PART 3
AC
AC CIRCUITS
I. BASIC AC THEORY
1. ALTERNATING CURRENT
A current that is constantly changing in amplitude and direction
Either as a voltage switching polarity or as a current switching direction back
and forth
Advantages of AC:
Magnitude can easily be changed (stepped-up or stepped-down) with the use
of a transformer
Can be produced either single phase for light loads, two phase for control
motors, three phase for power distribution and large motor loads or sixphase for large scale AC to DC conversion
2. AC WAVEFORMS
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AC CIRCUIT
Parameters of Alternating Signal
Period (T) the time of one complete cycle in seconds.
Frequency (f) the number of cycles per second (Hertz)
a. 1 cycle/second (cps) = 1 Hertz (Hz)
b. Proper operation of electrical equipment requires specific frequency.
c. Frequencies lower than 60 Hz would cause flicker when used in lighting.
Wavelength () the length of one complete cycle
Propagation Velocity (v) the speed of the signal
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AC CIRCUIT
Phase() - an angular measurement that specifies the position of a sine
wave relative to a reference
f=
1
T
v
f
The Sinusoidal Wave
is the most common AC waveform that is practically generated by
generators used in households and industries
General equation for sine wave:
a(t) =Amsin(t + )
where: a(t) instantaneous amplitude of voltage or current at a given time(t)
A m maximum voltage or current amplitude of the signal
angular velocity in rad/sec
- 2f
t time (sec)
phase shift ( + or in degrees)
Measurements of AC Magnitude
Amplitude - height as depicted on a graph over time. An amplitude
measurement can take the form of peak, peak-to-peak, average, or RMS
quantity.
Peak amplitude - the height of an AC waveform as measured from the zero
mark to the highest positive or lowest negative point on a graph. Also known
as the crest amplitude of a wave.
Peak-to-peak amplitude - the total height of an AC waveform as measured
from maximum positive to maximum negative peaks on a graph. Often
abbreviated as "P-P".
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AC CIRCUIT
Average amplitude - the mathematical "mean" of all a waveform's points over
the period of one cycle. Technically, the average amplitude of any waveform
with equal-area portions above and below the "zero" line on a graph is zero.
However, as a practical measure of amplitude, a waveform's average value is
often calculated as the mathematical mean of all the points' absolute values
(taking all the negative values and considering them as positive).
Avehalf wave = 0.636 Am
Avesine wave = 0
"RMS" stands for Root Mean Square (effective value) - a way of expressing
an AC quantity of voltage or current in terms functionally equivalent to DC.
RMSsine wave = 0.707 Am
Crest factor of an AC waveform - the ratio of its peak (crest) to its RMS value.
Form factor of an AC waveform - the ratio of its peak (crest) value to its
average value.
3. AC QUANTITIES
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AC CIRCUIT
a. Resistance(R)
Opposes the AC current similar to DC circuits
Opposition offered by resistors
b. Reactance(X)
Depends on the AC frequency of the AC source which is the opposition to
current due to inductance and capacitance
Inductive Reactance (XL)
the property of the inductor to oppose alternating current.
XL = 2fL
Inductive Susceptance (BL)
reciprocal of inductive reactance
BL=
1
XL
BL =
1
2fL
Capacitive Reactance (Xc)
the property of a capacitor to oppose alternating current
Xc =
1
2fC
Capacitive Susceptance (BC)
reciprocal of capacitive reactance
BC =
1
XC
BC = 2 fC
c. Impedance (Z)
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AC CIRCUIT
Total opposition to the flow of Alternating current
Combination of the resistance in a circuit and the reactances involved
Z = R + jXeq
Where: IZI = R 2 X eq 2
Z = IZI
X eq
= Arctan
Phasor Diagram of Impedance
If I = Im is the resulting current drawn by a passive, linear RLC circuit from a
source voltage V = Vm , then
Z=
Vm
V
= I
= Z
I
m
Zcos + jZsin = R + jX =
Where: Z=
Vm
=
Im
1
R 2 X 2 tan
X
R
R 2 X 2 = magnitude of the impedance
= tan-1
X
= phase angle of the impedance
R
R=Zcos =active or real component of the impedance
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X = Zsin = reactive or quadrature component of impedance
d. Admittance (Y)
The reciprocal of impedance
Expressed in siemens or mho (S)
Y=
Im
= Y y = Ycos y + jYsin y = G + jB
Vm
Y=
Where: Y=
Im
=
Vm
G2 B2
G2 B2
tan -1
B
G
1
=magnitude of the admittance
Z
B
=tan G = phase angle of the admittance
-1
G=Ycos y =conductive/conductance component
B = Ysin y = susceptive/susceptance component
II. AC CIRCUITS
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AC CIRCUIT
1. AC Resistor Circuit
Impedance(Z) = R
With an AC circuit like this which is purely resistive, the relationship of the voltage
and current is as shown:
Voltage (e) is in phase with the current(i)
Power is never a negative value. When the current is positive (above the
line), the voltage is also positive, resulting in a power (p=ie) of a positive
value
This consistent "polarity" of power tell us that the resistor is always
dissipating power, taking it from the source and releasing it in the form of
heat energy. Whether the current is positive or negative, a resistor still
dissipates energy.
2. AC Inductor Circuit
Impedance(Z) = jXL
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The most distinguishing electrical characteristics of an L circuit is that current
lags the voltage by 90 electrical degrees
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AC CIRCUIT
Because the current and voltage waves are 90 o out of phase, there are times
when one is positive while the other is negative, resulting in equally frequent
occurrences of negative instantaneous power.
Negative power means that the inductor is releasing power back to the
circuit, while a positive power means that it is absorbing power from the
circuit
The inductor releases just as much power back to the circuit as it absorbs
over the span of a complete cycle.
Review Question: Determine the opposition (Z) of the inductor in the circuit below
with respect to the phase angles of voltage and current:
Solution:
XL of 10mH at 60 Hz: XL=2 fL= 3.7699
I=
E
10
=
= 2.6526
XL
3.7699
Opposition =
Voltage
=
Current
10 90 .
2.6526 0 .
= 3.7699 90 .
REVIEW:
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AC CIRCUIT
Inductive reactance is the opposition that an inductor offers to alternating
current due to its phase-shifted storage and release of energy in its magnetic
field. Reactance is symbolized by the capital letter "X" and is measured in
ohms just like resistance (R).
Inductive reactance can be calculated using this formula: XL = 2fL
The angular velocity of an AC circuit is another way of expressing its
frequency, in units of electrical radians per second instead of cycles per
second. It is symbolized by the lower-case Greek letter "omega," or .
Inductive reactance increases with increasing frequency. In other words, the
higher the frequency, the more it opposes the AC flow of electrons.
3. AC Capacitor Circuit
Impedance(Z) = -jXC
The most distinguishing electrical characteristics of an L circuit is that current
leads the voltage by 90 electrical degrees
the current through a capacitor is a reaction against the change in voltage
across it
A capacitor's opposition to change in voltage translates to an opposition to
alternating voltage in general, which is by definition always changing in
instantaneous magnitude and direction. For any given magnitude of AC
voltage at a given frequency, a capacitor of given size will "conduct" a certain
magnitude of AC current
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AC CIRCUIT
capacitors "conduct" current in proportion to the rate of voltage change, they
will pass more current for faster-changing voltages (as they charge and
discharge to the same voltage peaks in less time), and less current for
slower-changing voltages.
the phase angle of a capacitor's opposition to current is -90 o, meaning that a
capacitor's opposition to current is a negative imaginary quantity
REVIEW:
Capacitive reactance is the opposition that a capacitor offers to alternating
current due to its phase-shifted storage and release of energy in its electric
field. Reactance is symbolized by the capital letter "X" and is measured in
ohms just like resistance (R).
Capacitive reactance can be calculated using this formula: X C = 1/(2fC)
Capacitive reactance decreases with increasing frequency. In other words,
the higher the frequency, the less it opposes (the more it "conducts") the AC
flow of electrons.
4. SERIES Resistor-Inductor Circuit
Impedance(Z) = R+jXL
Admittance(Y) =
R jX L
1
=
R jX L
R 2 XL 2
For a series resistor-inductor circuit, the voltage and current relation is
determine in its the phase shift. Thus , current lags the voltage by a phase
shift ( )
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AC CIRCUIT
Phase shift( ) = Arctan (
XL
e
) I Z I = R 2 XL 2 =
i
R
REVIEW:
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When resistors and inductors are mixed together in circuits, the total
impedance will have a phase angle somewhere between 0 o and +90o. The
circuit current will have a phase angle somewhere between 0 o and -90o.
Series AC circuits exhibit the same fundamental properties as series DC
circuits: current is uniform throughout
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AC CIRCUIT
5. SERIES Resistor- Capacitor Circuit
Impedance (Z) = R jXC
Admittance (Y) =
1
R jXc
= 2
R jX C
R Xc 2
For a series resistor capacitor circuit, the voltage and current relation is
determined by the phase shift. Thus the current leads the voltage by an
angle less than 90 degrees but greater than 0 degrees
Phase shift ( ) = Arctan (
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Xc
)
R
IZI=
e
R 2 Xc2 =
i
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AC CIRCUIT
6. PARALLEL Resistor Inductor
Z parallel =
1
Admit tan ce(Y )
Y = G - j L where: G conductance = 1/R L = susceptance = 1/XL
Z=
E
, by Ohms Law
I
The basic approach with regards to parallel circuits is using admittance
because it is additive
REVIEW
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When resistors and inductors are mixed together in parallel circuits (just as in
series circuits), the total impedance will have a phase angle somewhere
between 0o and +90o. The circuit current will have a phase angle somewhere
between 0o and -90o.
Parallel AC circuits exhibit the same fundamental properties as parallel DC
circuits: voltage is uniform throughout the circuit, branch currents add to form
the total current, and impedances diminish (through the reciprocal formula) to
form the total impedance
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AC CIRCUIT
7. PARALLEL Resistor Capacitor
Y = G + j c where: G conductance = 1/R
c = susceptance = 1/XC
REVIEW
When resistors and capacitors are mixed together in circuits, the total
impedance will have a phase angle somewhere between 0 o and -90o.
Review Question
1. Compute for the total impedance of the following SERIES RLC circuit?
Solution:
Convert individual components to their equivalent impedances:
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AC CIRCUIT
Since this is a series RLC, the total impedance (Ztotal) is the sum of individual
impedances
Z total= ZR + ZL + ZC
Z total = ( 250+ j0 ) + ( 0+j245.04 ) + (0 j1.7684K)
Z total = 250-j1.5233k = 1.5437k -80.680 . ohms
Analysis:
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Although impedances add in series, the total impedance for a circuit
containing both inductance and capacitance may be less than one or more of
the individual impedances, because series inductive and capacitive
impedances tend to cancel each other.
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AC CIRCUIT
III. POWER in AC Circuits
1. APPARENT POWER(S)
Represents the rate at which the total energy is supplied to the system
Measured in volt-amperes (VA)
S = VrmsIrms = Irms2 I Z I
It has two components, the Real Power and the Capacitive or Inductive
Reactive Power
Power Triangle
Complex Power
S = P jQ
2. REAL POWER (P)
The power consumed by the resistive component
Also called True Power, Useful Power and Productive Power
Measured in Watts (W)
It is equal to the product of the apparent power and the power factor
P = Scos
Power Factor
Cosine of the power factor angle( )
Measure of the power that is dissipated by the circuit in relation to the
apparent power and is usually given as a decimal or percentage
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Pf = cos
Ratio of the Real Power to the Apparent Power(
When:
Pf =1.0
Pf=lagging
Pf = leading
Pf=0.0 lag
Pf =0.0 lead
P
)
S
I is in phase with V; resistive system
I lags V by ; inductive system
I leads V by ; capacitive system
I lags V by 90; purely inductive
I leads V by 90; purely capacitive
Power factor Angle ( )
Let
The angle between the apparent power and the real power in the power
triangle
v(t) = Vmcos(t + v ) volts V = Vrms v
i(t) = Imcos (t + i ) A
I = Irms i
Instantaneous Power (watts)
P(t) = v(t) i(t)
P(t) =
1
1
VmImcos( v - i ) +
VmImcos(2t + v + i )
2
2
Average Power (watts)
Pave =
1
VmImcos( v - i ) = VrmsIrmscos
2
where: = phase shift between v(t) and i(t) or the phase angle of the
equivalent impedance
3. REACTIVE POWER (QL or QC)
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AC CIRCUIT
Represents the rate at which energy is stored or released in any of the
energy storing elements(the inductor or the capacitor)
Also called the imaginary power, non-productive or wattless power
Measured in volt-ampere reactive (Var)
When the capacitor and inductor are both present , the reactive power
associated with them take opposite signs since they do not store or release
energy at the same time
It is positive for inductive power(Q L) and negative for capacitive power(Q C)
Q = VrmsIrmssin
Reactive factor
Ratio of the reactive Power to the Apparent Power
Sine of the power factor angle( )
Rf = sin
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AC CIRCUIT
IV. BALANCED THREE PHASE SYSTEMS
Comprises of three identical single-phase systems operating at a 120 phase
displacement from one another. This means that a balance three-phase
system provides three voltages(and currents) that are equal in magnitude
and separated by 120 from each other
1. CLASSIFICATION:
Three-Phase, 3-wire systems
Provide only one type of voltage(line to line to both single phase and three
phase loads
Three-Phase, 4-wire systems
Provide two types of voltages (line to line and line to neutral) to both single
phase and three phase loads.
2. BALANCED Y-system
VLL = 3 VLN and IL = IP
VLL and VLN are out of phase by 30
3. BALANCED system
IL = 3 IP and VLL = VLN
IL and IP are out of phase by 30
Where: VLL or VL line to line or line voltage
VLN or VP line to neutral or phase voltage
IL - line current
IP phase current
Note: for balanced 3-phase systems:
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AC CIRCUIT
IA + IB + IC =0
VAN + VBN + VCN = 0
VAB +VBC + VCA = 0
4. Three - Phase Power
P = 3VPIPcos =
VLIL cos watts
Q = 3VPIPsin =
VLIL sin
S = 3VPIP =
VLIL
vars
va
TEST YOURSELF 3
Review Questions
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AC CIRCUIT
1.The description of two sine waves that are in step with each other going through their maximum
and minimum points at the same time and in the same direction (November, 1999)
a. Sine waves in phase
b. Stepped sine waves
c. Phased sine waves
d. Sine waves in coordination
Answer a. Sine waves in phase
2. Term used for the out of phase, non-productive power associated with inductors and capacitors
(November, 1996)
a. Effective power
b. True power
c. Reactive power
d. Peak envelope power
Answer c. Reactive power
3. Refers to reactive power. (November, 2001)
a. Wattless ,non productive power
b. Power consumed in circuit Q
c. Power loss because of capacitor leakage
d. Power consumed in wire resistance in an inductor
Answer a. Wattless, non productive power
4. Term used for an out-of-phase, non-productive power associated with inductors and capacitors.
a. effective power
b. reactive power
c. peak envelope power
d. true power
Answer b. reactive power
5. The product of current and voltage in an AC circuit refers to the
a. Real power
b. Useful power
c. Apparent power
d. Dc power
Answer c. Apparent power
6. The distance covered or traveled by a waveform during the time interval of one complete cycle
(April, 2005)
a. Frequency
b. Wavelength
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AC CIRCUIT
c.
d.
Time slot
Wave time
Answer b. Wavelength
7. The power dissipated across the resistance in an AC circuit (November, 1999)
a. real power
b. reactive power
c. apparent power
d. true power
Answer a. real power
8. It is the number of complete cycles of alternating voltage or current completed each second
(November, 2003)
a. Period
b. Frequency
c. Amplitude
d. Phase
Answer b. Frequency
9. How many degrees are there in one complete cycle? (November, 2000)
a. 720
b. 360
c. 180
d. 90
Answer b. 360
10. The impedance in the study of electronics is represented by resistance and ________ (April,
1998)
a. Reactance
b. Inductance and capacitance
c. Inductance
d. Capacitance
Answer a. Reactance
11. It is the current that is eliminated by a synchro capacitor? (April, 2004)
a. Magnetizing stator
b. Loss
c. Stator
d. Rotor
Answer a. Magnetizing stator
12. It is a rotating sector that represents either current or voltage in an AC circuit (November, 2003)
a. Resistance
b. Phasor
c. Solar diagram
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AC CIRCUIT
d.
Velocity
Answer b. Phasor
13. The relation of the voltage across an inductor to its current is described as
a. leading the current by 90 degrees
b. lagging the current by 90 degrees
c. leading the current by 180 degrees
d. in phase with the current
Answer a. leading the current by 90 degrees
14. Find the phase angle between the voltage across through the circuit when X C is 25 ohms, R is
100 ohms and XL is 50 ohms. (April, 2005)
a. 76 degrees with voltage leading the current
b. 14 degrees with the voltage lagging the current
c. 14 degrees with the voltage leading the current
d. 76 degrees with the voltage lagging the current
Answer c. 14 degrees with the voltage leading the current
Solution
= arctan (
50 25
) 14.04 deg rees , since the circuit is inductive voltage leads
100
current
15. Calculate the period of an alternating current having an equation of
a. 4.167 ms
b. 8.33 ms
c. 16.67 ms
d. 33.33 ms
I =20sin120t
Answer c. 16.67 ms
Solution
Period =
1
1
16.67 x10 3 s
f
60
16. What do you mean by root-mean-square (rms) value? (November, 1996)
a. It is the average value
b. It is the effective value
c. It is the value that causes the same heating effect as the dc voltage
d. b or c
Answer d. b or c
17. The maximum instantaneous value of a varying current, voltage or power equal to 1.414 times
the effective value of a sine wave. (November, 2000)
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AC CIRCUIT
a.
b.
c.
d.
RMS value
Peak value
EFFECTIVE VALUE
Peak to Peak value
Answer b. Peak value
18. If an AC signal has a peak voltage of 55 V, what is the average value? (April, 2000)
a. 34.98 V
b. 61.05 V
c. 86.34 V
d. 38.89 V
Answer a. 34.98 V
Solution
Ave =
2 Vpk
2(55 )
35.01
19. If an AC signal has an average voltage of 18 V, what is the rms voltage? (April, 2000)
a. 12.726 V
b. 19.980 V
c. 25.380 V
d. 16.213 V
Answer b. 19.980 V
Solution
RMS = 1.11Vave(AC signal) = 1.11(18) = 19.98 V
20. A 220-volt, 60Hz is driving a series RL circuit. Determine the current if R = 100 ohms and 20
mH inductance
a. 2.2A(lagging)
b. 2.0 A(lagging)
c. 2.2 A(leading)
d. 2.0 A(leading)
Answer a. 2.2 A(lagging)
Solution
I=
220
220
220
100 j2( 60 )(20m) 100 j7.54
100.284.311
=2.19 A (magnitude only), lagging because the circuit is inductive
21. Ignoring any inductive effects, what is the impedance of RC series capacitor made up of a 56
Kilo ohm resistor and a 0.33 F capacitor at a signal frequency of 4650 Hz (November, 1999)
a. 66730 ohms
b. 57019 ohms
c. 45270 ohms
d. 10730 ohms
Answer b. 57019 ohms
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AC CIRCUIT
Solution
Z = 56K - j
1
56K j103 .71 56000 .09 0.11 ohms
2( 4650 )(0.33 )
22. What is the time constant of a 500mH coil and a 3300 ohm resistor in series? (April, 2000)
a. 0.00015 sec
b. 6.6 sec
c. 0.0015 sec
d. 0.000015 sec
Answer a. 0.00015 sec
Solution
=
L
500 x10 3
0.1515 x10 3
R
3300
23. What is the relationship between frequency and the value of Xc? (November, 2001)
a. Frequency has no effect
b. XC varies inversely with frequency
c. XC varies indirectly with frequency
d. XC varies directly with frequency
Answer b. XC varies inversely with frequency
24. The reactance of a 25 mH coil at 5000 Hz is which of the following?
a. 785 ohms
b. 785000 ohms
c. 13 ohms
d. 0.0013 ohm
Answer a. 785 ohms
Solution
XL = 2(5000)(25x10-3) = 785.398 ohms
25. There are no transients in pure resistive circuits because they
a. Offer high resistance
b. Obey Ohms Law
c. Are linear circuits
d. Have no stored energy
Answer d. Have no stored energy
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