ALTERNATING CURRENT

18.1 Alternating Current through a Resistor

Objective : Students should be able to:
(a) explain the concept of the r.m.s. value of an alternating current, and

calculate its value for the sinusoidal case only

18.1.1 Root-mean-square (r.m.s.) Value of an Alternating Current

Alternating current

 An alternating current (a.c.) is a current in which the direction of the current changes
or reverses periodically.

 A direct current (d.c.) is a current in which the direction of the current is always the
same or in one direction only

Figure below shows how a sinusoidal alternating current changes with time.

Note:
1. A sinusoidal alternating current can be represented by a sine function or a

cosine function.

2. Alternating currents are produced by alternating voltages.

3. The peak value of an alternating current is Io. The peak value is the maximum
value of the amplitude of the a.c

4. Peak-to-peak value of an a.c. = 2 x peak value = 2 Io
5. Mean-square value of an a.c.

 Mathematically, to find the root-mean-square of an alternating current, we square
the current, then find the mean of the square of the current, and finally find the
square-root of the square of the current.

6. Definition: Root-mean-square value of an a.c.
The r.m.s. value of an a.c is equal to the value of a constant d.c
which can produce the same heating power as the a.c in a given resistor.

7. To deduce the relationship for r.m.s. for an sinusoidal A.C.
(a) The sinusoidal alternating voltage is given by the relationship,

I  Io sin  t . The variation of I with time,t is as shown below.

(b) Squaring, I2  I 2 sin 2  t . The variation of the I2 with time t is shown below.
o

(c ) The curves of the graph at the top and at the bottom are symmetrical. Note that if we

cut off the top half of the curve and invert it, it fits exactly the lower part of the curve.

Hence the mean value is the same as the value of the dotted straight line.

Hence, the mean square value for one cycle is given by I2  I 2
o

2

(d) Taking square-root, the r.m.s. current is given by I rms  Io
2

(e) Similarly, the r.m.s. value of the alternating voltage is V. rms  Vo
2

Notes: The r.m.s. value of an a.c. is also known as the d.c. equivalent of the a.c

Notes:
The number of oscillations of an a.c per second is called its frequency, f.
For one oscillation, the a.c changes its direction twice.
The time taken for one oscillation of the a.c is the period T, of the a.c.

Period , T  1
f

The angular frequency,  of an a.c

  2 f therefore , f 
2

8. Calculation of r.m.s value of other forms of A.C.

Mathematically, the r.m.s. value squaring, then find mean, and then find the root.

From I  Io sin  t ; ( t )

I2  I 2 sin2  t
o

Mean -square value of an a.c, < I2 > = < I 2 sin 2  t >
o

= I 2 sin 2  t
o

= I 2
o

2

Similarly the mean-square value of an a.c, voltage =  V 2   Vo2
2

Example:
Q1 . An a.c. is given by I = 8 sin (20π t), where I is in amperes. Find

(a) the peak value of the a.c.,
(b) the r.m.s. value of the a.c.,

(c) the frequency of the a.c.
Solution

Q2. Figure below shows how an a.c. voltage varies with time.

(a) What is the r.m.s. value of the a.c. voltage?
(b) What is the frequency of the a.c. voltage?
solution

18.1.2 Phasor Diagram

1. A phasor is a vector of constant magnitude and rotating in an anticlockwise direction

with a constant angular velocity,  . When a phasor rotates, it will generate a waveform.

The type of waveform generated will depend on the initial position of the phasor.
2. A sine function waveform is generated by a phasor having an initial position as shown
in Figure below

3. A cosine function waveform is generated by a phasor having an initial position as
shown in figure below

4. A negative sine function waveform is generated by a phasor having an initial
position as shown in Figure below.

5. A negative cosine function waveform is generated by a phasor having an
initial position as shown in Figure below.

6. Figure below shows phasor P and phasor Q rotating with the same angular



Velocity,  and with a phase difference of 2 rad. or 90°.

The phasor diagram above shows the positions of the two phasors at
time t = 0.

The sinusoidal waveform generated by phasor Q is represented v = r sin w t
The sinusoidal waveform generated by phasor P is represented by v = r cos wt.


Phasor P is leading phasor Q by 2 rad. or


phasor Q is lagging phasor P by 2 rad.

Objective (b): Candidates should be able to derive an expression for the current

from V  Vo sin  t

18.1.3Alternating Current through a Resistor

1. Figure below shows an alternating voltage connected to a pure resistor.

Let’s say that the a.c. voltage is given by the equation, V  Vo sin  t

Applying Ohm's law:

I V
R

I  Vo sin  t
R

I  Vo sin  t
R

Io  Vo
R
( Peak current )

Therefore, I  Io sin  t ****

In the R-circuit, if the voltage is given by V  Vo sin  t ,
Then the current will be I  Io sin  t

Notes: ****

Io  Vo I rms  Vrms
R R
and

Objective (c) explain the phase difference between the current and voltage for a pure
resistor;

2. Phase difference between the current, I and the voltage, V for a pure Resistor

 If the a.c. voltage across R is, V  Vo sin  t then. I  Io sin  t

 The result shows
the a.c. voltage across the resistor and
the a.c. current in the resistor are oscillating with the same
sine function,

 therefore they are oscillating in-phase.
( The phase difference between the current and the voltage is zero).

 The a.c. current and the a.c voltage in a resistor are in-phase.

3. Figure below shows two phasors,
 a current phasor and a voltage phasor which are in-phase.

When the current is maximum, the voltage is also maximum.
When the current zero, the voltage is also zero.

Objective :(d) derive and use the formula for the power in an a.c circuit which
consists only of a pure resistor.

4 .The instantaneous power and maximum power
The instantaneous power is the power at a particular moment of time.

Instantaneous power, P = V I

P = (Vo sin  t) (Io sin  t )
P = Vo Io sin2  t ***

Maximum power, Po  Vo Io ***

5. The power-time graph

Instantaneous power , P =Vo Io Sin2  t

From , P = Vo Io sin2  t ,
maximum power Po  Vo Io

Po  2VrmsIrms

Po  I 2 R
o

Po  Vo2
R

6. Mean power

From P = Vo Io sin2  t ,

Mean power (average power), <P> = <Vo Io sin2  t > ,

= Vo Io sin2  t ,

P  Vo Io sin2  t  1
2
Therefore, mean power, 2 Since

Mean power of an a.c. in a resistor is also the rate of heat dissipation in the resistor.

7. Maximum power Po  Vo Io

mean power, P  Vo Io
2

Maximum power = 2 x (mean power)

P  Vo Io I 2 R
2 = Vrms Irms rms
8. Mean power =

 Resistance is defined as the ratio of the peak value of voltage, Vo to the peak value

of current, Io

Hence R  Vo
Io

 Since the peak values of the voltage and current are directly proportional to their

root-mean-square values, hence the

R  Vrms
I rms

Questions:

1. An a.c. is given by I  8 sin20 t where I is in amperes. Find

(a) the peak value of the a.c.,
(b) the r.m.s. value of the a.c.,
(c) the frequency of the a.c.

2. Figure below shows how an a.c. voltage varies with time.

(a) What is the r.m.s. value of the a.c. voltage?
(b) What is the frequency of the a.c. voltage?

3. An a.c. power supply is connected across a resistor of 20 Ω as shown in Figure below. The a.c. voltage
is given by V= 15 sin 100 π t, where V is in volts.

Calculate
(a) the frequency of the a.c.,
(b) the maximum current in the circuit,
(c) the maximum power,
(d) the rate of heat dissipation in the resistor.

4. An a.c. power supply is connected across a resistor of 40 Ω. The a.c. voltage is given by
V= 5 sin 40 π t, where V is in volts.

(a) State the phase relation between the current and the voltage in the resistor.
(b) Write an expression for the current in the resistor.

5. The graph below shows the waveform of an alternating voltage which is connected to a resistor of
20 Ω

Calculate
(a) the peak-to-peak voltage and the r.m.s. voltage,
(b) the r.m.s. current through the resistor,
(c) the mean power dissipated in the resistor.

6. An a.c. voltage given by V = 30 sin 120 π t is connected across a resistor of 4 kΩ.
(a) Find the angular frequency of the a.c.
(b) Find the r.m.s. current.
(c) Determine the maximum power of the resistor.
(d) Determine the average power of the resistor.
(e) Determine the heat dissipated in the resistor in 5 minutes

12
34
5.
6.

END
(RESISTOR)

18.2 a.c. through a pure inductor (refer to the diagram)

Figure below shows an a.c. voltage applied to a pure inductor of inductance L.
A L-circuit refers to a pure inductor connected to an a.c. source.
When an a.c is oscillating in the inductor coil, an induced e.m.f. f is produced in the
inductor.
The induced e.m.f. due to self-induction is given by

  - L dI
dt

Apply Kirchhoff’s Second Law

V - L dI 0 then V  L dI
dt dt

Objective: (e) derive an expression for the current from V= Vo sin  t

2. Derivation of an expression for the a.c. current in an inductor starting from

V= Vo sin  t

V  L dI and V= Vo sin  t
dt

Vo sin  t  L dI
dt

Vo sin  t d t   L d I

 Vo cos  t  L I


I   Vo cos  t
L

I   Vo cos  t
L

Io  Vo therefore circuit current , I=  Io cos t
L
Let

If V= Vo sin  t then I =  Io cos  t and  L  Vo
Io

The result shows that the a.c. voltage, V in the inductor leads the a.c. current, I in the
inductor by 90°(  rad ) ***

2

Extra
3. Derivation of an expression for the a.c. voltage in an inductor, starting from

I = Io sin  t

From V  L dI and I = Io sin  t ;
dt

V  L d (Io sin  t)
dt

V  LIo cos  t

Let Vo  LIo

then V  Vo cos t


(V leads I by 90o or rad )****

2

Extra
4. Derivation of an expression for the a.c. current in an inductor, starting from

V  Vo cos t

V  L dI and V  Vo cos t
dt
From

Vo cos  t  L dI
dt

Vo cos t dt  L dI

 Vo cos t dt =  L dI

Vo sin  t  L I


I  Vo sin  t
L

Io  Vo then I  Io sin  t
L
Let

The result also shows that the a.c. voltage across an inductor leads the a.c. current in the


inductor by 90° ( 2 rad )


(V leads I by 90o or 2 rad )****

Objective (f): explain the phase difference between the current, I and voltage, V for
a pure inductor;

5. Phasor diagram for the alternating current and the alternating voltage in a
pure Inductor

 Figure below shows two phasors, a voltage phasor which generates a cosine
waveform and a current phasor which generates a sine waveform.

Phasor V generates the waveform V  Vo cos t and

phasor I generates I waveform I  Io sin  t

 From the phasor diagram or the waveform diagram, we can see that:

(a) Inductor voltage,V leads inductor current, I by 90o.

(b) Inductor current, I lags inductor voltage, V by 90°.

(c) There is a phase difference between voltage and current of  rad
2

or 90°.

Note:

(a) When the instantaneous voltage, V is at its peak, the current, I is zero.

(b) When the instantaneous current, I is at its peak, the voltage, V is zero.

(c) Looking at the waveforms, one should also be able to see that the inductor
voltage , V leads the inductor current , I by 90°.

Objective : (g) define the reactance of a pure inductor;

6. Definition: Reactance of a pure inductor

The reactance is the opposition to the current by an inductor or a capacitor in an a.c. circuit.

Reactance of an inductor X L is defined as:

XL  Peak voltage of inductor  Vo Unit for reactance is ohm (Ω)
Peak current of inductor Io

XL  r.m.s voltage of inductor  Vr.m.s
r.m.s current of inductor I r.m.s

Objective: (h) Use the formula X L   L

7. Equation for the reactance of a pure inductor

Vo  LIo  L  Vo
Io

XL  Vo X L   L  2 f L
Io
Since therefore

8. Relationship between inductive reactance and frequency

XL  L is directly proportional to  or

XL is directly proportional to frequency, f .

The graph of XL against f is shown in Figure

9. Relationship between peak current and frequency

Io  Vo  Vo  Vo L
XL L 2 f
Io, is inversely proportional to frequency, f.

The graph of Io against f is shown in Figure

As the frequency increases, the reactance increases, and the peak current, Io decreases.

I r.m.s  Io
2

Objective :(h) derive and use the formula for the power in an alternating current
circuit which consists only of a pure inductor;

10. Derivation of the instantaneous power and maximum power of inductor

In a L-circuit, if I  Io cos  t , then V  Vo sin  t

Instantaneous power, P = VI

= (Vo sin  t)(  Io cos  t)  (Vo cos  t)( Iosin  t)

= VoIo sin  t cos  t
Vo Io sin 2 t

=2

Maximum power, Po = Vo Io  Vr.m.s.  I r.m.s
2

11. Power-time graph of inductor in a L-circuit
Figure below shows the variation of power in an inductor with time.

The frequency of the power waveform is twice the frequency of the voltage waveform
or the current waveform.

12. Average power of inductor

Vo Io sin 2 t Vo Io 2 t
2
Average power, < P > = < >= 2 <sin > = 0

 Since the average power is zero, there is no heat

dissipation in a pure inductor.

 From the power-time graph:

(a) Positive power means that the inductor is absorbing power from the circuit.
Energy absorbed is stored in the inductor as magnetic field.

(b) Negative power means that the inductor is releasing power to the circuit.
Energy from the inductor is being released to the circuit.

 The inductor releases just as much power back to the circuit as it absorbs over
the span of a complete cycle.

 Therefore, the net energy absorbed by the inductor is zero during one complete
cycle.

 Hence an inductor is suitable to use to control the flow of alternating current, such
as in the fan regulator to control the speed of the fans.

13. Energy stored in an inductor

Energy stored in an inductor, U

U  1LI2 ( If we let I  Iosin  t)
2

U  1 L I o sin  t2
2

U  1 L I 2 sin 2  t
2 o

Energy stored in an inductor varies with time.

The maximum energy stored is U  1 L I2 , when
2

 current, I is at peak value and

 voltage,V is at zero value.

End
(Inductor)

18.3 Alternating Current through a Capacitor

1. Figure below shows an a.c voltage applied to a pure capacitor. The C- circuit refers to
a pure capacitor connected to an a.c. source.

 When an a.c. voltage is applied across a capacitor, an a.c flows in the circuit.
 When the current is oscillating in the circuit, the capacitor is undergoing charging

and discharging processes.
 Let Q = charge on the capacitor, I = capacitor current and V = capacitor voltage.

These three quantities, V, Q and I vary with time and oscillate sinusoidally.

CQ

 The capacitance, C of the capacitor is a constant, and V and Q = CV.
 The source voltage and the capacitor voltage are equal in magnitude, so

V is the source voltage and also the capacitor voltage.

Objective (i) derive an expression for the current from V  Vo sin  t

2. Derivation of an expression for the current in the circuit from V  Vo sin  t

In a C-circuit, Q = CV dQ  C dV
dt dt

I  C dV
dt

Let the a.c. voltage be V  Vo sin  t (starting)

dV   Vo cos  t
dt

I  C dV =  CVo cos  t
dt

From equation, I   CVo cos t

 Let peak current, Io   CVo Therefore current, I  Io cos  t

 In this circuit, if we let V  Vo sin  t then the current is I  Io cos  t and

Io   CVo .

 This result shows that the a.c. voltage, V in the capacitor lags the a.c. current, I in


the capacitor by 90° ( rad ).
2

Objective: (j) explain the phase difference between the current and voltage for a
pure capasitor;

3. Phasor diagram for alternating current and alternating voltage in a capacitor

In a C-circuit, if V  Vo sin  t then I  Io cos  t

Figure below shows two phasors,
 a voltage phasor which generates a sine waveform and
 a current phasor which generates a cosine waveform.

From the phasor diagram or the waveform diagram, we can see that:

(a) Capacitor current is leading capacitor voltage by a phase angle of  rad
2


(b) Capacitor voltage is lagging behind capacitor current by a phase angle of rad
2

 There is a phase difference of  rad , between the capacitor voltage and the
2

capacitor current.

Note:
(a) When the voltage is maximum, the current is zero.
(b) When the current is maximum, the voltage is zero.

V  Vo sin  t then I  Io cos  t

or or
V  Vo cos  t then I  Io sin  t
V  Vo sin t then I  Io cos  t

II
V

V

or
I  Io sin  t then V  Vo cos  t

(refer to the phasor diagram of an ac }
through a resistor note’s

I IV
V

(A phasor is a vector of constant magnitude and rotating in an anticlockwise direction with

a constant angular velocity,  . When a phasor rotates, it will generate a waveform. The

type of waveform generated will depend on the initial position of the phasor.)

Objective: (k) define the reactance of a pure capasitor

4. Definition: Reactance of a pure capacitor

The reactance is the opposition to the current flow by a capacitor.
Reactance of a capacitor, X c , is defined as:

XC  Peak voltage of cap acit or  Vo I r.m.s  Io Vr.m.s  Vo
Peak current of cap acit or Io 2 2

XC  r.m.s voltage of capacitor  Vr.m.s
r.m.s current of capacitor I r.m.s

X C  1
C
Objective: (l) use the formula

5. Equation for the reactance of a pure capacitor

Io   CVo and X C  Vo 1
Io C

X C 1  1 C
C 2 f
Unit for reactance is ohm (Ω).

6 . Relation between capacitive reactance and frequency

X C  1  1 C
C 2 f
Capacitive reactance ,

X C is inversely proportional to f

Figure below shows the graphs of (a) Xc against f (b) Xc against 1

f

(a) Xc against f (b) Xc against 1

f

7. Relation between peak current and frequency

In a C-circuit, Io   CVo

= 2 f C Vo

I r.m.s  2 f C Vr.m.s

The peak current, Io is directly proportional to frequency f, or angular frequency,  . Figure
below shows the graph of Io against  .

From Io   CVo the gradient of the
Io -  graph is equal to CVo

Objective:(d) derive and use the formula for the power in an a.c circuit which consists
only of a pure capasitor;

8 .Derivation of the instantaneous power and maximum power of capacitor

In a C-circuit, if V  Vo sin  t then I  Io cos  t

Instantaneous power, P = VI = (Vo sin  t)( Io cos  t)

=VoIo sin  t cos  t

Vo Io sin 2 t
P= 2

The angular frequency of the power waveform is twice the angular frequency of the
voltage waveform or the current waveform.

Maximum power, Po  Vo Io
2

9. Power-time graph of capacitor in a C-circuit

10 .Average power of capacitor

 VoIo sin 2 t   VoIo  sin 2 t   0
Average power,< P > = 2 2

Since the average power is zero, there is no heat dissipation in a pure capacitor.

(b) From the power-time graph:
(i) Positive power means that the capacitor is absorbing power from the circuit.

Energy absorbed is stored in the capacitor as electric field.
(ii) Negative power means that the capacitor is releasing power to the circuit.
(iii) The capacitor releases just as much power back to the circuit as it absorbs

over the span of a complete cycle.
Therefore, the net energy absorbed by the capacitor is zero during one complete
cycle.

11. Energy stored in a capacitor,U

U  1 CV 2 and V  Vo sin  t
2

U  1 CVo2 sin 2  t
2
Therefore, U varies with time t.

U  1 CVo2
2
Maximum energy stored, and this occurs when the voltage is at the

peak value Vo and current is zero.

Example:
1. An A.C. generator with peak voltage of 120 V is connected across a pure capacitor of

300 µF. Calculate the peak value and root-mean-square value of the current in the circuit
for the generator of frequency.
(a) 50 Hz
(b) 2500 Hz

2 .A sinusoidal alternating voltage represented by the equation V = 140 sin (1500 π t) is
supplied to a capacitor. The root-mean-square current in the circuit is 3.0 A.
Calculate
(a) the capacitance of the capacitor and
(b) the maximum charge stored in the capacitor.

Exercise: AC through a capacitor

1. An a.c. voltage is connected to a pure capacitor of capacitance I 200 µF. Calculate

(a) the reactance of the capacitor in this circuit,
(b) the r.m.s. current,
(c) the maximum power,
(d) the average power,
(e) the maximum energy stored in the capacitor.

2 An u.c. voltage given by V = 16 sin 50 π t is connected to a capacitor of 500 µF
Calculate

(a) the peak current in the circuit,
(b) the maximum power of the capacitor.

3 An a.c. voltage given by V = 50 sin 300 t is connected to a capacitor of capacitance C.
If the maximum power in the capacitor is 40 W, what is the capacitance of the
capacitor?

4. An a.c. voltage of maximum voltage 40 V and 100 Hz is connected to a capacitor of 200 µF

(a) What is the maximum current in the capacitor?
(b) What is the maximum power?

5. An a.c. source which has an angular frequency of 60  rad. s-1 is connected to a pure
capacitor. The maximum output voltage of the source is 80 V and the r.m.s. output
current is 0.45 A.

(a) What is the capacitance of the capacitor?
(b) What is the maximum power?

6. The alternating current flowing through a pure capacitor of capacitance C is
represented by the following equation.

I  Io sin  t

Derive an expression for the voltage across the capacitor.

7. A 200 µF capacitor is connected to an alternating voltage of 12 V (r.m.s.) and a
frequency of 50 Hz. Calculate
(a) the reactance of the capacitor,
(b) the r.m.s. current in the circuit.

8. The graph below shows the variation of a.c. voltage with time in a pure capacitor

(a) Draw on the same diagram, to show how the a.c. in the capacitor varies with time.
(b) Draw on the same diagram, to show how the power of the capacitor varies with

time.