INTERACTIVE LEARNING PLANS ALTERNATING CURRENT

Program: II IIT - JEE Achiever II-IIT JEE PHYSICS

Chapter: Alternating Current ILP # 48 Year 2023 - 24

Chapter/Topic Alternating Current/Alternating Circuits

Syllabus Alternating Current (AC), Generation of AC − AC parameters − Mean and RMS
values of AC
Concept list Direct Current (DC) and Alternating current (AC) − Generation of AC − AC parameters −
Mean values of AC − RMS values of AC
Learning 1. To understand the distinction between DC and AC.
Objects 2. To understand the generation of AC.
3. To understand the characteristics of AC.
4. To determine the mean value of AC.
5. To determine the RMS value of AC.

Concept map

Introduction
Electric current flows only in one direction through an automobile light bulb, since the polarity of the emf of
the driving battery never changes. Such a current is called a direct current (DC) and a battery is called a dc
source. In contrast to this the electric current in the bulb of our houses reverses its direction periodically.
Such a current whose magnitude and direction varies periodically with time is called an alternating current
(AC) and the driving generator is called an ac source.
Till the discovery of the phenomenon of electromagnetic induction by Michael Faraday in 1821, the only
source of electric current were Galvanic cells or accumulators. Because of the low efficiency, the power
output of these chemical cells was small and hence DC currents were only of experimental interest without
any technological impact on society. But the phenomenon of electromagnetic induction led to a large scale
generation of AC power, by simply rotating a coil of wire in a magnetic field. This heralded a second stage
of industrial revolution through electricity.
During 1880s in US, there was a heated debate over the best method of power generation and distribution.
Thomas Alva Edison favoured DC power while George Westinghouse and Nicolas Tesla favoured AC

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2I2324PILP48(AC)

power. Though AC generators are simple in design than DC generators, the complexity of mathematics
needed to describe AC was a deterrent. Finally AC power won the race because it could be easily
transformed and transported without much loss of power.

(a) Sinusoidal waveform (b) Rectangular waveform (c) Sawtooth waveform

Though AC with sinusoidal waveforms are most popular in large scale use, other forms of AC like
saw-tooth, square, triangular are also used in certain specialized applications such as radar, computer etc.
The modern electronic devices use complex AC wave forms, which are limited only by our ingenuity and
imagination. The whole of modern signal processing and digital electronics are built around the design and
analysis of complex wave forms.

Generation of AC
The basic elements of an AC generator are shown in
figure. It consists of a coil mounted on a rotor shaft. The
axis of rotation of the coil is perpendicular to the
direction of the magnetic field. The coil (called armature)
is mechanically rotated in the uniform magnetic field by
some external means. The rotation of the coil causes a
change of magnetic flux through the coil, so that an emf
is induced in the coil. The ends of the coil are connected
to an external circuit by means of slip rings and brushes.
When the coil is rotated with a constant angular speed ,
the angle  between the magnetic field vector B and the
area vector A of the coil at any instant t is  = t
(assuming  = 0 at t = 0). As a result, the effective area of the coil exposed to the magnetic field lines
changes with time.
The flux at any time t is φB = BA cos = BA cos t
From Faraday’s law, the induced emf for the rotating coil of N turns is then,
d d
e = − N B = − NBA (cos t)
dt dt ... (1)

Thus, the instantaneous value of the emf is e = NBA  sin t ... (2)
where NBA is the maximum value of the emf, which occurs when sin t = ± 1. If we denote NBAω as e0,
then e = e0 sin t ... (3)
Since the value of the sine function varies between +1 and –1, the sign, or polarity of the emf changes with
time. Note from figure that the emf has its extremum value when  = 90 or  = 270, as the change of flux
is greatest at these positions.

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 2I2324PILP48(AC)

The direction of the current changes periodically and therefore the current is called alternating current (AC).
Since ω = 2π, Eq (3) can be written as e = e0 sin 2t ... (4)
where ν is the frequency of revolution of the generator’s coil.
Note that Eq. (3) and (4) give the instantaneous value of the emf and e varies between +e0 and –e0
periodically.

AC Parameters
AC emf and current are represented by the equations
 = 0 sin t … (1)
and
I = I0 sin t … (2)

(a) Instantaneous values ( and I)

The equations (1) and (2) gives the values of AC emf and AC current at any instant t. Hence,  and I are
called the instantaneous values of AC emf and AC current. These values vary sinusoidally as shown in
the figure
(b) Peak values (0 and I0)
The maximum value of an AC emf and AC current can take in a cycle are called peak values.
(c) Angular frequency ()
 is the angular frequency of rotation of the coil in the AC generator.
(d) Frequency of AC (f)
The angular frequency  can also be written as  = 2f, where f is called the frequency of AC. This is
the frequency of rotation of the coil in the generator producing AC. In India AC is generated at a
frequency of 50 Hz.
(e) Period of AC (T)
1
Period of AC is the time taken by the coil of the generator to complete one rotation. T = .
f
(f) Phase of AC ()
The state of variation of AC emf or current at any instant is called the phase of AC at that instant. If
 = 0 sin t and I = I0 sin(t + ),  is the phase difference between  and I.
(g) Mean value of AC (mean or Imean)
Mean value of AC emf or current is mean of the instantaneous value calculated over one half cycle of
AC.
T T
2 2

 dt  dt
T T T
2 2
22
2  cos t  2 20 40
  mean = 0
= 0
=   0 sin t dt = 0  sin t dt = 0  −  = (1 + 1) =
T
T T0 T 0 T    0 T  
2
2
 
 dt
0
2 

2
= 0 = (0.6366)0


Similarly, Imean = (0.6366)I0

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2I2324PILP48(AC)

(h) RMS value of AC (rms or Irms)

RMS value of AC emf or current is the square root of the mean of instantaneous values over one
complete cycle.
  rms =  mean
2

  2rms =  mean
2


2
dt T T
1 2 1
 = =   dt =  2 sin 2 t dt
2 0
rms T
T0 T0
 dt
0

2  1 − cos 2t  02 T 02

T
 2rms = 0    dt = =
T 0 2  T 2 2

  rms = 0 = (0.707)0
2
I
Similarly, Irms = 0 = (0.707)I0
2
1. The ratio of the rms value to mean value is called the form factor.
0
 rms 
 Form factor = = 2 = = 1.11
 mean 2  2 2

0

The form factor is 1.11 only for sinusoidal AC. For other types of AC (square or triangular)
 form factors are different.
2. While calculating mean and rms values, the following standard results of integration are useful.
T T

 sin n t dt =  cos nt dt = 0, where n is an integer

0 0
T T
T
 sin nt dt =  cos 2 nt dt =
2
, where n is an integer
0 0
2

Irreducible Set of Problems

1. If the voltage of ac circuit is represented by the equation V = 220 2 sin(314t − ) the peak value and
rms value of the voltage respectively are
(A) 311 V, 220 V (B) 220 V, 311 V (C) 622 V, 220 V (D) 220 V, 622V
Ans (A)
Given equation is V = 220 2 sin(314t − ) .
Comparing this with the equation, V = V0 sin (t – ), peak value is V0 = 220 2 V = 311 V
V0 220 2
Vrms = = = 220 V
2 2

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 2I2324PILP48(AC)

Think Further
(a) In the above problem, calculate average value of voltage and frequency of AC
Ans
2 2
Vav = V0 =  311 = 198.17 V
 
 314
 = 2   = = = 50 Hz
2 2  
(b) The plate on the back of a personal computer says that it draws 2.7 A current from 120 V,
60 Hz line. For this computer, what is (i) the average of the square of the current (ii) current
amplitude (iii) the average current?
Ans
(i) 7.3 A (ii) 3.82 A (iii) 2.432 A
i rms = 2.7 A (given)
(i) irms =  i2 one cycle
  i 2 = i 2rms = (2.7) 2 = 7.3 A
(ii) i0 = irms 2 = (2.7) ( 2 ) = 3.82 A
2 2
(iii) iav = i0 = (3.82) = 2.432A
 
 
(c) An alternating current is given by the equation i = 3 2 100t −  . Calculate the frequency and
 4
mean value of the current.
Ans
 = 100  2 = 100   = 50 Hz i0 = 3 2 A ; iav = i0 = ( 3 2 ) = 2.7 A
2 2
 
2. If a direct current of value ‘A’ ampere is superimposed on an alternating current i = B sin t flowing
through a wire, the effective value of resulting current in the circuit is

1/2 1/2
 B2   B2  A+B
2
(A) [A + B ] 2 1/2
(B)  A 2 +  (C)  A 2 +  (D)
 2  2  2
Ans (C)
Current at any instant in the circuit is i = iDC + iAC = A + B sin t
1/2
T 2 
  i dt  1/2 1/2
1   1  
1/2 T 1/2 T

i eff = i rms =  0T  =     ( A + Bsin t ) dt  =     ( A2 + B2 sin 2 t + 2ABsin t ) dt 

2

   T  0   T  0 
  dt 
 0 
1/2 1/2
1  B2 T   B2 
T T 1/2
1 1 1
But  sin 2 tdt = and  sin tdt = 0  i eff =    A 2 + = A 2 + 
T0 2 T0 T  2   2 

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2I2324PILP48(AC)

Think Further
(a) Find the average and rms values of the saw tooth wave form shown in figure.

Ans zero, 1.15 V

2
V(t) = t − 2  V(t) = 4t − 2
0.5
1

 (4t − 2)dt
 V(t)  one cycle = 0
1

 dt
0
1
 2t − 2t 
2

 V(t) one cycle =  0

=0
1
1
1
16t 3 16t 2 
0 − +  3 − 2 + 4t 
2
(16t 16t 4)dt
V2 (t) = 16t 2 + 4 −16t ;  V2 (t) one cycle = 1
= 0
1
 dt 0

16 16
 V 2 (t) one cycle = − +4
3 2
4 4
= ; V(t) rms = = 1.15 V
3 3
(b) Calculate the average and the rms value of the voltage wave shown in figure.

Ans 1 V, 3.27 V
From 0 to 2 ms is one cycle
during 0 to 1 ms → V = 4
−4
during 1 ms to 2 ms → V = t + 4 = 4 − 4t
1
1 2

 4dt +  (4 − 4t) dt  4t 0 +  4t − 2t 2 1

1 2
4 +  (8 − 8) − (4 − 2) 
 V one cycle = 0 1
2
= = =1 V
2 2
 dt
0

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 2I2324PILP48(AC)

1 2

16dt +  (16 + 16t − 32t)dt

2

 V 2 one cycle = 0 1
2

 dt
0
2
 16t 3 
16t 0 + 16t + − 16t 2 
1

=  3 1
2
  128   16  
16 +  32 + − 64  − 16 + − 16   
    
=
3 3
= 10.67 V2
2
Vrms =  V2 one cycle = 10.67 = 3.27 V
(c) Find the average value of current shown graphically in figure from t = 0 to t = 2s.

Ans 5 A
(d) Show graphically that the average of sinusoidally varying current in half cycle may or may not
be zero.
Ans

3. The average value of saw-tooth voltage of peak value V0 (shown in the figure) is ________

V0 V0 V0 2
(A) (B) (C) (D) V0
3 2 2 3
Ans (C)
2V0
The equation of saw-tooth wave shown in the figure is V = t − V0
T
T

 Vdt 2   2V0
T/2 
  2   2V0  t  
2
T/2

=  t − V0  dt  = −   
T/2
Vav = 0
    V0 t
T 0  T   T   T  2   0 
T/2 0

 0
dt

2  2V0 1  T 2   T   2V  T 2  V V
=  .  − 0  − V0  − 1  = 20  − 0 − V0  Vav = 0 − V0 = 0
T  T 2 4   2  T  4  2 2

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2I2324PILP48(AC)

Think Further
(a) In the above problem what is the rms voltage?
V
Ans 0
3
1/2
T 2 
  V dt 
Vrms =0T 
 
  dt 
 0 0
1/2
 T  4V02 2 4V02  
   2 .t + V0 −
2
t  dt 
0
 T T  
=
 T 
1/2
1  4V 2 T 3 4V 2 T 2 
= 1/2  20 . + V02T − 0
T 2 
or Vrms
T  T 3
1/2
V0  4  V0 T V0
= T + T − 2T  = =
T  3  T 3 3
t
−
(b) The current in a discharging LR circuit is given by i = i0 e  , where  is the time constant of the
circuit. What is the rms value of current for the period t = 0 to t =  ?
i0 e2 − 1
Ans
e 2
t 2t
− −
i = i0e   i 2 = i 02e 

 2t
− 
i 02  e 
dt  − 2t 
2 2
=   = − i0 [e−2 − e−0 ]
i e
 i2  = 0 0
   2   2
 −  
    0
i 02  1  i 02 2 i e2 − 1
 i2  =  1 − 2 
= 2 (e − 1) i rms =  i 2  = 0
2  e  2e e 2
4. The average current of a sinusoidally varying alternating current of peak value 5 A and initial phase
T T
zero, between the instants t = to t = is
8 4
14.14 17.32 12.33 10.00
(A) A (B) A (C) A (D) A
   
Ans (A)
T T
4 4

 i dt  5sin t dt
40  1 
T T T
i = 8
= 8
=  −  cos t  T
4
T
4
T T
− T   8

 dt
T
4 8
8

40 T  2 T 2 T  20  1  10 2 14.14
i =− .  cos . − cos .  = − 0 − = = A
T 2  T 4 T 8  2   

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 2I2324PILP48(AC)

Think Further
 t 1
An alternating current i is given by the equation i = i 0 sin 2  +  . What is the average current in
T 4
the first one quarter of time period?
2i
Ans 0

T

 2 
4

i sin  .t +  dt T

 2 
0
 T 2 4i 4
i = 0
= 0  cos  .t  dt
 T 
T
4
T 0
 dt
0

 T

4i 0 T   2  4  2i 0    2i
i = . sin .t  =  sin − sin 0  = 0
T 2   T 0    2  
 
5. Show that the average heat produced during cycle of AC is the same as that produced by DC with
i = irms.
Solution
DC  i 2 R = i rms
2
R
T

 H dt 1
T
R
T
R T i 2R  i 
2

AC  H av = 0
T
=  i 2 R dt =  i 02sm 2t dt = i02 = 0 =  0  R = i rms
2
R
T0 T0 T 2 2  2
 dt
0

6. An alternating current has a frequency of 60 Hz. Find the minimum time elapsed between the instants
1
when the current attains times its peak value from half its peak value.
2
Solution
i = i0 sin t
i0 i0 
At t = t1;i = ; = i0 sin t1  t1 = i
2 2 6 i0
i0
  1
2 f t1 =  (120)t1 =  t1 = s. 2
6 6 720
i i  1 i0
At t = t 2 ;i = 0 ; 0 = i 0 sin t 2  t 2 = , t 2 = s t1 t2
t
2 2 4 480 2
t
1 1 −3
t = t 2 − t1 = − ~ 0.7 10 s.
480 720
7. A circuit carries a DC of 3 A and an AC of 4 sin t A. Find the rms current.
Solution
i = iDC + iAC = 3 + 4 sin t
i 2rms = i 2 = ( 3 + 4sin t ) = 9 + 24sin t + 16sin 2 t = 9 + 24 sin t + 16 sin 2 t
2

1
= 9 + 24(0) + 16   = 9 + 8 = 17.
2
irms = 17 A

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2I2324PILP48(AC)

i0 t
8. In a certain circuit, the electric current i varies with t as i = where 0  t  t0.
2t 0 − t
Find the rms value of current between the instants t = 0 to t = t0
Solution
t0

 i dt
2

i 2rms = i 2 = 0
t0

 dt
0
t t i
1 0 i 02 t 2 i 02 0 t2
= 
t 0 0 (2t 0 − t) 2
dt = dt.
t 0 0 (2t 0 − t) 2
2 2
i 02 0  t  i 02 0  2t 0 
t t

=  
t 0 0  2t 0 − t 
 dt =   − 1 dt
t 0 0  2t 0 − t  O t0
t

i02 0  4t 02 
t
4t 02
=  − + 1 dt
t 0 0  ( 2t 0 − t )2 2t 0 − t 
t0
i 02  4t 02 
=  + 4t 0 ln (2t 0 − t) + t 
t0  2t 0 − t 0
i02
= 4t 0 − 2t 0 + 4t 0 ln t 0 − 4 t 0 ln 2t 0 + t 0 
t0
= i02 3 − 4ln 2 = i02 (1.33)  irms = (0.48)i0

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