UEUNEEG002B
Electrical, Electronics
Engineering Department
Solve Problems in
Single and Three
Phase Circuits
UEUNEEG002B
Vol 1 of 2
Revision
Date
Contact
1
03/09
RC
2
05/10
DK
3
4
Chisholm Institute
Stud Road
Dandenong 3175
2
Solve problems in single and three phase low voltage
circuits
Required Skills and Knowledge
Evidence shall show that knowledge has been acquired of safe working practices
and solving problems in electromagnetic circuits.
The extent of the essential knowledge and associated skills (EKAS) required is
given in Volume 2 – Part 2.2 EKAS. It forms an integral part of this unit.
2.8.2.2 Alternating current principles – power
Evidence shall show an understanding of alternating currents principles used in
power circuits to an extent indicated by the following aspects:
a) Sinusoidal alternating voltage and current encompassing:
• Generation of a sinusoidal voltage with a single turn coil rotated in a
uniform magnetic field.
• The terms ‘period’, ‘maximum value’, ‘peak-to-peak value’,
‘instantaneous value’, ‘average value’, ‘root-mean-square (r.m.s.)
value’, ‘crest factor’, and ‘form factor’ in relation to a sinusoidal
waveform.
• The instantaneous value of induced voltage of a generated
sinusoidal waveform.
• Measurement of the instantaneous, peak, peak-to-peak values and
the period of a sinusoidal waveform.
• The root-mean-square (r.m.s) value and frequency of a sinusoidal
waveform.
• Phase relationship between two or more sinusoidal waveforms.
b) Phasors encompassing:
• The terms ‘in-phase’, ‘out-of-phase’, ‘phase angle’, ‘lead’, and ‘lag’.
• The phase angle between two or more alternating quantities from a
given sinusoidal waveform diagram.
• Convention for representing voltage, current and the reference
quantity in a phasor diagram
• Phasor diagrams two or more a.c. values of voltage and / or current.
c) Resistance in a.c. circuits encompassing:
• Connection of a single-source a.c. circuit to take resistance, voltage
and current measurements.
• The voltage, current, resistances or power dissipated from
measured or given values of any two of these quantities.
• The relationship between voltage drops and current in a resistive
a.c. circuit.
3
d) Inductive in a.c. circuits encompassing:
• Definition of ‘inductive reactance’.
• The inductive reactance of a given inductor and show the
relationship between inductive reactance and frequency.
• Equivalent inductive reactance in an a.c. circuit or any part of a
circuit.
• Application of Ohm’s Law to determine voltage, current or inductive
reactance in a purely inductive a.c. circuit given any two of these
quantities.
• Examples of inductive components in power circuits and systems
and describe their effect on the phase relationship between voltage
and current.
• The comparative current limiting characteristics of inductors and
resisters.
e) Capacitance in a.c. circuits encompassing:
• Definition of ‘capacitive reactance’.
• The capacitive reactance of a given capacitor and the relationship
between capacitive reactance frequency.
• Equivalent capacitive reactance in an a.c. circuit or any part of a
circuit.
• Application of Ohm’s Law to determine voltage, current or capacitive
reactance on a purely capacitive a.c. circuit given any two of these
quantities.
• Examples of capacitive components in power circuits and systems
and describe their effect on the phase relationship between voltage
and current.
f) Impedance encompassing:
• Definition of ‘impedance’.
• Impedance of series, parallel and series-parallel circuits and
diagrams showing the relationship between resistive, inductive and
capacitive components (impedance triangle).
• Connection of a single-source a.c. circuit and take resistance,
voltage and current measurements.
• Voltage, current or impedance values from measured or given
values of any two of these quantities.
• Phasor diagram usage to solve problems and show the relationship
between voltages and currents in a.c. circuits.
4
g) Resonance encompassing:
• Conditions in a circuit that produce resonance.
• The relationship between resonance and frequency.
• The effect on the current of series resonance and parallel resonance
conditions.
• Applications where resonance is applied.
h) Power and power factor encompassing:
• Difference between true power, apparent power and reactive power
and the units.
• Definition of the term “power factor”.
• The effects of low power factor.
i) Multiphase systems encompassing:
• Features of a multiphase system.
• Voltages generated by single and multiphase alternators.
• Reasons for the adoption of three-phases for power systems.
j) Three-phase principles encompassing:
• Generation of three-phases in a single alternator.
• r.m.s. value of voltage generated in each phase.
• The relationship between the phase voltages generated in a three
phase alternator and the conventions for identifying each.
• The term ‘phase sequence’ (also referred to as ‘phase rotation”).
• Determination of the phase sequence of a three-phase supply.
k) Three-phase star-connections encompassing:
• Connection of three-phase system.
• The phase relationship between line and phase voltages and line
and phase currents of a star-connected system.
• The r.m.s. value of line and phase voltage given any one of these
quantities.
• The r.m.s. value of line and phase current given any one of these
quantities.
• The terms ‘balanced load’ and ‘unbalanced load’.
• Example of balanced and unbalanced load in a typical power
system.
5
l) Three-phased four wire systems encompassing:
• Purpose of the neutral conductor in a three-phase four wire system.
• Effects of a high impedance in the neutral conductor of a threephase four wire system supplying an unbalanced load where MEN
earthing is employed.
• The value and phase relationship of neutral current in an
unbalanced three-phase four wire system given line currents and
power factors.
m) Three-phase Delta systems encompassing:
• Connection of three-phase Delta systems.
• Phase relationship between line and phase voltages and currents in
a Delta connected system.
• The r.m.s. value of line and phase voltage given any one of these
quantities.
• The r.m.s. value of line and phase currents given any one of these
quantities.
• Example of Delta-connected loads in typical power systems.
n) Interconnected star and Delta systems encompassing:
• Relationship between line and phase voltages and currents in a
system with a star supply and a Delta load
• Relationship between line and phase voltages and currents in a
system in a Delta connected supply and a star connected load.
o) Energy and power requirements in a.c. systems encompassing:
• The purposes for measuring power, energy, power factor and
maximum demand of a.c. power systems and loads.
• Methods used to measure power, energy, power factor and
maximum demand.
• Power factor improvement of a three-phase installation.
p) Harmonics encompassing:
• The term ‘harmonic’ in relationship to the sinusoidal waveform of an
a.c. power system.
• Sources in a.c. systems that produce harmonics.
• Problems that may arise in a.c. circuits as a result of harmonics and
how these are overcome.
6
Contents
Contents .......................................................................................................................... 7
ALTERNATING CURRENT THEORY........................................................................ 8
GENERATING A SINE WAVE ................................................................................ 8
Exercises ................................................................................................................... 10
ALTERNATING CURRENT TERMINOLOGY ......................................................... 11
Exercises ................................................................................................................... 14
AVERAGE VALUE ...................................................................................................... 17
ROOT MEAN SQUARE (R.M.S.) VALUE................................................................ 18
INSTANTANEOUS VALUE ........................................................................................ 20
CREST FACTOR ......................................................................................................... 20
FORM FACTOR ........................................................................................................... 22
TYPES OF WAVEFORMS ......................................................................................... 23
PHASE RELATIONSHIPS AND WAVEFORMS .................................................... 24
Exercises ................................................................................................................... 26
VECTOR (OR PHASOR) QUANTITIES .................................................................. 31
WAVEFORMS AND PHASORS................................................................................ 34
Exercises ................................................................................................................... 38
RESISTANCE IN AC CIRCUITS ............................................................................... 49
Exercises ................................................................................................................... 52
Practical Exercise .................................................................................................... 56
INDUCTANCE IN AC CIRCUITS .............................................................................. 57
COMPONENTS IN AC CIRCUITS............................................................................ 58
INDUCTORS IN AN AC CIRCUITS ......................................................................... 58
Exercises ................................................................................................................... 62
INDUCTORS IN SERIES ........................................................................................... 64
INDUCTORS IN PARALLEL ...................................................................................... 64
PHASOR RELATIONSHIP OF INDUCTORS IN SERIES .................................... 66
PHASOR RELATIONSHIP OF INDUCTORS IN PARALLEL ............................... 68
POWER IN AN INDUCTIVE CIRCUIT ..................................................................... 70
Exercises ................................................................................................................... 71
CAPACITANCE IN AC CIRCUITS ............................................................................ 77
CAPACITORS IN AN AC CIRCUIT .......................................................................... 78
Circuit Diagram, Waveform Diagram and Phasor Diagram .............................. 79
CAPACITORS IN SERIES AND CAPACITIVE REACTANCE ............................. 81
CAPACITORS IN PARALLEL AND CAPACITIVE REACTANCE ....................... 82
Exercises ................................................................................................................... 83
POWER IN A CAPACITIVE CIRCUIT ...................................................................... 85
Exercises ................................................................................................................... 86
AC CIRCUITS .............................................................................................................. 91
IMPEDANCE ................................................................................................................ 92
RESISTANCE AND INDUCTANCE IN SERIES ..................................................... 93
Exercises ................................................................................................................... 97
RESISTANCE AND CAPACITANCE IN SERIES ................................................ 100
Exercises ................................................................................................................. 104
7
ALTERNATING CURRENT THEORY
GENERATING A SINE WAVE
If single loop conductor, connected to slip rings, is uniformly rotated through a
magnetic field, it will have an electro motive force (e.m.f.) induced into the loop as
it cuts the magnetic lines of force.
Maximum induced e.m.f. will occur at the time the loop is at right angles to the
lines of force and minimum induction occurs at the time that the loop is parallel to
the lines of force.
V+
At 0
Intended waveshape
N
S
D.O.R.
V-
Loop at minimum cutting effect
8
V+
At 90
N
S
D.O.R.
V-
Loop at maximum (+ ve) effect
V+
At 180
N
S
D.O.R.
V-
Loop at minimum cutting effect
V+
At 270
N
S
D.O.R.
V-
Loop at maximum (-ve) cutting effect
V+
At 360
N
S
D.O.R.
V-
Loop back at starting point with no cutting effect
9
EXERCISES
Using the rotating line method, sketch the resulting wave shape in the space
provided.
10
ALTERNATING CURRENT TERMINOLOGY
Frequency
Symbol
-
f
Unit
-
Hertz (Hz)
This is the number of complete cycles per second.
Our mains supply is generated at 50 Hz. Multiples of Hz is common, such as
KHz, MHZ and GHz (used in communications)
Period
Symbol
-
T
Unit
-
Seconds (s)
This is the time taken for an AC waveform to pass through one complete cycle.
V+
1 Cycle = Period
T
V-
A mathematical relationship exists between f and T.
1
f =
T
where -
f
=
frequency in
=
time in
Hz
T
seconds
11
EXAMPLE 1
If the period of a sine wave is 100ms, how many cycles per second?
1
f=
T
=
1
100ms
= 10 Hz
EXAMPLE 2
What is the frequency of a sine wave whose period is 50µs?
f=
=
1
T
1
50μ0
= 20,000 Hz
= 20 KHz
EXAMPLE 3
A sine wave has a frequency of 1 KHz. How long would it take for one cycle?
T=
=
1
f
1
1000
= 1 ms
12
PEAK VALUE
The peak value is the maximum value of current or voltage. It occurs twice in one
cycle. There is a +ve peak and a -ve peak.
V+
Positive Peak
T
Negative Peak
V-
PEAK TO PEAK VALUE
This is twice peak value and is measured from the +ve peak to the -ve peak.
V+
T
V-
13
EXERCISES
For the following questions identify the response you consider best answers the
question by placing the corresponding letter in the bracket provided.
1.
The recurring period of an AC waveform is called the:
a.
b.
c.
d.
e.
2.
cycle
period
frequency
mechanical degrees
electrical degrees
(
)
cycle
period
frequency
mechanical degrees
electrical degrees
(
)
When a coil or armature conductor makes one complete revolution it
passes through 360:
a.
b.
c.
d.
e.
5.
)
The number of cycles occurring in a unit of time is called the:
a.
b.
c.
d.
e.
4.
(
The time required for one complete cycle is called the:
a.
b.
c.
d.
e.
3.
cycle
period
frequency
mechanical degrees
electrical degrees
cycle
period
frequency
mechanical degrees
electrical degrees
(
)
When an electromotive force or an alternating current passes through 1
cycle, it passes through 360:
a.
b.
c.
d.
e.
cycle
period
frequency
mechanical degrees
electrical degrees
(
)
14
6.
A period of 4ms would have a frequency of:
a.
b.
c.
d.
e.
7.
(
)
250Hz
2000Hz
2.5Hz
0.5Hz
4Hz
(
)
0.02 seconds
0.001 seconds
0.00004 seconds
0.2 seconds
0.004 seconds
(
)
The time taken for each cycle if the frequency was 1kHz would be:
a.
b.
c.
d.
e.
11.
250Hz
2000Hz
2.5Hz
0.5Hz
4Hz
A waveform which has 50 cycles would have a period of:
a.
b.
c.
d.
e.
10.
)
A period of 2s would have a frequency of:
a.
b.
c.
d.
e.
9.
(
A period of 400ms would have a frequency of:
a.
b.
c.
d.
e.
8.
250Hz
2000Hz
2.5Hz
0.5Hz
4Hz
0.02 seconds
1 x 10-3 seconds
4 x 10-5 seconds
0.2 seconds
0.004 seconds
(
)
(
)
The time for each cycle if the frequency was 25kHz would be:
a.
b.
c.
d.
e.
0.02 seconds
0.001 seconds
0.00004 seconds
0.2 seconds
0.004 seconds
15
12.
The peak to peak value of voltage can be calculated by:
a.
b.
c.
d.
Adding the +ve maximum value and the -ve maximum value
Find the difference between the maximum +ve value and the zero
voltage line
Find the difference between the maximum -ve value and the zero
voltage line
Summing the instantaneous values at 0o and 180o
( )
16
AVERAGE VALUE
Mathematically the average value of either voltage or current can be found by
using the formula:
Av = 0.637 × Vmax
for V or I:
VAV = 0.637 × Vmax
or
IAV = 0.637 × Imax
[Maximum (max) is another term similar to peak]
On a sine wave, the average value can be shown:
V+
0.637
Average
T
V-
EXAMPLE 1
What is the average value of a peak voltage of 340V?
VAv
=
=
=
0.637 x Vmax
0.637 x 340
216.5 V
EXAMPLE 2
Find the maximum current if an average current of 100A flows in a circuit:
Imax =
=
IAV
0.637
100
0.637
= 156.98A
17
ROOT MEAN SQUARE (R.M.S.) VALUE
The Root Mean Square (R.M.S.) value is based on the power-producing ability of
an alternating current. Another term used instead of RMS is “effective” value.
It is the value of the AC that has the same effect as a DC in producing power in a
circuit.
Mathematically expressed:
RMS = 0.707 x max
for V or I
V RMS = 0.707 x V max or IRMS = 0.707 x Imax
On a sine wave the RMS value can be shown:
V+
RMS Value
0.707
T
V-
EXAMPLE 1
What is the RMS value of a 60A maximum current flow?
IRMS
=
=
=
0.707 x Imax
0.707 x 60
42.42A
18
EXAMPLE 2
A cable is designed to withstand 660V. What is the maximum voltage?
V max =
=
V RMS
0.707
660
0.707
= 933V
Showing the previous two values on a sine wave:
V+
0.707
RMS Value
0.637
Average Value
T
V-
19
INSTANTANEOUS VALUE
This is the value of V or I at any point along the time axis. It can have a +ve or ve value.
The instantaneous value can be mathematically found by:
Inst. Value = Max. Value x Sin Θ
CREST FACTOR
This term is the ratio relationship between the maximum (peak) value and the rms
value of a waveform.
i.e.
crest factor =
V MAX
V RMS
There is no unit for this term only a number.
If the RMS value of any waveform is divided into the maximum value, the number
1.414 should always be obtained. Its primary use is to find the insulation value (in
volts) of conductors.
EXAMPLE
What is the crest value of 240V RMS?
First find the max value:
now crest factor:
max value =
V MAX
V RMS
=
339
240
RMS
0.707
=
240
0.707
= 339 V
= 1.414
V MAX = V RMS x 1.414
The formula can also be transposed:
and V RMS =
NOTE -
1.414 is also √2
V MAX
1.414
20
EXAMPLE 1
If the maximum voltage of a supply is 415V, what is the RMS value?
V RMS =
=
V MAX
2
415
2
= 293.5V
EXAMPLE 2
Given the RMS value of 115V, find the value of insulation required for the
conductors of the supply.
VMAX =
=
=
V RMS x 1.414
115 x 1.414
162.6 V
21
FORM FACTOR
This term is given to the ratio of the RMS value to the average value.
i.e.
form factor
=
=
RMS Value x Max Value
Av. Value x Max Value
0.707
0.637
= 1.11
As the waveform becomes “flat topped” the number tends toward a value of <
1.11. When the waveform becomes more “peaked” the value rises above 1.11.
Its primary use is in calibrating the scales of AC moving coil meters.
V+
V+
T
T
V-
V-
Form Factor of >1.11
Form Factor of <1.11
22
TYPES OF WAVEFORMS
V+
(A) DIRECT CURRENT
v+
v-
(B) SINUSOIDAL
v+
v-
(C) SAWTOOTH
v+
v-
(D) TRIANGULAR
v+
v-
(E) SQUARE
v+
v-
(F) PULSE
23
PHASE RELATIONSHIPS AND WAVEFORMS
V+
Voltage & Current
V
I
V-
T
I
V
Current and Voltage in Phase
V+
Voltage & Current
V
I
T
V-
I
Φ
V
Current Leading the Voltage
24
V+
Voltage & Current
V
I
T
V-
V
Φ
I
Current Lagging the Voltage
120o
Three Phase Line and Phase Voltage
25
EXERCISES
For the following questions identify the response you consider best answers the
question by placing the corresponding letter in the bracket provided.
1.
The instantaneous value of alternating current is given by:
a.
b.
c.
d.
2.
(
)
The peak or maximum value of alternating current is given by:
a.
b.
c.
d.
3.
VINST = VMAX x sin Θ
VINST = VMAX
sin Θ
VINST = (VMAX)2 x sin Θ
VINST = VMAX x (sin Θ)2
VMAX = VRMS
0.707
VMAX = VRMS
1.414
VMAX = 0.707
VRMS
VMAX = √2 VRMS
(
)
The average value of voltage in an AC waveform can be calculated using:
a.
b.
c.
d.
VAVE = VMAX
0.707
VAVE = VMAX
0.637
VAVE = 0.637 x VMAX
VAVE = 0.707 x VMAX
(
)
26
4.
On the following diagram, indicate the five (5) missing terms:
(b)
(e)
(c)
(a)
(d)
a)
b)
c)
d)
e)
5.
The R.M.S. voltage or effective voltage of an AC circuit can be calculated
using:
a.
b.
c.
d.
VRMS = VMAX x 1.413
VRMS = VMAX x √2
VRMS = VMAX x 0.707
VRMS = VMAX ÷ .707
6.
What is the term used to denote cycles per second?
7.
What do the initials R.M.S. stand for?
(
)
27
For the following exercises, show all formulae and calculations
8.
Calculate the R.M.S. values of the following peak values:
a.
707V
b.
4.5 kV
c.
6A
d.
339V
9.
Calculate the peak values from the following:
a.
110V R.M.S.
b.
345 VAV
28
10.
Calculate the average values of the following peak values:
a.
10V
b.
50A
c.
600V
d.
339V
11.
Define the term “crest factor”
12.
Define the term “form factor”
29
13.
Sketch the waveform (and label appropriately) of
a.
Form factor of < 1.11
b.
Form factor of >1.11.
30
VECTOR (OR PHASOR) QUANTITIES
A vector (or phasor) shows the magnitude and angular displacement from a given
reference point.
The phasor has magnitude and direction and is represented as a straight line
measured to a given scale.
The phasor is commonly used in AC theory calculations to represent values such
as:
•
voltages
•
currents
•
impedances
•
powers
•
current in the neutral
By using appropriate scales, solutions can be found by solving two, three or more
phasors at a time.
The importance of using phasors is:
•
a rule is required
•
a protractor is used
•
a compass is needed
•
the task is completed using a sharp, light pencil
31
In phase, phase relationship
This occurs when the V and current rise and fall at the same time.
V+
Voltage & Current
V
I
T
V-
Current and Voltage in Phase
This is shown by phasors as:
I
V
Out of phase (lagging), phase relationship
This occurs in inductive circuits when the current lags the voltage by Ø:
V+
Voltage & Current
V
I
T
V-
By phasors it is shown as:
V
Ø
I
It can lag up to 90
o
32
Out of phase (leading), phase relationship
This occurs in capacitive circuits when the current leads the voltage by Ø:
V+
Voltage & Current
V
I
T
V-
By phasors it is shown as:
I
Ø
V
It can lead up to 90o
33
WAVEFORMS AND PHASORS
Two voltages in phase
V+
VT
V2
Voltage & Current
V1
I
T
V-
The above is shown by phasors as:
200V
V2
100V
V1
I
300V
VT
I
The circuit representation is:
100V
200V
R1
R2
300V
AC
The total voltage is 300V.
34
Two voltages out of phase
VT
V+
200V
VC
100V
VR
I
0
V-
The above is shown by phasors as:
VR 100V
VC
200V
I
VT
223V
The circuit representation is:
VR
VC
R1
C
VT
AC
35
Two voltages out of phase
VT
V+
200V
VR
100V
VL
I
0
V-
The above is shown by phasors as:
VL
100V
VT
180V
VR
150V
I
The circuit representation is:
VR
VL
R1
L
VT
AC
36
Three voltages out of phase (R, L & C)
VT
V+
VR
VC
VL
I
V-
The above is shown by phasors as:
VL
100V
VT 158V
I
VR 150V
VC
50V
The circuit representation is:
VR
R
VL
L
VT
AC
VC
C
I
37
EXERCISES
For each of the following questions (1 - 6), identify the response you consider best
answers the question by placing a tick () in the relevant box.
1.
A closed arrowhead on a phasor diagram represents a current phasor.
 TRUE
2.
A phasor and a vector mean the same thing.
 TRUE
3.
 FALSE
Current and voltage phasors cannot be added together.
 TRUE
5.
 FALSE
Phasors are always drawn to represent peak values.
 TRUE
4.
 FALSE
 FALSE
Both the direction and magnitude must be taken into consideration during
the process of adding phasors together.
 TRUE
 FALSE
38
6.
a.
Complete the following phasors.
I1
I2
b.
I1
I2
c.
V1
V2
d.
V1
V2
39
e.
f.
g.
h.
40
In the following questions you are required to provide a phasor diagram, scaled as
indicated, in the space provided under the question.
7.
Two voltages, one of 340V and the other of 240V are 90o to each other.
Using a phasor diagram, determine the magnitude and the angle of the
resultant voltage with respect to the 340V value.
Scale: 10mm = 20V
41
8.
Find the magnitude and direction of the resultant phasor if phasor 1 is 40V
and leads by 90o, phasor 2 is 30V and lags by 20o. (Use a horizontal line
as a reference)
Scale: 10mm = 5 volts
42
9.
Find the magnitude and direction of the resultant phasor if phasor 1 is 15
units long and phasor 2 is leading by 180o and is 10 units long.
Scale: 10mm = 1.5 units
43
10.
Find the resultant phasor and give its magnitude and direction if phasor 1
leads by 120o and is 20 units long and phasor 2 is lagging by 60o and is 20
units long. (Use a horizontal line as a reference).
Scale: 10mm = 2 units
44
11.
Find the magnitude and direction of the resultant phasor if phasor 1 is 90
units long and leading by 60o and phasor 2 is 50 units long and leads by
30o. (Use a horizontal line as a reference).
Scale: 10 mm = 10 units
45
12.
Given 3 phasors, find the resultant size and direction.
Phasor 1
Phasor 2
Phasor 3
=
=
=
30A
70A
50A
leading by 45o
lagging by 45o
leading by 135o
Scale: 10 mm = 7.5A
46
13.
Find the size and direction of the resultant phasor when:
Phasor 1
Phasor 2
Phasor 3
Phasor 4
=
=
=
=
6A
15A
10A
20A
lagging by 20o
leading by 40o
lagging by 100o
leading by 110o
Scale: 10mm = 2A
47
14.
Using appropriate scales, solve the following:
a.
25V
35V
b.
100V
120V
135o
c.
90V
120V
60V
80V
d.
110V
60o
e.
70V
100V
60o
8V
80o
120o
15V
10V
NOTES:
Further exercises, self-testing problems and summary on this topic can be found in
“Electrical Principles for Electrical Trades”, J. R. Jenneson, 5th Ed. on page 18.
48
RESISTANCE IN AC CIRCUITS
On completion of this topic the learner will be able to:
Work with single-source resistive ac circuits and solve problems related to
voltages, currents and power dissipated in such circuit.
Assessment Criteria
3.1
Set up and connect a single-source ac circuit and take resistance,
voltage and current measurements.
3.2
Determine the voltage, current, resistances or power dissipated from
measured or given values of any two of these quantities.
3.3
Show the relationship between voltage drops and current in a
resistance ac circuit.
Reference “Electrical Principles for Electrical Trades”,Jenneson, J.R. (5th
Edition) - chapter 8 page 171
49
When pure resistance is placed in an AC circuit, it behaves with the same
characteristics as though it is placed in a DC circuit.
R
I
V
AC
Waveform Diagram
V
V+
I
Phase angle 0
V-
Phasor Diagram
I
V
Power consumed in a circuit containing pure resistance is:
P= I xV
The power consumed is the instantaneous value of voltage and current. The
electrical energy is converted into heat at the same rate as in an equivalent DC
circuit.
50
From the above waveform:
Since
P=IxV
then all +ve values of I and all +ve values of V when multiplied = +ve values of
power (in watts).
AND -ve values of I and all -ve values of V when multiplied = +ve values of
power (in watts)
P =I R
2
NOTE:
The other two power formulae
-
and
P=
V
2
R
can be used in AC circuits but only with regard to the resistive component.
51
EXERCISES
In the following questions you are required to provide a sketch or identify the
response you consider best answers the question by placing the corresponding
letter in the brackets provided.
1.
Sketch the circuit diagram for a purely resistive AC circuit.
52
2.
Sketch the waveform diagram showing current and voltage for a resistive
AC circuit.
3.
Sketch the phasor diagram representing current and voltage for a resistive
AC circuit.
4.
Sketch on one set of axis waveform diagrams that show the relationship
between voltage, current and power for a resistive circuit.
5.
In a resistive AC circuit voltage and current are:
a.
b.
c.
d.
90o out of phase with each other
leading by 90o
in phase with each other
lagging by 90o
53
6.
In a resistive AC circuit, the value of power is always:
a.
b
c.
d.
7.
8.
positive
negative
greater than 10
positive and negative.
The formula to find power in a resistive circuit is;
a.
P =I xV
b.
P =I R
c.
P=
d.
All of the above
2
V
2
R
From the circuits below, determine the unknown quantity.
a.
R=?
5A
R
120V
AC
b.
R=?
5A
R
120V
AC
c.
R = 10R
I=12A
R
?V
AC
54
d.
P=?
I=25A
R
100
V
AC
e.
P=?
I=2A
R
100
V
AC
f.
P=?
I = 5A
R = 30R
V
AC
55
Practical Exercise
1.
Connect the following circuit to an AC supply.
A
W
V
R
1
V1
R
2
V2
VS
2.
Have your circuit checked before turning supply on.
3.
From the meter readings, complete the table below:
VS
Volts
I
Amps
W
Watts
V1
Volts
V2
Volts
4.
From the above readings, answer the following:
a.
Adding V1 + V2 =
Volts
b.
Multiplying I x VS =
Watts
c.
Using P = I2R =
Watts
d.
Using I = V /R =
Amps
From this exercise state what you have noticed about resistance in an AC circuit.
NOTES:
Further exercises, self-testing problems and summary on this topic can be found in
“Electrical Principles for Electrical Trades”, J. R. Jenneson, 5th Ed. on page 191
56
INDUCTANCE IN AC CIRCUITS
On completion of this topic the learner will be able to:
Explain how inductance behaves in an ac circuit.
Assessment Criteria:
4.1
Define inductive reactance.
4.2
Calculate the inductive reactance of a given
inductor and show the relationship between
inductive reactance and frequency.
4.3
Apply series and parallel circuit rules to
determine the equivalent inductive reactance in
an ac circuit or any part of a circuit.
4.4
Apply Ohm’s Law to determine voltage, current
or inductive reactance in a purely inductive ac
circuit given any two of these quantities.
4.5
Give examples of inductive components in
power circuits and systems and describe their
effect on the phase relationship between voltage
and current.
4.6
Compare the limiting characteristics of inductors
and resistors.
57
COMPONENTS IN AC CIRCUITS
When components such as capacitors and inductors are placed into AC circuits,
they “behave” in a different manner because of the frequency of the supply.
When inductors and capacitors are in DC circuits they also will have a different
effect.
This topic is on components such as resistors, inductors and capacitors when
placed in series and in parallel circuits and the effect that AC has on each. Both
types of circuits will be analysed to determine voltages and currents and their
relationship.
This topic is mathematically oriented so an understanding of trigonometry,
phasors and Pythagoras’ theorem is essential.
INDUCTORS IN AN AC CIRCUITS
When an AC current passes through a coil, a magnetic field is set up around that
coil. Any change in the value of current will alter the strength of the field and an
e.m.f. is induced into the coil.
If the current is increased, the field expands and the induced e.m.f. acts in
opposition to the e.m.f. applied to the coil, thus opposing the rise in current.
If the current is decreased, the field contracts and induces an e.m.f. in the same
direction as the applied e.m.f., thus opposing the fall in current.
For the purposes of analysis, the inductor is considered to be “pure” and
possessing no resistance.
Inductance is the property of a circuit to generate an e.m.f. of self-induction which
opposes change in current and is measured in Henry.
Circuit Diagram
L
V
AC
I
58
Waveform Diagram
V+
Induced EMF
Applied Voltage
I
V-
Phasor Diagram
V
90o
I
In a pure inductive circuit (no resistance) the current lags the voltage by 90o
Inductive reactance is measured in ohms. (It is opposition to current flow).
Symbol
-
XL
For pure inductors, Ohm’s Law is applicable.
IL=
That is
-
where
-
IL
VL
XL
VL
XL
=>
=>
=>
Current through the inductor
Voltage to the inductor
Inductive reactance
The value of inductive reactance in a circuit depends on the inductance of the
inductor and the rate of change of current flow which is in turn dependent on the
supply frequency.
59
That is
-
XL
is proportional to
L
XL
is proportional to
f
XL = 2πfL
where:
f
L
2π
XL
=>
=>
=>
=>
Frequency in Hertz
Inductance in Henrys
Constant for cycle
Inductive reactance in ohms
From the formula on the previous page, we can deduce:
*
If
L increases or decreases
then
XL
increases or decreases
*
If
f
then
XL
increases or decreases
increases or decreases
EXAMPLE 1
What is the inductive reactance of a coil having an inductance of 4H when
connected to a 50Hz supply?
4H
L
I
50 Hz
XL
=
=
=
=
2πfL
2π x 50 x 4
1256.6 Ω
1.256 kΩ
60
EXAMPLE 2
In the circuit below, determine
a.
Inductive reactance
b.
Current flow
0.5 H
L
200V
50 Hz
a.
I
= 2πfL
= 2π x 50 x 0.5
= 157.079 Ω
XL
b..
IL =
=
VL
XL
200
157.079
= 1.273 A
61
EXERCISES
1.
a.
b.
A coil has an inductance of 0.05 H, what would be the inductive
reactance at a frequency of:
i.
25 Hz
ii.
50 Hz
iii.
100 Hz
At what frequency would it have a reactance of 10Ω?
62
2.
A current of 2.5A flows through a coil when connected to a 240 V 50
Hz supply. Determine the inductance of the coil.
3.
A 500V, 50 Hz supply is applied to a coil of 0.15H. Determine the current
flow.
63
INDUCTORS IN SERIES
L2
L1
XL2
XL1
I
V
LTOTAL = L1 + L2 ..........
XLTOTAL = XL1 + XL2 ..........
INDUCTORS IN PARALLEL
IT
IL2
IL1
V
AC
L1
XL1
L2
XL2
EXAMPLE
Two inductors having an inductive reactance of 10Ω and the other 20Ω are in
parallel to a 240V, 50 Hz supply. What is the total inductance?
IL1 =
=
V1
XL1
240
10
= 24 A
IL2 =
=
V2
XL2
240
20
= 12 A
64
Each current lags the voltage by 90o; therefore they are in phase with each other:
IL1 = 12A
V
IL2 = 24A
ILT = 36A
ITOTAL = IL1 + IL2
= 24 + 12
= 36A
XLTOTAL =
=
V
ITOTAL
240
36
= 6.666 Ω
Another method is:
1
XLTOTAL
=
=
1
XL1
1
10
+
+
1
XL2
1
20
= 6.666R
65
PHASOR RELATIONSHIP OF INDUCTORS IN SERIES
0.4H
0.8H
L1
L2
I
6V
200Hz
When inductors are connected in series, the current is the same through both
inductors and becomes the reference vector.
XL1
=
=
=
XLTOTAL
2πfL
2π x 200 x 0.4
500 Ω
=
=
=
XL2
=
=
=
2πfL
2π x 200 x 0.8
1000 Ω
XL1 + XL2
500 + 1000
1500 Ω
ITOTAL =
=
VL
XLTOTAL
6
1500
= 4 mA
By Ohm’s Law
-
Thus:
VL1
=
=
=
IL1 x XL1
4 mA x 500
2V
VL2
=
=
=
IL2 x XL2
4 mA x 1000
4V
66
This is represented:
VT = 6V
VL2 = 4V
Scale: 10mm
= 1V
= 1mA
VL1 = 2V
IT = 4mA
67
PHASOR RELATIONSHIP OF INDUCTORS IN PARALLEL
When connected in parallel, the voltage is the same across the inductors. This
becomes the reference phasor.
IT
IL2
IL1
15V
160Hz
XL1
L1
2πfL
2π x 160 x 0.05
50 Ω
=
=
=
1
XLTOTAL
=
=
1
XL!
1
50
+
+
L1
50
mH
XL2
L2
=
=
=
L2
75
mH
2πfL
2π x 160 x 0.075
75 Ω
1
XL2
1
75
= 30 Ω
By Ohm’s Law
IL1 =
=
V1
XL1
IL2 =
15
=
30
= 300 mA
Thus -
V2
XL2
15
75
= 200 mA
ITOTAL =
=
=
IL1 + IL2
300 + 200
500 mA
68
V = 15V
IL2 = 200mA
Scale: 10mm
IT
500mA
=
3V
= 100mA
IL1 = 300mA
In both series and parallel circuits, the current lags the voltage.
69
POWER IN AN INDUCTIVE CIRCUIT
The power input to a “pure” inductor at any instant of time is equal to the produce
of the instantaneous values of V and I.
As the current rises, energy is used to produce the magnetic field.
As the current falls, the magnetic field collapses and energy is returned to the
supply.
Over a complete cycle, the positive and negative powers of the waveform cancel
each other out. Therefore the average power consumed by a “pure inductor” is
zero.
70
EXERCISES
In the following questions, you are required to provide a sketch or identify the
response you consider best answers the question by placing the corresponding
letter in the brackets provided.
1.
Sketch the circuit diagram for a purely inductive AC circuit.
2.
Sketch the waveform diagram showing current and voltage for a purely
inductive circuit.
71
3.
Sketch the phasor diagram representing current and voltage for a purely
inductive circuit.
4.
In a purely inductive circuit, current lags the voltage by:
a.
b.
c.
d.
5.
7.
(
)
The average power consumed in a purely inductive circuit is:
a.
b.
c.
d.
6.
30o
90o
45o
60o
proportional to current
proportional to the voltage
equal to the phasor sum of voltage and current
equal to zero
(
)
Inductors connected in series must be added using the formula:
a.
b.
LTOTAL =
LTOTAL =
c.
d.
LTOTAL =
LTOTAL =
½L1 + ½L2 ............... + Ln
1 + 1 + .............. . + Ln
L1
L2
L1 + L2 + .............. .+ Ln
L12 + L22 + ............ + Ln
(
)
Inductors connected in parallel must be added using the formula:
a.
b.
c.
d.
1
LTOTAL
1
LTOTAL
1
LTOTAL
1
LTOTAL
=
½L1 + ½L2 + .............. + ½Ln
=
1 + 1 + ................. + 1
L1
L2
Ln
L1 + L2 + ................ + Ln
=
=
L12 + L22 + ................... + Ln2
( )
72
In the following questions you are required to provide a calculation in which you
must show all working.
8.
What would be the inductive reactance of a coil, given that it has a supply
frequency of 50 Hz and inductance of 0.03H?
9.
What would be the supply frequency if a coil has an inductive reactance of
20Ω and its inductance is 0.06H?
10.
What would be the inductance of a coil if the inductive reactance is 25Ω
and the frequency of supply was 50Hz?
11.
A 240V, 50Hz supply is applied to a coil with an inductance of 0.02H.
Determine the current in the circuit.
73
12.
A 500V, 50Hz supply is applied to an inductive load of negligible resistance
and the current through the load is 5A. Determine the inductance of the
load.
13.
A 240V, 50Hz supply is applied to a coil of negligible resistance and the
current through the coil is 2.5 Amps. Determine the inductance of the coil.
74
14.
Calculate the total current when two inductors, one of 160 mH and the
other 240 mH when connected in parallel to a 30V AC supply of 100 Hz
15.
Calculate the total voltage of the above two inductors when connected in
series.
75
16.
An inductor (coil) draws a current of 20A on 240V dc and 10A on 240 50
Hz ac.
a.
Determine the resistance of the coil.
b.
Determine the impedance of the coil.
c.
Compare the two currents and state why it has a lower value when
the coil is connected to an ac supply.
NOTES:
Further exercises, self-testing problems and summary on this topic can be found in
“Electrical Principles for Electrical Trades”, J. R. Jenneson, 5 th Ed. on pages 190 to 192.
76
CAPACITANCE IN AC CIRCUITS
On completion of this topic the learner will be able to:

Explain how capacitance behaves in an ac circuit.
Assessment Criteria:
5.1
Define capacitive reactance.
5.2
Calculate the capacitive reactance of a given capacitor and show
the relationship between capacitive reactance and frequency.
5.3
Apply series and parallel circuit rules to determine the equivalent
capacitive reactance in an ac circuit or any part of a circuit.
5.4
Apply Ohm’s Law to determine voltage, current or capacitive
reactance in a purely capacitive ac circuit given any two of these
quantities.
5.5
Give examples of capacitive components in power circuits and
systems and describe their effect on the phase relationship between
voltage and current.
Reference: “Electrical Principles for Electrical Trades”, Jenneson, J.R. (5th
Edition) - chapter 8 page 175
77
CAPACITORS IN AN AC CIRCUIT
When a capacitor is placed into an AC circuit, it will be in a manner similar to an
inductor, that is, there will be a reactance produced that will oppose current flow.
As the applied AC voltage rises from zero, a charging current will begin to flow.
This flow of current sets up a charge in the capacitor and produces a back or
counter e.m.f. which in turn reduces the charging current.
As the applied e.m.f. rises to a maximum (b) the charge on the capacitor also
increases to a maximum and causes the charging current to fall to a minimum (b).
The capacitor is now fully charged (b) and the current is zero (b).
The next quarter of a cycle shows the applied voltage falling to zero (c) and at the
same time the capacitor discharges. At the same time current rises to a
maximum (c) the capacitor now is discharged.
The capacitor then charges and discharges in a similar manner as in the first half
cycle (c d) and (d
e)
** This process repeats itself every cycle **
Note: No current flows through the capacitor. It is charging and discharging
Because of this effect, the current leads the voltage by 90o (for a pure capacitor
only)
A capacitor connected in a circuit with an ammeter on a 50 Hz supply would not
show such fast changes in current flow, but would indicate a steady current flow
(the charging and discharging currents in the circuit).
This phenomenon is referred to as “capacitive reactance” and has the ability to
oppose current flow in a circuit with an AC supply.
Capacitive Reactance is the opposition to the flow of current in an AC
circuit due to the effect of capacitance of the capacitor, the frequency of the
circuit and is measured in ohms.
78
Circuit Diagram, Waveform Diagram and Phasor Diagram
Symbol
-
XC
The value of Capacitive Reactance depends on the rate of change (f ) and the
capacitance.
Thus
-
XC
=
1/ f
and
-
XC
=
1/ C
That is
XC =
where
-
XC
f
C
2π
=
=
=
=
1
2πfC
Capacitive reactance in ohms
Frequency in Hertz
Capacitance in Farads
Constant for a cycle
From the above formula:
*
If
f
increases/decreases
then
XC
decreases/increases
*
If
C
increases/decreases
then
XC
decreases/increases
Ohm’s Law is also applicable to capacitive circuits:
That is
IC =
where
-
IC
VC
XC
=
=
=
VC
XC
Current “through” the capacitor
Voltage applied to capacitor
Capacitive reactance
79
EXAMPLE
Find the capacitive reactance and current flow of a 40μF capacitor when
connected to 200V in a 50Hz circuit.
To find XC
XC =
=
To find IC
1
2πfC
IC =
VC
XC
200
79.577
1
2π x 50 x 40 µF
=
= 79.577Ω
= 2.513 A
80
CAPACITORS IN SERIES AND CAPACITIVE REACTANCE
Consider this circuit:
C1
C2
V AC
50Hz
XC1 =
=
1
2πfC
XC2 =
1
2π x 50 x 10 µF
=
= 318.309R
1
2πfC
1
2π x 50 x 20 µF
= 159.155R
Thus:
XC(total) = XC1 + XC2
= 318.309 + 159.155
= 477.464R
Alternatively, the total capacitance of the circuit can be found:
i.e.
1
CT
=
=
1
C1
+
1
C2
1
1
+
10 20
= 6.666μ.
Thus XC can be found:
XC =
=
1
2πfC
1
2π x 50 x 6.666µF
= 477.464R
81
CAPACITORS IN PARALLEL AND CAPACITIVE REACTANCE
Consider this circuit:
AC
50Hz
C1
C2
10uF
20uF
Thus:
1
XC(total)
=
=
1
XC1
+
1
XC2
1
1
+
318.309 159.155
(Refer previous question for XC1 & XC2)
= 106.103R
Again alternatively, the total capacitance of the circuit can be found and then
calculate for XC(total) :
CT = C1 + C2
= 10 + 20
i.e.
= 30 µF
Therefore
XC(total) =
=
1
2πfC
1
2π x 50 x 30 µF
= 106.103R
NOTE:
and
*
*
As C (increases) XC  (decreases)
As C  (decreases) XC  (increases)
(providing f remains the same)
82
EXERCISES
NOTE - a circuit is to be drawn for each exercise
1.
What is the capacitive reactance of a 10µF capacitor on a 50Hz supply and
what current will flow from an 110V supply?
2.
A capacitor of 25µF is connected to a 240V, 60Hz supply. Calculate the
current flow.
3.
What would be the reactance of a 16µF capacitor at a frequency of 40Hz
and at what frequency would its reactance be 300Ω?
83
4.
At a frequency of 40 Hz, what voltage would be required for a current of 2A
to pass through a 40μF capacitor?
5.
What would be the capacitance of a capacitor, which would have the same
reactance at 50Hz as a coil with an inductance of 1H?
6.
Calculate the reactance of a 50pF capacitor on a 100Hz supply.
84
POWER IN A CAPACITIVE CIRCUIT
The power consumed at any instant of time by a pure capacitor is equal to the
product of the instantaneous values of V and I.
Power is taken from the supply to build up the electrostatic field in the first quarter
cycle and returned in the next quarter cycle.
Thus the average power is zero.
85
EXERCISES
In the following questions you are required to provide a sketch or identify the
response you consider best answers the question by placing the corresponding
letter in the brackets provided.
1.
Sketch the circuit diagram for a capacitive circuit.
2.
Sketch the waveform diagram for a capacitive circuit.
3.
Sketch the phasor diagram representing current and voltage for a purely
capacitive circuit.
86
4.
5.
6.
7.
8.
The charge on a capacitor can be calculated using the formula:
a.
Q
b.
c.
d.
Q
Q
Q
(
)
Energy stored in a capacitor can be calculated using the formula:
a.
W
b.
c.
d.
W
W
W
=
V
XC
=
½CV2
=
VC
=
1
2πfC
(
)
(
)
Capacitive reactance can be calculated using the formula:
a.
XC
b.
c.
d.
XC
XC
XC
=
V
XC
=
½CV2
=
VC
=
1
2πfC
Which formula would be used to calculate the current drawn in a capacitive
circuit?
a.
I
b.
c.
d.
I
I
I
V
=
XC
=
½CV2
=
VC
=
1
2πfC
(
)
In a purely capacitive circuit, the average power drawn from the supply is:
a.
b.
c.
d.
9.
=
V
XC
=
½CV2
=
VC
=
1
2πfC
proportional to the current
zero
proportional to the voltage
half the maximum power
(
)
(
)
In a purely capacitive circuit:
a.
b.
c.
d.
current leads the voltage by 90o
voltage leads the current by 90o
current lags the voltage by 90o
voltage is in phase with the current
87
In the following questions you are required to provide a sketch and calculations in
which you must show all working.
10.
Find the capacitive reactance of a 40µF capacitor when connected to a
50Hz supply.
11.
An 8µF capacitor is connected to a 240V 50Hz supply. Draw a phasor
diagram to show the angle of displacement between V and I and calculate:
a.
b.
capacitor reactance
circuit current
88
12.
What current would flow when a 360µF capacitor is connected across a
single phase 415V 2 wire 50Hz supply?
13.
A capacitor takes 2.413A from a 50Hz 240V supply. Determine the
capacitance.
14.
Find the total capacitive reactance of the following circuits:
a.
C1
C2
C3
5uF
10uF
15uF
V AC
50Hz
89
b.
C1
AC
50Hz
15.
C2
20uF
40uF
Determine the values of the capacitors of the following circuits:
a.
C1
C2
XC1 47R
XC2 85R
V AC
50Hz
b.
AC
50Hz
C1
XC1
20R
C2
XC2
30R
NOTES:
Further exercises, self-testing problems and summary on this topic can be found in
“Electrical Principles for Electrical Trades”, J. R. Jenneson, 5th Ed. on pages 192.
90
AC CIRCUITS
On completion of this topic the learner will be able to:

Work with single-source ac circuits and solve problems related to voltages,
currents and impedance in such circuit.
Assessment Criteria
6.1
Define impedance.
6.2
Determine the impedance of series, parallel and series/parallel circuits and
draw diagrams showing the relationship between resistive, inductive and
capacitive components (impedance triangle).
6.3
Set up and connect a single-source ac circuit and take resistance, voltage
and current measurements.
6.4
Determine the voltage, current or impedance from measured or given
values of any two of these quantities.
6.5
Use phasor diagrams to solve problems and show the relationship between
voltages and currents in ac circuits.
91
IMPEDANCE
Reactance is the opposition to current flow due to the effect of inductance or
capacitance or both in an AC circuit.
The combination of these reactances (XL and XC) and resistance (R) is known as
the impedance of a circuit.
Impedance is the total opposition to current flow in
an AC circuit due to the presence of XL, XC and R in
either series or parallel circuits.
Symbol
-
Unit
Z
Ohm
Ohm’s Law for Z
IZ =
NOTE where
VZ
Z
this formula is used only in a series circuit
IZ
VZ
Z
=
=
=
Total current in the circuit
Voltage applied to circuit
Total impedance
92
RESISTANCE AND INDUCTANCE IN SERIES
VR
VL
R
L
I
VT
AC
The phasor diagram of the above circuit is represented:
VL
VT
OR
VZ
VR
Φo
•
•
•
•
•
•

I
The current is common to R and L
VL and VR are 90o apart
IR is in phase with VR
IR lags VL at 90o
VL is the voltage across L
VR is the voltage across R
The applied voltage (VZ) can be found by solving the phasors (or using
Pythagoras’ Theorem)
That is
-
2
2
2
V Z = VR + V L
Also, the inductor produces XL and the combination of XL and R becomes
impedance (Z).
~
~
~
VR
VL
VZ
=
=
=
IR
IL
IZ
x
x
x
R
XL
Z
93
The impedance of the circuit can be found by either Ohm’s Law or using:
2
2
2
Z = R + XL
The previous circuit is represented by the waveform:
V+
Applied Voltage
I
V-
and by phasor diagram
V
υ
I
That is, the current (IZ) lags the voltage (VZ) by θ0.
94
Example
1200R
0.32H
R
L
I =100mA
AC
450Hz
From the above circuit, determine;
a.
VR
b.
VL
c.
VZ (supply)
a.
b.
To Find VR
VR =
=
IR x R
100mA x 1200
i.
XL
=
2πfL
=
2π x 450 x
=
904.778Ω
=
=
=
IL x XL
100mA x 904.778
90.478V
0.32
=
120V
ii.
VZ
c.
VZ
2
=
VL
To Find VL
2
2
VR + V L
2
= 1202 + (90.478 )
= 150.2288V
95
Example
From the previous problem find the impedance (Z) of the circuit.
XL
904R
Z=?
R
1200R
2
Z =
2
2
R + XL
2
Z =
2
(1200 ) + (904 )
= 1502.4Ω
as a check
-
VZ
=
=
=
IZ x Z
100 mA x 15024
150V
The previous examples can be shown (and solved by phasors)
VL
90V
VZ
150V
I
VR
120V
96
EXERCISES
In the following questions you are required to identify the response you consider
best answers the question by placing the corresponding letter in the brackets
provided.
1.
In a series RL circuit:
a.
b.
c.
d.
2.
The voltage drop across the resistive component of a series RL circuit is:
a.
b.
c.
d.
3.
voltage is common to all parts
the resistance is always greater than the inductive reactance
the current is common to all parts
the current varies throughout the circuit.
( )
in phase with the current
leading the current by 90o
lagging the current by 90o
none of the above
(
)
The voltage drop across the inductive component of a series RL circuit is
(assume pure inductor):
a.
b.
c.
d.
in phase with the current
leading the current by 90o
lagging the current by 90o
none of the above
(
)
97
Provide a circuit diagram and show all calculations for the following:
4.
A winding has 10Ω resistance and 0.15H inductance. It is to be connected
to a 240V, 50Hz supply. Calculate the:
a.
b.
c.
d.
inductive reactance
impedance
circuit current
angle by which the current lags the voltage (by phasors)
98
5.
A coil has a resistance of 12Ω and an inductance of 0.1H and is connected
across a 100V, 50Hz supply. Calculate:
a.
b.
c
d.
6.
reactance of the coil
impedance of the coil
current drawn from supply
the phase angle between V and I
What value of resistance must be connected in series with an inductance of
0.2H in order that the phase angle between V and I will be 45o when a
60Hz supply is applied.
99
RESISTANCE AND CAPACITANCE IN SERIES
VR
VC
R
C
VT
AC
I
The phasor diagram of the above circuit is represented:
VR
VT (VZ)
VC
*
*
*
*
*
*
I
The current is common to R and C
VC and VR are 90o apart
IR is in phase with VR
IR leads VC by 90o
VR is the voltage across R
VC is the voltage across C
The applied voltage (VZ) can be found by solving the phasors (or using
Pythagoras’ Theorem)
That is
-
2
2
2
V Z = VR + V C
100
Also, the capacitor produces XC and the combination of XC and R becomes
impedance (Z).
~
~
~
VR
VC
VZ
=
=
=
IR
IC
IZ
x
x
x
R
XC
Z
The impedance of the circuit can be found by either Ohm’s Law or using:
2
2
2
Z = R + XC
The above circuit is represented by the waveform:
V+
VC
I
V-
and by phasor diagram
IZ
VZ
That is, the current (IZ) leads the voltage (VZ) by θo.
101
Example 1
R
C
2uF
60R
100mA
VT
AC
From the above circuit, determine -
a.
b.
c.
VR
VC
VZ (supply volts)
a.
b.
To Find VC
To Find VR
i.
XC =
V R = IR x R
1
2πfC
1
=
= 100 mA x 60
2π x 1000 x 2 x10
-6
=6V
= 79.578Ω
ii.
c.
2
VZ =
VC
=
=
=
IC x XC
100mA x 79.578
7.958V
2
2
VR + V C
2
2
V Z = 6 + 7. 958
= 9.966V
102
EXAMPLE 2
From the previous problem find the impedance (Z) of the circuit.
R = 60R
Φ
XC = 79R
2
Z =
Z =
2
R + XC
Z
2
2
2
60 + 79. 578
= 99.662Ω
as a check
-
VZ
=
=
=
IZ x Z
100 mA x 99.662
9.966V
The previous examples can be shown (and solved by phasors)
R 6V
VC
8V
IZ
VZ 10V
103
EXERCISES
In the following questions you are required to identify the response you consider
best answers the question by placing the corresponding letter in the brackets
provided.
1.
In a series RC circuit:
a.
b.
c.
d.
2.
The voltage drop across the resistive component of a series RC circuit is:
a.
b.
c.
d.
3.
the voltage is common to all parts
the resistance is always greater than the capacitive reactance
the current is common to all parts
the current varies throughout the circuit
( )
in phase with the current
leading the current by 90o
lagging the current by 90o
none of the above
(
)
The voltage drop across the capacitive component of a series RC circuit is:
a.
b.
c.
d.
in phase with the current
leading the current by 90o
lagging the current by 90o
none of the above
(
)
104
Provide a circuit diagram and show all calculations for the following:
4.
Calculate the value of resistance that must be connected in series with a
48µF capacitor to give a phase angle of 60o between the applied voltage
and the current when connected to a 50Hz supply.
105