UNIT 21 : ALTERNATING CURRENT
(5 Hours)
21.1 Alternating current
21.2 Root mean square (rms)
21.3 Resistance, reactance and
impedance
21.4 Power and power factor
1
SUBTOPIC :
21.1 Alternating Current (1 hour)
LEARNING OUTCOMES :
At the end of this lesson, students should
be able to :
a) Define alternating current (AC).
b) Sketch and interpret sinusoidal AC waveform.
c) Write and use sinusoidal voltage and current equations.
2
21.1 Alternating current
• An alternating current (ac) is the electrical
current which varies periodically with time
in direction and magnitude.
• An ac circuit and ac generator, provide an
alternating current.
• The usual circuit-diagram symbol for an ac
source is
.
3
• The output of an ac generator is sinusoidal
and varies with time.
• Current
where:
I
I : instantaneous
current @ current at
time t (in Ampere)
I0
0
 I0
1
T
2
T
3
T
2
2T
t
I  I o sin t
I o : peak current
T : period
 : angular frequency
4
• voltage
V
where:
V : instantaneous
voltage @ voltage at
time t (in Volt)
V0
0
 V0
1
T
2
T
3
T
2
2T
V  Vo sin t
t
Vo : peak volta
ge
T : period
 : angular frequency
5
Vo
Io
T/2
T
• The output of an ac generator is sinusoidal
and varies with time.
Equation for the current ( I ) :
Equation for the voltage ( V ) :
I  I o sin t
V  Vo sin t
6
Terminology in a.c.
• Frequency ( f )
– Definition: Number of complete cycle in one second.
– Unit: Hertz (Hz) or s-1
• Period ( T )
– Definition: Time taken for one complete cycle.
– Unit: second (s)
– Equation : T  1
f
• Peak (maximum) current ( Io )
– Definition: Magnitude of the maximum current.
• Peak (maximum) voltage ( Vo )
– Definition: Magnitude of the maximum voltage.
• Angular frequency (  )
– Equation:
  2f
– Unit: radian per second (rads-1)
7
SUBTOPIC :
21.2 Root Mean Square (rms) (1 hour)
LEARNING OUTCOMES :
At the end of this lesson, students should
be able to :
a) Define root mean square (rms), current and
voltage for AC source.
b) Use I rms 
Io
2
, Vrms 
Vo
2
8
21.2 Root mean square (rms)
Root mean square current (Irms) is defined as the
effective value of a.c. which produces the same
power (mean/average power) as the steady d.c.
when the current passes through the same resistor.
the average or
mean value of
current in a halfpow erdc  average pow erac
cycle flows of
2
2
I R  I ave R
current in a certain
I  I 2 ave
direction
 square root of the average
2Io Io
value of the current
I av 
 π
π
2
 I rms

I  I 2 ave  I rms
• The r.m.s (root mean square) current means the
square root of the average value of the current.
Root mean square voltage/p.d (Vrms ) is defined as
the value of the steady direct voltage which when
applied across a resistor, produces the same power
as the mean (average) power produced by the
alternating voltage across the same resistor.
V
P  Pave
V 2 V 2 ave

R
R
V=Vo sin ωt V  V 2 ave  V
rms
10
I rms
I0

2
Vrms
V0

2
only for a sinusoidal
alternating current and voltage
• The average power, P  I V
ave
rms rms
2
V
2
 I rms
R  rms
R
I o Vo 1
Po
Pave 
 I oVo 
2
2 2 2
• The peak power, Po  I oVo
• Most household electricity is 240 V AC which
means that Vrms is 240 V.
11
Example 21.2.1
A sinusoidal, 60.0 Hz, ac voltage is read to be 120 V
by an ordinary voltmeter.
a) What is the maximum value the voltage takes on
during a cycle?
b) What is the equation for the voltage ?
a)
Vrms
Vo

2
Vo  2 (Vrms )  2 (120)  170 V
b) V  Vo sin t
V  170 sin 120t
12
Example 21.2.2
A voltage V= 60 sin 120πt is applied across a 20 Ω
resistor.
a) What will an ac ammeter in series with the
resistor read ?
b) Calculate the peak current and mean power.
a) V  Vo  60  42.4V I  Vrms  42.4  2.12 A
rms
rms
2
b)
I rms
I0

2
20
R
2
Pav  I rms  R
2
 (2.12) (20)  90 W
I o  2 ( I rms )  2 (2.12)  3.0 A
2
13
Example 21.2.3
V (Volt)
200
0
 200
0.02
0.04
0.06 0.08
t (second)
and
The alternating potential difference shown above
is connected across a resistor of 10 k. Calculate
a. the r.m.s. current,
b. the frequency,
c. the mean power dissipated in the resistor.
14
Solution 21.2.3
R =10 x 10-3 , V0 = 200 V and T = 0.04 s
a) I rms
I rms
b)
Vrms

R
and Vrms
V0

R 2
1
f 
T
V0

2
I rms  0.014 A
f  25.0 Hz
c) Pav  I rms 2 R
Pav  1.96 W
15
Exercise 21.2
An ac current is given as I = 5 sin (200t) where the
clockwise direction of the current is positive. Find
a)The peak current
b)The current when t = 1/100 s
c) The frequency and period of the oscillation.
5 A , 4.55 A, 31.88 Hz, 0.0314 s
16
SUBTOPIC :
21.3 Resistance, reactance and
impedance (2 hours)
LEARNING OUTCOMES :
At the end of this lesson, students should
be able to :
a) Sketch and use phasor diagram and sinusoidal
waveform to show the phase relationship
between current and voltage for a single
component circuit consisting of
i) Pure resistor
ii) Pure capacitor
iii) Pure inductor
17
b) Define and use:
i) capacitive reactance, X c 
ii)
1
2fC
inductive reactance, X L  2fL
iii) impedance, Z  R2  ( X L  X C )2 , and
( X L  XC )
phase angle,   tan
R
1
c) Use phasor diagram to analyse voltage, current, and
impedance of series circuit of:
i) RC
ii) RL
iii) RLC
18
21.3 Resistance, reactance and impedance
Phasor diagram
• Phasor is defined as a vector that rotate
anticlockwise about its axis with constant angular
velocity.
• A diagram containing phasor is called phasor
diagram.
• It is used to represent a sinusoidal alternating
quantity such as current and voltage.
• It also being used to determine the phase
difference between current and voltage in ac
circuit.
19
Phasor diagram
y
y
Ao
N
O
P
A  Ao sin t
ω
0
1
T
2
T
3
T
2
2T
t
• The projection of OP on the vertical axis (Oy) is ON,
represents the instantaneous value.
• Ao is the peak value of the quantity.
20
Resistance, reactance and impedance
Key Term/Ω
Meaning
Resistance,R Opposition to current flow in purely
resistive circuit.
Reactance,X Opposition to current flow resulting from
inductance or capacitance in ac
circuit.
Capacitive
Opposition of a capacitor to ac.
reactance,Xc
Inductive
Opposition of an inductor to ac.
reactance,XL
Impedance, Z Total opposition to ac.
(Resistance and reactance combine
21
to form impedance)
i) Pure Resistor in the AC Circuit
I
VR
ω
I
V
Phasor diagram
22
i) Pure Resistor in the AC Circuit
• The current flows in the resistor is
I  I 0 sin ωt
• The voltage across the resistor VR at any instant is
VR  IR
I 0 R  V0
VR  I 0 R sin ωt
V : Supply voltage
VR  V0 sin ωt  V
• The phase difference between V and I is
Δ  ωt  ωt
  0
• In pure resistor, the voltage V is in phase with the
current I and constant with time.(the current and the
23
voltage reach their maximum values at the same time).
i) Pure Resistor in the AC Circuit
• The resistance in a pure resistor is R  Vrms  Vo
I rms I o
• The instantaneous power,
• The average power,
2
V
P  IV  I 2 R 
R
P  I o sin t Vo sin t 
2
Pave  I rms
R
1 2
 Io R
2
1
 Vo I o
2
1
 Po
2
P  I oVo sin t
2
Power(P)
P0
0
1
T
2
T
3
T
2
2T
t
24
A resistor in ac circuit dissipates energy in the form of heat
ii) Pure Capacitor in the AC Circuit
• Pure capacitor means that no resistance and
self-inductance effect in the a.c. circuit.
I
 

2
VR
I
rad
ω
I
V
Phasor diagram
25
ii) Pure Capacitor in the AC Circuit
• When an alternating voltage is applied across a
capacitor, the voltage reaches its maximum value
one quarter of a cycle after the current reaches its
maximum value,(     rad )
2
• The voltage across the capacitor VC at any instant
is equal to the supply voltage V and is given by

V  V0 sin( ωt  )  VC
2
• The charge accumulates on the plates of the
capacitor is
Q  CV

Q  CV0 sin( ωt  )
2
C
• The current flows in the ac circuit is
dQ
I
dt
26
ii) Pure Capacitor in the AC Circuit
d
 

I   CV0 sin  ωt   
dt 
2 

d 
 
I  CV0  sin  ωt   
dt  
2 

I  CV0 cos(ωt  ) and
2
or
CV0  I 0
I  I 0 sin ωt
• The phase difference between V and I is


Δ   ωt    ωt 

Δ  
2

2
27
ii) Pure Capacitor in the AC Circuit
• In pure capacitor,
the voltage V lags behind the current I by /2 radians
or the current I leads the voltage V by /2 radians.
• The capacitive reactance in a pure capacitor is
Vrms Vo
Vo
XC 


I rms I o CVo
1
1
XC 

C 2fC
• The capacitive reactance is defined as
1
1
XC 

C 2fC
28
ii) Pure Capacitor in the AC Circuit
• The instantaneous power,
• The average power,
2
V
P  IV  I 2 R 
Power(P)
R
P0
P  I o sin t  Vo cost  2
1

P   I oVo  sin 2t 
0
1
2

T
2
P0
1


P   Po  sin 2t 
2
2

Pave  0
T
3
2T
T
2
• For the first half of the cycle where the power is
negative, the power is returned to the circuit. For
the second half cycle where the power is positive,
the capacitor is saving the power.
29
t
Example 21.3.1
ii) Pure Capacitor in the AC Circuit
An 8.00 μF capacitor is connected to the terminals of
an AC generator with an rms voltage of 150 V and a
frequency of 60.0 Hz. Find the capacitive reactance
rms current and the peak current in the circuit.
Capacitive reactance,
1
1
Xc 

 332 Ω
C 2fC
Rms current,
Peak current ?
I rms
Vrms

 0.452 A
XC
30
iii) Pure Inductor in the AC Circuit
• Pure inductor means
that no resistance and
capacitance effect in the
a.c. circuit.
I
VL
I
ω
V
I
Phasor diagram
 

2
rad
31
iii) Pure Inductor in the AC Circuit
• When a sinusoidal voltage is applied across a
inductor, the voltage reaches its maximum value
one quarter of a cycle before the current reaches its
maximum value,(     rad )
2
• The current flows in the ac circuit is I  I 0 sin ωt
• When the current flows in the inductor, the back emf
caused by the self induction is produced and given by
dI
εB   L
dt
d
 B   L I 0 sin ωt 
dt
 B   LI 0 ω cos ωt
32
iii) Pure Inductor in the AC Circuit
• At each instant the supply voltage V must be equal
to the back e.m.f B (voltage across the inductor)
but the back e.m.f always oppose the supply voltage V.
• Hence, the magnitude of V and B ,
V   B  LI0 ω cos ωt
or


V  LI 0 ω sin  ωt  
2

0
V   B  IR
V  B


V  Vo sin  ωt  
2

where Vo  LI o
33
iii) Pure Inductor in the AC Circuit
• The phase difference between V and I is


Δ   ωt    ωt
2

Δ 

2
• In pure inductor,
the voltage V leads the current I by /2 radians or the
current I lags behind the voltage V by /2 radians.
• The inductive reactance in a pure inductor is
Vrms Vo LI o
XL 
I rms

Io

X L   L  2 fL
Io
34
iii) Pure Inductor in the AC Circuit
• The inductive reactance is defined as X L   L  2 fL
• The instantaneous power,
2
V
P  IV  I 2 R 
R
P  I o sin t Vo cost 
1

P  I oVo  sin 2t 
2

1

P  Po  sin 2t 
2

• The average power,
Pave  0
Power(P)
P0
2
0
P0

2
1
T
2
T
3
T
2
2T
• For the first half of the cycle where the power is
positive, the inductor is saving the power. For the
second half cycle where the power is negative, the
35
power is returned to the circuit.
t
Example 21.3.2
iii) Pure Inductor in the AC Circuit
A coil having an inductance of 0.5 H is connected to
a 120 V, 60 Hz power source. If the resistance of the
coil is neglected, what is the effective current
through the coil.
I rms
Vrms Vrms


 0.64 A
X L 2fL
Example 21.3.3
A 240 V supply with a frequency of 50 Hz causes a
current of 3.0 A to flow through an pure inductor.
Calculate the inductance of the inductor.
V
X L   80 
I
X L  2fL  L  0.26 H
36
i) RC in series circuit
C
R
VC
VR
I
V
VR
I

VC
 : phase angle
V
ω
V  supply voltage
Phasor diagram
In the circuit diagram :
• VR and VC represent the instantaneous voltage
across the resistor and the capacitor.
In the phasor diagram :
• VR and VC represent the peak voltage across the
resistor and the capacitor.
37
i) RC in series circuit
VR
Note
Vo  VRo  VCo
2
2
Vo  I o R   I o X L 
2
Vo 
2
2
ω
Phasor diagram
2
R    X C 
2
2
...divide both side by 2
R    X C 
Vrms  I rms
V I
V
2
I o R   I o X L 
Vo  I o

VC
2
I
2
2
R    X C 
2
2
 1 
V  I R    2 2 
 C 
2
2
38
i) RC in series circuit
VR
C
R

VC
VC
VR
I
 : phase angle
V
V  supply voltage
I
ω
V
Phasor diagram
• The total p.d (supply voltage), V across R and C is
equal to the vector sum of VR and VC as shown in
the phasor diagram.
VR  IR
VC  IX C
V 2  VR2  VC2
V  IR   IX C 
2
2

V 2  I 2 R 2  X C2
and
1
V I R  2 2
ωC
2
2

1
XC 
ωC
39
i) RC in series circuit
VR
I

VC
R
XC

V
ω
Phasor diagram
• The impedance in RC
circuit,
1
I R  2 2
C

I
Z
ω
Impedance diagram
• From the phasor diagrams,
I leads V by Φ
2
V
Z  rms
I rms
1
Z R  2 2
C
VC
tan  
VR
XC
tan  
R
or
2
IX C
tan  
IR
1
tan  
ωCR40
i) RC in series circuit
Z
1
XC 
2fC
R
f
0
Graph of Z against f
41
i) RC in series circuit
Example 21.3.4
An alternating current of angular frequency of
1.0 x 104 rad s-1 flows through a 10 k resistor
and a 0.10 F capacitor which are connected in
series. Calculate the rms voltage across the
capacitor if the rms voltage across the resistor is
20 V.
From
XC
tan  
R
and
XC
1
tan  

 0.1
R CR
VC
tan  
VR
VC
0.1 
VR
VC  200.1  2.0 V
42
ii) RL in series circuit
R
L
VR
VL
I
V
ω
VL
 : phase angle
V

VR
V  supply voltage
I
Phasor diagram
• The voltage across the resistor VR and the capacitor
VL are
VR  IR
VL  IX L
43
ω
ii) RL in series circuit
L
R
VR
VL
VL
I
V
V
 : phase angle

VR
V  supply voltage
I
Phasor diagram
• The total p.d (supply voltage), V across R and L is
equal to the vector sum of VR and VL as shown in
the phasor diagram.
V 2  VR2  VL2
V 2  IR   IX L 
2

V 2  I 2 R 2  X L2
and
X L  ωL
2

V  I R 2  ω2 L2
44
ii) RL in series circuit
VL
ω
XL
V

VR
I
Z

R
Impedance diagram
Phasor diagram
• The impedance in RC
circuit,
Vrms I R   L
Z

I rms
I
2
Z  R 2   2 L2
ω
2 2
• From the phasor diagrams,
V leads I by Φ
XL
tan  
R
VL
tan  
VR
or
IX L
tan  
IR
ωL
tan  
R
45
ii) RL in series circuit
Z
X L  2fL
R
f
0
Graph of Z against f
46
iii) RLC in series circuit
L
R
C
VL
VR
VC
I
V
47
VL
iii) RLC in series circuit
L
R
C
VL
VR
VC
VL  VC 
I
V
VC
ω
V

VR
I
Phasor diagram
48
iii) RLC in series circuit
L
R
C
VL
VL
VR
VC
VL  VC 
I
V
VC
ω
V

VR
I
Phasor diagram
• The voltage across the inductor VL , resistor VR and
capacitor VC are V  IX V  IR V  IX
L
L
R
C
C
49
iii) RLC in series circuit
L
R
C
VL
VR
VC
VL
ω
VL  VC 
V

VR
I
VC
V
I
Phasor diagram
• The total p.d (supply voltage), V across L, R and C
is equal to the vector sum of VL ,VR and VC as
shown in the phasor diagram.
V 2  VR2  VL  VC 
2
V  IR   IX L  IX C 
2
2

2
V  I R  X L  X C 
2
2
2
2
V  I R  X L  X C 
2

2
50
XL
iii) RLC in series circuit
VL
ω
VL  VC 
V

VR
VC
X L  X C 
I
XC
ω
Z

R
Impedance diagram
Phasor diagram
• The impedance in RLC
circuit,
Vrms I R   X L  X C 
Z

I rms
I
2
2
Z  R2  X L  X C 
2
• From the phasor diagrams,
V leads I by Φ
I X L  X C 
tan  
IR
1 

 ωL 

X L  XC
ωC 

tan  
tan  
R
R 51
VL  VC
tan  
VR
Resonance in RLC series circuit
• Resonance is defined as the phenomenon that
occurs when the frequency of the applied voltage
is equal to the frequency of the LRC series circuit.
X C , X L , R, Z
Z
The series resonance
circuit is used for
tuning a radio receiver.
XL  f
R
0
fr
1
XC 
f
f
Graph of impedance Z, inductive reactance
XL, capacitive reactance XC and resistance
52R
with frequency.
Resonance in RLC circuit
The graph shows that :
X C , X L , R, Z
• at low frequency, impedance Z
is large because 1/ωC is large.
• at high frequency, impedance Z
is high because ωL is large.
Z
XL  f
R
0
fr
1
XC 
f
f
• at resonance, impedance Z is minimum (Z=R)
which is
2
2
X L  XC
1
2f r L 
2f r C
resonant
frequency
fr 
1
2 LC
and I is maximum
Z  R  X L  X C 
Z min  R 2  0
Z min  R
Vrms Vrms
I rms 

Z
R
53
54
iii) RLC in series circuit
EXERCISE
Example 21.3.5
A series circuit contains a 50 Ω resistor adjacent
to a 200 mH inductor attached to a 0.050μF
capacitor, all connected across an ac generator
with a terminal sinusoidal voltage of 150 V
effective.
a)What is the resonant frequency ? (1.59 kHz)
b)What voltages will be measured by voltmeters
across each element at resonance ? (150V,6kV)
c) What is the voltage across the series
combination of the inductor and capacitor ?
a)Write the equation for the supply voltage at fr.
55
Example 21.3.6
iii) RLC in series circuit
A 200  resistor, a 0.75 H inductor and a capacitor
of capacitance C are connected in series to an
alternating source 250 V, fr = 600 Hz.
Calculate
a. the inductive reactance and capacitive
reactance when resonance is occurred.
b. the capacitance C.
c. the impedance of the circuit at resonance.
d. the current flows through the circuit at
resonance.
e. Sketch the phasor diagram.
56
iii) RLC in series circuit
Solution 21.3.6
R = 200  , L = 0.75 H ,Vrms = 250 V, f = 600 Hz.
a) X L  L  2.83 k
X C  2.83 k
1
b) 2.83 x10 
, C  93.9 nF
2fC
3
VL
c) Z = R = 200 
e)
d) I rms
Vrms Vrms


 1.25 A
Z
R
VR
I
VC
57
Exercise 21.3
iii) RLC in series circuit
A series RLC circuit has a resistance of 25.0 Ω, a
capacitance of 50.0 μF, and an inductance of
0.300 H. If the circuit is driven by a 120 V, 60 Hz
source, calculate
a)The total impedance of the circuit
b)The rms current in the circuit
c) The phase angle between the voltage and the
current.
64.9 Ω , 1.85 A, 67.3o
58
SUBTOPIC :
21.4 Power and power factor (1 hour)
LEARNING OUTCOMES :
At the end of this lesson, students should
be able to :
a) Apply
i) average power, Pav  I rmsVrms cos 
ii)
instantaneous power, P  IV
iii) power factor, cos  
Pav
Pr

Pa I rmsVrms
in AC circuit consisting of R, RC, RL and RLC in
series
59
21.4 Power and power factor
• In an ac circuit , the power is only dissipated by
a resistance, none is dissipated by inductance or
capacitance.
• Therefore, the real power (Pr) that is used or gone
is given by the average power (Pave) i.e :
Pave  I rms R  Pr
2
Pave  I rms VR rms … (1)
rms voltage
across resistor
60
(for RLC circuit)
VL
VL  VC 
VC
ω
V

VR
XL
X L  X C 
I
XC
Phasor diagram
ω
Z

R
Impedance diagram
• From the diagrams above,
VR
R
cos 
and cos 
V
Z
(2) into (1)
Pave  I rms Z cos
….. (2)
2
V=rms
supply
voltage
or
Pave  I rms V cos
Papparent  Pa
Pave
cos 
Pa
61
• The term cos  is called the power factor.
• The power factor (cos  ) can vary from a
maximum of +1 to a minimum of 0.
• When  = 0o (cos  =+1) ,the circuit is completely
resistive or when the circuit is in resonance (RCL).
• When  = +90o (cos  =0) ,the circuit is completely
inductive.
• When  = -90o (cos  =0) ,the circuit is completely
capacitive.
62
• The power factor can be expressed either as a
percentage or a decimal.
• A typical circuit has a power factor of less than 1
(less than 100%).
• Example :
A motor has a power factor of 80% and the motor
consumes 800 W to operate. In order to operate
properly, the motor must be supplied with more power
than it consumes i.e.1000 W .
power factor
(80%)
Pave
cos 
Pa
800 W (consume)
1000 W (supply)
63
Example 21.4.1
An oscillator set for 500 Hz puts out a sinusoidal
voltage of 100 V effective. A 24.0 Ω resistor, a
10.0μF capacitor, and a 50.0 mH inductor in
series are wired across the terminals of the
oscillator.
a) What will an ammeter in the circuit read ?
b) What will a voltmeter read across each
element ?
c) What is the real power dissipated in the
circuit?
d) Calculate the power supply.
e) Find the power factor.
f) What is the phase angle?
64
Solution 21.4.1
f=500 Hz , V=100 V , R=24.0 Ω , C=10.0μF,
L=50.0 mH.
Vrms
a) I rms 
 784 mA
Z
VR  IR  18.8 V
b)
VL  IX L  123 V
VC  IX C  24.9 V
c) Real power ?
Pave  I rms V cos
Pave  0.7841000.188
Pave  14.7 W  Pr
65
d) Pow er supply,Pa  I rms Vrms
 0.784(100)
 78.4 W
e) Power factor,
R
cos    0.188
Z
f)
cos  0.188
  cos (0.188)
1
  79.16
o
66
Exercise 21.4
1.
A coil having inductance 0.14 H and
resistance of 12  is connected to an
alternating source 110 V, 25 Hz. Calculate
a. the rms current flows in the coil.
b. the phase angle between the current and
supply voltage.
c. the power factor of the circuit.
d. the average power loss in the coil.
4.4 A, 61.3o , 0.48, 0.23 kW
67
Exercise 21.4
2.
A series RCL circuit contains a 5.10 μF
capacitor and a generator whose voltage is
11.0 V. At a resonant frequency of 1.30 kHz t
he power dissipated in the circuit is 25.0 W.
Calculate
a. the inductance
b. the resistance
c. the power factor when the generator
frequency is 2.31 kHz.
2.94 x 10-3 H , 4.84 Ω , 0.163
68