram (figure 6.3.2.1) and set a sinusoidal voltage Urms = 6 V; f = 1 kHz.
A
116
Alternating Current Technology Transformers
Primary coil
•
Secondary coil
Measure the secondary voltage U 2 at the secondary coil with 300 and 900 windings and enter the values in the table 6.3.2.1.
•
Calculate the transformation ratio of the voltages with the formula r = U1 / U2.
•
Then measure the primary and secondary currents on the secondary coils with 300 and 900 windings as shown in the circuit (figure 6.3.2.2) and enter the values in table 6.3.2.2. The voltage U 1 of 6 V (rms) should be kept constant.
N2 300 900
N1 900
Figure 6.3.2.3
•
Calculate the transformation ratio r of the currents with the formula:
r =
I2 I1
=
N1
N2
900
300
900
900
U[V] 1
U
2[V]
N1
N2
6
900
300
6
900
900
Table 6.3.2.1
r
I[mA] 1
I
2[mA]
Table 6.3.2.2
Transformation ratios of the voltages:
Transformation ratios of the currents:
When N=1 900;
When N=1 900;
r =
U1 U2
U1 U2
2
= 300
r =
=
When N=1 900; r =
N
=
N
2
= 900
I2 I1
I2 I1
2
= 300
N
2
= 900
=
When N=1 900; r =
N
=
r
Alternating Current Technology
117
Transformers
6.4
Resistance Transformation
6.4.1 General If a transformer is not loaded (no-load) the transformation ratio between primary and secondary voltages corresponds approximately to the ratio between the number of windings on the primary and secondary coils (transformation ratio). r =
U1
=
U2
N1 N2
When there is a load on the secondary coil, a current I 2 flows through the load resistor R L. This current is active, multiplied by the reciprocal of the transformation ratio, in the primary coil of the transformer. I1 = I 2 ⋅
1
r =
r
I2 I1
Primary current and primary voltage together form the input resistance of the transformer. R1 =
U1 I1
The output or load resistance is given by: R2 = R L =
U2 I2
If both resistances are put in relationship, the following resistance transformation is achieved:
U1 R1 RL
=
I1 U2
=
U1 ⋅ I2 U 2 ⋅ I1
=r⋅r =r
2
or:
R1 = R L ⋅ r
2
I2
This means the secondary load R L is transformed to the primary side of the transformer with the square of the transformation ratio N1 / N2.
6.4.2 Experiments Experiment Establish the resistances R1 and R2 by measuring the current and voltage on the primary and secondary sides of a transformer. Do this for different transformation ratios of the number of primary windings to the number of secondary windings and for different loads R L. Then form the ratio between R1 and R2 and compare with the square of the respective transformation ratio.
118
Alternating Current Technology Transformers
Experiment procedure
I1
I2
mA
mA
•
Put together the transformer with the split tape core and the coils (N1 = 900; N 2 = 300) as shown in the
Urms = 6 V
G ~
N1
f = 1 kHz
N2
U2
diagram (figure 6.4.2.2).
RL •
Connect the function generator as shown in the circuit diagram (figure 6.4.2.1) and set a sinusoidal
0V
voltage of
Figure 6.4.2.1
Urms = 6 V; f = 1 kHz (at R L = 10
RL = 100 Ω / 1 kΩ , N1 = 900, N 2 = 300 / 900
•
Ω).
Measure current and voltage at the winding ratios and loads RL specified in table 6.4.2.3.
Primary coil
Secondary coil
•
Calculate the resistances R1 and R2 according to Ohm’s Law with the formula R = U / I.
•
Then calculate the transformation ratios of the resistances according to the specifications and compare with the winding ratios.
N2 300 900
N1 300
Figure 6.4.2.2
N1
N2
r
RL [ ]
900
300
0.33
100
900
900
1
1000
U [V] 1
U [V] 2
I [mA] 1
I [mA] 2
R
1
[ ]
Table 6.4.2.3
Ratio between the resistances:
When N 1 = 900; N 2 = 300; R L = 100 r
2
=
R1
Ω
When N 1 = 900; N 2 = 900; R L = 1000 r
=
2
R2
r =
r
2
=
R1
=
R2
=
=
r =
r
2
=
=
Ω
R2 [ ]
