CHAPTER 18 ALTERNATING CURRENT CIRCUIT

CHAPTER 18 : ALTERNATING CURRENT

18.1 Alternating Current Through a Resistor
18.2 Alternating Current Through an Inductor
18.3 Alternating Current Through a Capacitor
18.4 R-C and R-L Circuits in Series

1

SUBTOPIC :
18.1 Alternating Current
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

18.1 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

I0 I : instantaneous

current @ current at

0 1 T T 3 T 2T t time t (in Ampere)
2 2 Io : peak current

− I0 T : period

 : angular frequency

I = Io sin t

4

• voltage

V

V0

where:

0 1T T 3T 2T t V : instantaneous
2 voltage @ voltage at
− V0 2
time t (in Volt)

V = Vo sin t Vo : peak voltage

T : period

 : angular frequency

5

Vo T
Io

T/2

• The output of an ac generator is sinusoidal
and varies with time.

Equation for the current ( I ) : I = Io sin t
Equation for the voltage ( V ) :
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 :
Root Mean Square (rms)

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 Irms = Io , Vrms = Vo
2 2

8

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

powerdc = average powerac current in a half-
cycle flows of
I 2R = I 2 R current in a certain
ave direction

I= I 2
ave

= square root of the average

value of the current ( )Iav=2Io= Io
= Irms π
π 9
2

I = I 2ave = Irms

• 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

V2 = V 2
ave

RR

V=Vo sin ωt V= V 2 = Vrms 10
ave

Irms = I0 Vrm s = V0
2 2

only for a sinusoidal
alternating current and voltage

• The average power, Pave = I rm sVrm s = I 2 sR = V2
rm rm s
R

Pave = Io Vo = 1 IoVo = Po
2 2 2 2

• The peak power, Po = IoVo

• 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.

13

Example 21.2.3

V (Volt)

200

0 0.02 0.04 0.06 0.08 t(second)

− 200 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

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 :
Resistance, reactance and
impedance

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: 1
2fC
i) capacitive reactance, Xc =

ii) inductive reactance, XL = 2fL

iii) impedance, Z = R2 + ( XL − XC )2 , and

phase angle,  = tan−1 ( X L − XC ) 18

R

c) Use phasor diagram to analyse voltage,
current, and impedance of series circuit of:
i) RC
ii) RL
iii) RLC

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 A = Ao sin t

yy

Aωo

NP

O 0 1T T 3T 2T t
2 2

• 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 to

form impedance) 21

i) Pure Resistor in the AC Circuit 22

I VR

ω

IV

Phasor diagram

i) Pure Resistor in the AC Circuit

• The current flows in the resistor is

I = I0 sin ωt

• The voltage across the resistor VR at any instant is

VR = IR I0 R = V0

VR = I0 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
voltage reach their maximum values at the same tim23e).

i) Pure Resistor in the AC Circuit
• The resistance in a pure resistor is R = Vrms = Vo
Irms Io
• The instantaneous power,

P = IV = I 2R = V 2 • The average power,
R
Pave = I 2 sR
P = Io sin t(Vo sin t) rm

P = IoVo sin 2 t = 1 I 2 R
2 o
Power(P)
= 1
P0 2 Vo Io

= 1 Po
2

0 1T T 3 T 2T t

22

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.

 =  rad I VR I

2 ω

I

V 25

Phasor diagram

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 VC at any instant
capacitor

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 = C VC
Q = CV0
sin( ωt −  )

2 I = dQ
dt
• The current flows in the ac circuit is 26

ii) Pure Capacitor in the AC Circuit

I = d  CV0 sin  ωt −   
dt  
2

I = CV0 d  sin  ωt −   
dt  
2

I = CV0 cos(ωt −  ) and CV0 = I0

or 2

I = I0 sin ωt

• The phase difference between V and I is

Δ = ωt −   − (ωt )

 2

Δ = −  27

2

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

XC = Vrms = Vo = Vo
I rms Io
CVo

XC = 1 = 1

C 2fC

• The capacitive reactance is defined as

XC = 1 = 1

C 2fC

28

ii) Pure Capacitor in the AC Circuit

• The instantaneous power, • The average power,

P = IV = I 2R = V 2 Power(P) Pave = 0
R
P0 3 T 2T
P = Io sin t(−Vo cost) 2
2
P = − I oVo  1 sin 2t  0 1T t
 2  2
− P0 T
2
P = − Po  1 s in 2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

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,

rms current, Peak current ?

30

iii) Pure Inductor in the AC Circuit

• Pure inductor means I VL I
that no resistance
and capacitance

effect in the a.c. circuit.

ω

V
I

Phasor diagram  =  rad

2 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 = I0 sin ωt

• When the current flows in the inductor, the back

emf caused by the self induction is produced and

given by εB = −L dI
dt

B = −L d (I0 sin ωt )
dt

B = −LI0ω 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 , 0
V −  B = IR
V =  B = LI0ω cos ωt or V =B

V = LI 0ω sin  ωt +  
 
2

V = Vo sin  ωt +  
 
2

where Vo = LIo

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

XL = Vrms = Vo = LIo
I rms Io
Io

X L = L = 2 fL 34

iii) Pure Inductor in the AC Circuit

• The inductive reactance is defined as X L = L = 2 fL

• The instantaneous power, • The average power,

P = IV = I 2R = V 2 Power(P) Pave = 0
R
P0
P = Io sin t(Vo cost)

P = I oVo  1 sin 2t  2
 2 
0 t
 1  1T 3T 2T
 2  − P0 2 T
2t
P = Po sin 2

• For the first half of th2e cycle where the power is

positive, the inductor is saving the power. For

the second half cycle where the power is

negative, the power is returned to the circuit35.

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.

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.

36

i) RC in series circuit

RC VR  : phase angle
I
VR VC VC V V = supply voltage
I
ω
V
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
I
Note VC V

Vo2 = VRo2 + VCo 2 ω

Vo2 = (IoR)2 + (Io X L )2 Phasor diagram

Vo = (IoR)2 + (Io X L )2

Vo = Io (R)2 + (X C )2 ...divide both side by 2

( ) ( )Vrms = Irms R 2 + X C 2

V = I (R)2 + (X C )2

V =I (R)2 +   1 2 2
 2C 
38

i) RC in series circuit VR  : phase angle
I
RC VC V V = supply voltage

VR VC ω
I
Phasor diagram
V

• 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.

V 2 = V 2 + VC2 R2 1
R ω2C 2
VR = IR V =I +
VC = IX C
V 2 = (IR )2 + (IX C )2
( )V 2
= I2 R2 + X 2
C

and XC = 1 39
ωC

i) RC in series circuit R

VR 
I
VC V XC Z

ω ω

Phasor diagram Impedance diagram

• The impedance in RC • From the phasor diagrams,

circuit, I leads V by Φ

Z = Vrms = I R2 +  1 tan = VC tan = X C
I rms 2C 2
VR R
I

Z= R2 + 1 or
2C
 2 tan = IX C 1
ωCR40
IR tan =

i) RC in series circuit

Z

XC = 1

2fC

R

0f
Graph of Z against f

41

Example 21.3.4 i) RC in series circuit

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.

42

ii) RL in series circuit ω

RL

VR VL V  : phase angle
I V VL 
V = supply voltage
VR
I

Phasor diagram

• The voltage across the resistor VR and the capacitor
VL are VR = IR

VL = IX L

43

ii) RL in series circuit ω

RL

VR VL V  : phase angle
I VL 
I V = supply voltage
VR

V 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 = V 2 + V 2
R L

( )V 2 = (IR )2 + (IX L )2 V = I R 2 + ω 2 L2

V2 = I2 R2 + X 2
L

and X L = ωL 44

ωii) RL in series circuit ω

V I Z
VL  XL 

VR R

Phasor diagram Impedance diagram

• The impedance in RC • From the phasor diagrams,

circuit, V leads I by Φ

Z = Vrms = I R2 +  2 L2 tan = VL tan = X L
VR R
Irms I
or
Z = R2 +  2L2
tan = IX L tan = ωL

IR R

45

ii) RL in series circuit

Z

X L = 2fL

R
0f

Graph of Z against f

46

iii) RLC in series circuit
L RC
VL VR VC

I
V

47

iii) RLC in series circuit VL ω

L RC (VL − VC ) V I

VL VR VC
I VC VR

V Phasor diagram

48

iii) RLC in series circuit

L RC VL V ω
VR VC 
VL (VL − VC ) I
I V VR
VC

Phasor diagram

• The voltage across the inductor VL , resistor VR and

capacitor VC are VL = IX L VR = IR VC = IX C

49

iii) RLC in series circuit VL V ω

L RC (VL − VC ) I
VL VR VC VR
I VC

V 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 R 2 + (X L − X C )2

V 2 = (IR )2 + (IX L − IX C )2 V = I 50

 V 2 = I 2 R 2 + (X L − X C )2