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# Alternating Current JEE Notes | EduRev

## JEE : Alternating Current JEE Notes | EduRev

``` Page 1

J E E - P h y s i c s
3 2
E
ALTERNATING CURRENT
ALTERNATING CURRENT AND VOLTAGE
Voltage or current is said to be alternating if it is change continously in magnitude and perodically in direction
with time. It can be represented by a sine curve or cosine curve
I = I
0
sin ? ?t or I = I
0
cos ? ?t
where I = Instantaneous value of current at time t, I
0
= Amplitude or peak value
? ?= Angular frequency  ? ? =
2
T
?
= 2 ?f T = time period f = frequency
I
0
–I
0
T
4
T
2
3
4
T
T
t
I
I as a sine function of t

I
0
–I
0
T
4
T
2
3
4
T
T
t
I
I as a cosine function of t
AMPLITUDE OF AC
The maximum value of current in either direction is called peak value or the amplitude of current. It is represented
by I
0
. Peak to peak value = 2I
0
PERIODIC TIME
The time taken by alternating current to complete one cycle of variation is called periodic time or time period
of the current.
FREQUENCY
The number of cycle completed by an alternating current in one second is called the frequency of the current.
UNIT : cycle/s ; (Hz)
In India : f = 50 Hz , supply voltage = 220 volt In USA  :f = 60 Hz ,supply voltage = 110 volt
CONDITION REQUIRED FOR CURRENT/ VOLTAGE TO BE ALTERNATING
• Amplitude is constant • Alternate half cycle is positive and half negative
The alternating current continuously varies in magnitude and periodically reverses its direction.
I
sinosudial AC
I
t
+
–
triangular AC
+
–
I
t
square wave AC
saw tooth wave
I
t
mixture of AC and DC
I
0
t
Not AC (direction not change)
I
0
I
0
t t
Not AC (not periodic)
JEEMAIN.GURU
Page 2

J E E - P h y s i c s
3 2
E
ALTERNATING CURRENT
ALTERNATING CURRENT AND VOLTAGE
Voltage or current is said to be alternating if it is change continously in magnitude and perodically in direction
with time. It can be represented by a sine curve or cosine curve
I = I
0
sin ? ?t or I = I
0
cos ? ?t
where I = Instantaneous value of current at time t, I
0
= Amplitude or peak value
? ?= Angular frequency  ? ? =
2
T
?
= 2 ?f T = time period f = frequency
I
0
–I
0
T
4
T
2
3
4
T
T
t
I
I as a sine function of t

I
0
–I
0
T
4
T
2
3
4
T
T
t
I
I as a cosine function of t
AMPLITUDE OF AC
The maximum value of current in either direction is called peak value or the amplitude of current. It is represented
by I
0
. Peak to peak value = 2I
0
PERIODIC TIME
The time taken by alternating current to complete one cycle of variation is called periodic time or time period
of the current.
FREQUENCY
The number of cycle completed by an alternating current in one second is called the frequency of the current.
UNIT : cycle/s ; (Hz)
In India : f = 50 Hz , supply voltage = 220 volt In USA  :f = 60 Hz ,supply voltage = 110 volt
CONDITION REQUIRED FOR CURRENT/ VOLTAGE TO BE ALTERNATING
• Amplitude is constant • Alternate half cycle is positive and half negative
The alternating current continuously varies in magnitude and periodically reverses its direction.
I
sinosudial AC
I
t
+
–
triangular AC
+
–
I
t
square wave AC
saw tooth wave
I
t
mixture of AC and DC
I
0
t
Not AC (direction not change)
I
0
I
0
t t
Not AC (not periodic)
JEEMAIN.GURU
J E E - P h y s i c s
E
3 3
AVERAGE VALUE OR MEAN VALUE
The mean value of A.C over any half cycle (either positive or negative) is that value of DC which would send
same amount of charge through a circuit as is sent by the AC through same circuit in the same time.
average value of current for half cycle < I  > =
T / 2
0
T / 2
0
Idt
dt
?
?
Average value of I = I
0
sin ?t over the positive half cycle :
< sin > = < sin 2 >=0
< cos >= < cos 2 >= 0
< sin  cos > = 0
< sin > = < cos >=
?

?
? ?
? ?
? ?
2 2 1
2
T
2
0
0
av T
2
0
I sin t dt
I
dt
?
?
?
?
= ? ?
T
0
2
0
2 I
cos t
T
? ?
?

0
2 I
?
?
• For symmetric AC, average value over full cycle = 0,
Average value of sinusoidal AC
Full cycle (+ve) half cycle (–ve) half cycle
0
2I
0
?
–2I
0
?
As the average value of AC over a
com plete cycle is zero, it is always
defined over a half cycle which must
be either positive or negative
MAXIMUM VALUE
• I = a sin ? ? I
Max.
= a • I = a + b sin ? ? ? I
Max.
= a + b ( if a and b > 0)
• I = a sin ? + b cos ? ? ? I
Max.
=
2 2
a b ?
• I = a sin
2
? ? I
Max.
= a (a > 0)
ROOT MEAN SQUARE (rms) VALUE
It is value of DC which would produce same heat in  given resistance in given time as is done by the alternating
current when passed through the same resistance for the same time.
T
2
0
rms T
0
I dt
I
dt
?
?
?
rms value = Virtual value = Apparent value
rms value of   I = I
0
sin ?t :
T
2
0
0
rms T
0
(I sin t) dt
I
dt
?
?
?
?
=
2
T
2 0
0
I
sin t dt
T
?
?
= 0
T
0
1 1 cos2 t
I dt
T 2
? ? ? ?
? ?
? ?
?
T
0
0
1 t sin 2 t
I
T 2 2 2
? ? ?
? ?
? ?
? ?
? ?
=
0
I
2
• If nothing is mentioned then values printed in a.c circuit on electrical appliances, any given or unknown
values, reading of AC meters are assumed to be RMS.
Current Average P e a k R M S Angular fequency
I
1
= I
0
sin ?t 0 I
0
0
I
2
?
I
2
= I
0
sin ?t cos ?t =
0
I
sin 2 t
2
? 0
0
I
2
0
I
2 2
2 ?
I
3
= I
0
sin ?t + I
0
cos ?t 0
0
2 I I
0
?
• For above varieties of current rms =
Peak value
2
JEEMAIN.GURU
Page 3

J E E - P h y s i c s
3 2
E
ALTERNATING CURRENT
ALTERNATING CURRENT AND VOLTAGE
Voltage or current is said to be alternating if it is change continously in magnitude and perodically in direction
with time. It can be represented by a sine curve or cosine curve
I = I
0
sin ? ?t or I = I
0
cos ? ?t
where I = Instantaneous value of current at time t, I
0
= Amplitude or peak value
? ?= Angular frequency  ? ? =
2
T
?
= 2 ?f T = time period f = frequency
I
0
–I
0
T
4
T
2
3
4
T
T
t
I
I as a sine function of t

I
0
–I
0
T
4
T
2
3
4
T
T
t
I
I as a cosine function of t
AMPLITUDE OF AC
The maximum value of current in either direction is called peak value or the amplitude of current. It is represented
by I
0
. Peak to peak value = 2I
0
PERIODIC TIME
The time taken by alternating current to complete one cycle of variation is called periodic time or time period
of the current.
FREQUENCY
The number of cycle completed by an alternating current in one second is called the frequency of the current.
UNIT : cycle/s ; (Hz)
In India : f = 50 Hz , supply voltage = 220 volt In USA  :f = 60 Hz ,supply voltage = 110 volt
CONDITION REQUIRED FOR CURRENT/ VOLTAGE TO BE ALTERNATING
• Amplitude is constant • Alternate half cycle is positive and half negative
The alternating current continuously varies in magnitude and periodically reverses its direction.
I
sinosudial AC
I
t
+
–
triangular AC
+
–
I
t
square wave AC
saw tooth wave
I
t
mixture of AC and DC
I
0
t
Not AC (direction not change)
I
0
I
0
t t
Not AC (not periodic)
JEEMAIN.GURU
J E E - P h y s i c s
E
3 3
AVERAGE VALUE OR MEAN VALUE
The mean value of A.C over any half cycle (either positive or negative) is that value of DC which would send
same amount of charge through a circuit as is sent by the AC through same circuit in the same time.
average value of current for half cycle < I  > =
T / 2
0
T / 2
0
Idt
dt
?
?
Average value of I = I
0
sin ?t over the positive half cycle :
< sin > = < sin 2 >=0
< cos >= < cos 2 >= 0
< sin  cos > = 0
< sin > = < cos >=
?

?
? ?
? ?
? ?
2 2 1
2
T
2
0
0
av T
2
0
I sin t dt
I
dt
?
?
?
?
= ? ?
T
0
2
0
2 I
cos t
T
? ?
?

0
2 I
?
?
• For symmetric AC, average value over full cycle = 0,
Average value of sinusoidal AC
Full cycle (+ve) half cycle (–ve) half cycle
0
2I
0
?
–2I
0
?
As the average value of AC over a
com plete cycle is zero, it is always
defined over a half cycle which must
be either positive or negative
MAXIMUM VALUE
• I = a sin ? ? I
Max.
= a • I = a + b sin ? ? ? I
Max.
= a + b ( if a and b > 0)
• I = a sin ? + b cos ? ? ? I
Max.
=
2 2
a b ?
• I = a sin
2
? ? I
Max.
= a (a > 0)
ROOT MEAN SQUARE (rms) VALUE
It is value of DC which would produce same heat in  given resistance in given time as is done by the alternating
current when passed through the same resistance for the same time.
T
2
0
rms T
0
I dt
I
dt
?
?
?
rms value = Virtual value = Apparent value
rms value of   I = I
0
sin ?t :
T
2
0
0
rms T
0
(I sin t) dt
I
dt
?
?
?
?
=
2
T
2 0
0
I
sin t dt
T
?
?
= 0
T
0
1 1 cos2 t
I dt
T 2
? ? ? ?
? ?
? ?
?
T
0
0
1 t sin 2 t
I
T 2 2 2
? ? ?
? ?
? ?
? ?
? ?
=
0
I
2
• If nothing is mentioned then values printed in a.c circuit on electrical appliances, any given or unknown
values, reading of AC meters are assumed to be RMS.
Current Average P e a k R M S Angular fequency
I
1
= I
0
sin ?t 0 I
0
0
I
2
?
I
2
= I
0
sin ?t cos ?t =
0
I
sin 2 t
2
? 0
0
I
2
0
I
2 2
2 ?
I
3
= I
0
sin ?t + I
0
cos ?t 0
0
2 I I
0
?
• For above varieties of current rms =
Peak value
2
JEEMAIN.GURU
J E E - P h y s i c s
3 4
E
Example
If I = 2 t ampere then calculate average and rms values over t = 2 to 4 s
Solution
< I > =
4
3
4
2
2 2
4 4
2
2
2 t.dt
(t ) 4 2
8 2 2
3 3 (t)
dt
? ?
? ? ?
? ?
?
?
and I
rms
=
4 4
2
2 2
4
2
(2 t ) dt 4t dt
2
dt
?
? ?
?
=
4
2
2
t
2 2 3 A
2
? ?
?
? ?
? ?
Example
Find the time required for 50Hz alternating current to change its value from zero to rms value.
Solution
? I = I
0
sin ? ?t ?
0
0
I
I sin t
2
? ?
?
1
sin t
2
? ?
t
4
?
? ? ?
2
t
T 4
? ? ? ?
?
? ?
? ?
?
T
t
8
?
1
2.5
8 50
? ?
?
ms
Example
If  E = 20 sin (100 ? ?t) volt then calculate value of E at t =
1
600
s
Solution
At t =
1
600
s  E = 20 Sin
1
100
600
? ?
? ?
? ?
? ?
= 20 sin
6
? ? ?
? ?
? ?
= 20 ×
1
2
= 10V
Example
A periodic voltage wave form has been shown in fig.
Determine. (a) Frequency of the wave form.
(b) Average value.
Solution
(a) After 100 ms wave is repeated so time period is
T = 100 ms. ?   f =
1
T
= 10 Hz
(b) Average value = Area/time period  =
( / )
( )
1 2 100 10
100
? ?
= 5 volt
Example
Explain why A.C. is more dangerous than D.C. ?
Solution
There are two reasons for it :
? ? A.C. attracts while D.C. repels.
? A.C. gives a huge and sudden shock.
O
-E
E
t
+E = 311.08
0
-E = -311.08
0
0.01 sec
for 220 V ac  V
rms
= 220 V
Hence, V
0
=
rms
2.V = 12.414 × 220 = 311.08 V
Voltage change from +V
0
(positive peak) to –V
0
(negative peak) = 311.08 – (–311.08) = 622.16 V
This change takes place in half cycle i.e., in
1
100
s (for a 50 Hz A.C.)
A shock of 622.16 within 0.01 s is huge and sudden, hence fatal.
JEEMAIN.GURU
Page 4

J E E - P h y s i c s
3 2
E
ALTERNATING CURRENT
ALTERNATING CURRENT AND VOLTAGE
Voltage or current is said to be alternating if it is change continously in magnitude and perodically in direction
with time. It can be represented by a sine curve or cosine curve
I = I
0
sin ? ?t or I = I
0
cos ? ?t
where I = Instantaneous value of current at time t, I
0
= Amplitude or peak value
? ?= Angular frequency  ? ? =
2
T
?
= 2 ?f T = time period f = frequency
I
0
–I
0
T
4
T
2
3
4
T
T
t
I
I as a sine function of t

I
0
–I
0
T
4
T
2
3
4
T
T
t
I
I as a cosine function of t
AMPLITUDE OF AC
The maximum value of current in either direction is called peak value or the amplitude of current. It is represented
by I
0
. Peak to peak value = 2I
0
PERIODIC TIME
The time taken by alternating current to complete one cycle of variation is called periodic time or time period
of the current.
FREQUENCY
The number of cycle completed by an alternating current in one second is called the frequency of the current.
UNIT : cycle/s ; (Hz)
In India : f = 50 Hz , supply voltage = 220 volt In USA  :f = 60 Hz ,supply voltage = 110 volt
CONDITION REQUIRED FOR CURRENT/ VOLTAGE TO BE ALTERNATING
• Amplitude is constant • Alternate half cycle is positive and half negative
The alternating current continuously varies in magnitude and periodically reverses its direction.
I
sinosudial AC
I
t
+
–
triangular AC
+
–
I
t
square wave AC
saw tooth wave
I
t
mixture of AC and DC
I
0
t
Not AC (direction not change)
I
0
I
0
t t
Not AC (not periodic)
JEEMAIN.GURU
J E E - P h y s i c s
E
3 3
AVERAGE VALUE OR MEAN VALUE
The mean value of A.C over any half cycle (either positive or negative) is that value of DC which would send
same amount of charge through a circuit as is sent by the AC through same circuit in the same time.
average value of current for half cycle < I  > =
T / 2
0
T / 2
0
Idt
dt
?
?
Average value of I = I
0
sin ?t over the positive half cycle :
< sin > = < sin 2 >=0
< cos >= < cos 2 >= 0
< sin  cos > = 0
< sin > = < cos >=
?

?
? ?
? ?
? ?
2 2 1
2
T
2
0
0
av T
2
0
I sin t dt
I
dt
?
?
?
?
= ? ?
T
0
2
0
2 I
cos t
T
? ?
?

0
2 I
?
?
• For symmetric AC, average value over full cycle = 0,
Average value of sinusoidal AC
Full cycle (+ve) half cycle (–ve) half cycle
0
2I
0
?
–2I
0
?
As the average value of AC over a
com plete cycle is zero, it is always
defined over a half cycle which must
be either positive or negative
MAXIMUM VALUE
• I = a sin ? ? I
Max.
= a • I = a + b sin ? ? ? I
Max.
= a + b ( if a and b > 0)
• I = a sin ? + b cos ? ? ? I
Max.
=
2 2
a b ?
• I = a sin
2
? ? I
Max.
= a (a > 0)
ROOT MEAN SQUARE (rms) VALUE
It is value of DC which would produce same heat in  given resistance in given time as is done by the alternating
current when passed through the same resistance for the same time.
T
2
0
rms T
0
I dt
I
dt
?
?
?
rms value = Virtual value = Apparent value
rms value of   I = I
0
sin ?t :
T
2
0
0
rms T
0
(I sin t) dt
I
dt
?
?
?
?
=
2
T
2 0
0
I
sin t dt
T
?
?
= 0
T
0
1 1 cos2 t
I dt
T 2
? ? ? ?
? ?
? ?
?
T
0
0
1 t sin 2 t
I
T 2 2 2
? ? ?
? ?
? ?
? ?
? ?
=
0
I
2
• If nothing is mentioned then values printed in a.c circuit on electrical appliances, any given or unknown
values, reading of AC meters are assumed to be RMS.
Current Average P e a k R M S Angular fequency
I
1
= I
0
sin ?t 0 I
0
0
I
2
?
I
2
= I
0
sin ?t cos ?t =
0
I
sin 2 t
2
? 0
0
I
2
0
I
2 2
2 ?
I
3
= I
0
sin ?t + I
0
cos ?t 0
0
2 I I
0
?
• For above varieties of current rms =
Peak value
2
JEEMAIN.GURU
J E E - P h y s i c s
3 4
E
Example
If I = 2 t ampere then calculate average and rms values over t = 2 to 4 s
Solution
< I > =
4
3
4
2
2 2
4 4
2
2
2 t.dt
(t ) 4 2
8 2 2
3 3 (t)
dt
? ?
? ? ?
? ?
?
?
and I
rms
=
4 4
2
2 2
4
2
(2 t ) dt 4t dt
2
dt
?
? ?
?
=
4
2
2
t
2 2 3 A
2
? ?
?
? ?
? ?
Example
Find the time required for 50Hz alternating current to change its value from zero to rms value.
Solution
? I = I
0
sin ? ?t ?
0
0
I
I sin t
2
? ?
?
1
sin t
2
? ?
t
4
?
? ? ?
2
t
T 4
? ? ? ?
?
? ?
? ?
?
T
t
8
?
1
2.5
8 50
? ?
?
ms
Example
If  E = 20 sin (100 ? ?t) volt then calculate value of E at t =
1
600
s
Solution
At t =
1
600
s  E = 20 Sin
1
100
600
? ?
? ?
? ?
? ?
= 20 sin
6
? ? ?
? ?
? ?
= 20 ×
1
2
= 10V
Example
A periodic voltage wave form has been shown in fig.
Determine. (a) Frequency of the wave form.
(b) Average value.
Solution
(a) After 100 ms wave is repeated so time period is
T = 100 ms. ?   f =
1
T
= 10 Hz
(b) Average value = Area/time period  =
( / )
( )
1 2 100 10
100
? ?
= 5 volt
Example
Explain why A.C. is more dangerous than D.C. ?
Solution
There are two reasons for it :
? ? A.C. attracts while D.C. repels.
? A.C. gives a huge and sudden shock.
O
-E
E
t
+E = 311.08
0
-E = -311.08
0
0.01 sec
for 220 V ac  V
rms
= 220 V
Hence, V
0
=
rms
2.V = 12.414 × 220 = 311.08 V
Voltage change from +V
0
(positive peak) to –V
0
(negative peak) = 311.08 – (–311.08) = 622.16 V
This change takes place in half cycle i.e., in
1
100
s (for a 50 Hz A.C.)
A shock of 622.16 within 0.01 s is huge and sudden, hence fatal.
JEEMAIN.GURU
J E E - P h y s i c s
E
3 5
Example
If a direct current of value a ampere is superimposed on
an alternating current I = b sin ?t flowing through a wire,
what is the
effective value of the resulting current in the circuit ?
I
t
AC
b
=? I
t
DC a
+
Solution
As current at any instant in the circuit will be,
I = I
DC
+ I
AC
= a + b sin ?t
?
T T
2 2
eff
0 0
1 1
I I dt (a b sin t) dt
T T
? ? ? ?
? ?
T
2 2 2
0
1
(a 2ab sin t b sin t)dt
T
? ? ? ? ?
?
but as
T
0
1
sin tdt 0
T
? ?
?
and
T
2
0
1 1
sin tdt
T 2
? ?
?
?
2 2
eff
1
I a b
2
? ?
SOME IMPORTANT WAVE FORMS AND THEIR RMS AND AVER AGE VALUE
Nature of Wave–for m RMS Value Average or mean
wave form Val ue
Sinusoidal
0
+
–
?
2 ?
0
I
2
0
2I
?
= 0.707 I
0
= 0.637 I
0
Half wave
0
?
2 ?
0
I
2
0
I
?
rectifired = 0.5 I
0
= 0.318 I
0
Full wave
0
?
2 ?
0
I
2
0
2I
?
rectifired = 0.707 I
0
0.637 I
0
Square or
+
–
I
0
I
0
Rectangular
Saw Tooth wave
2 ? ?
0
0
I
3
0
I
2
JEEMAIN.GURU
Page 5

J E E - P h y s i c s
3 2
E
ALTERNATING CURRENT
ALTERNATING CURRENT AND VOLTAGE
Voltage or current is said to be alternating if it is change continously in magnitude and perodically in direction
with time. It can be represented by a sine curve or cosine curve
I = I
0
sin ? ?t or I = I
0
cos ? ?t
where I = Instantaneous value of current at time t, I
0
= Amplitude or peak value
? ?= Angular frequency  ? ? =
2
T
?
= 2 ?f T = time period f = frequency
I
0
–I
0
T
4
T
2
3
4
T
T
t
I
I as a sine function of t

I
0
–I
0
T
4
T
2
3
4
T
T
t
I
I as a cosine function of t
AMPLITUDE OF AC
The maximum value of current in either direction is called peak value or the amplitude of current. It is represented
by I
0
. Peak to peak value = 2I
0
PERIODIC TIME
The time taken by alternating current to complete one cycle of variation is called periodic time or time period
of the current.
FREQUENCY
The number of cycle completed by an alternating current in one second is called the frequency of the current.
UNIT : cycle/s ; (Hz)
In India : f = 50 Hz , supply voltage = 220 volt In USA  :f = 60 Hz ,supply voltage = 110 volt
CONDITION REQUIRED FOR CURRENT/ VOLTAGE TO BE ALTERNATING
• Amplitude is constant • Alternate half cycle is positive and half negative
The alternating current continuously varies in magnitude and periodically reverses its direction.
I
sinosudial AC
I
t
+
–
triangular AC
+
–
I
t
square wave AC
saw tooth wave
I
t
mixture of AC and DC
I
0
t
Not AC (direction not change)
I
0
I
0
t t
Not AC (not periodic)
JEEMAIN.GURU
J E E - P h y s i c s
E
3 3
AVERAGE VALUE OR MEAN VALUE
The mean value of A.C over any half cycle (either positive or negative) is that value of DC which would send
same amount of charge through a circuit as is sent by the AC through same circuit in the same time.
average value of current for half cycle < I  > =
T / 2
0
T / 2
0
Idt
dt
?
?
Average value of I = I
0
sin ?t over the positive half cycle :
< sin > = < sin 2 >=0
< cos >= < cos 2 >= 0
< sin  cos > = 0
< sin > = < cos >=
?

?
? ?
? ?
? ?
2 2 1
2
T
2
0
0
av T
2
0
I sin t dt
I
dt
?
?
?
?
= ? ?
T
0
2
0
2 I
cos t
T
? ?
?

0
2 I
?
?
• For symmetric AC, average value over full cycle = 0,
Average value of sinusoidal AC
Full cycle (+ve) half cycle (–ve) half cycle
0
2I
0
?
–2I
0
?
As the average value of AC over a
com plete cycle is zero, it is always
defined over a half cycle which must
be either positive or negative
MAXIMUM VALUE
• I = a sin ? ? I
Max.
= a • I = a + b sin ? ? ? I
Max.
= a + b ( if a and b > 0)
• I = a sin ? + b cos ? ? ? I
Max.
=
2 2
a b ?
• I = a sin
2
? ? I
Max.
= a (a > 0)
ROOT MEAN SQUARE (rms) VALUE
It is value of DC which would produce same heat in  given resistance in given time as is done by the alternating
current when passed through the same resistance for the same time.
T
2
0
rms T
0
I dt
I
dt
?
?
?
rms value = Virtual value = Apparent value
rms value of   I = I
0
sin ?t :
T
2
0
0
rms T
0
(I sin t) dt
I
dt
?
?
?
?
=
2
T
2 0
0
I
sin t dt
T
?
?
= 0
T
0
1 1 cos2 t
I dt
T 2
? ? ? ?
? ?
? ?
?
T
0
0
1 t sin 2 t
I
T 2 2 2
? ? ?
? ?
? ?
? ?
? ?
=
0
I
2
• If nothing is mentioned then values printed in a.c circuit on electrical appliances, any given or unknown
values, reading of AC meters are assumed to be RMS.
Current Average P e a k R M S Angular fequency
I
1
= I
0
sin ?t 0 I
0
0
I
2
?
I
2
= I
0
sin ?t cos ?t =
0
I
sin 2 t
2
? 0
0
I
2
0
I
2 2
2 ?
I
3
= I
0
sin ?t + I
0
cos ?t 0
0
2 I I
0
?
• For above varieties of current rms =
Peak value
2
JEEMAIN.GURU
J E E - P h y s i c s
3 4
E
Example
If I = 2 t ampere then calculate average and rms values over t = 2 to 4 s
Solution
< I > =
4
3
4
2
2 2
4 4
2
2
2 t.dt
(t ) 4 2
8 2 2
3 3 (t)
dt
? ?
? ? ?
? ?
?
?
and I
rms
=
4 4
2
2 2
4
2
(2 t ) dt 4t dt
2
dt
?
? ?
?
=
4
2
2
t
2 2 3 A
2
? ?
?
? ?
? ?
Example
Find the time required for 50Hz alternating current to change its value from zero to rms value.
Solution
? I = I
0
sin ? ?t ?
0
0
I
I sin t
2
? ?
?
1
sin t
2
? ?
t
4
?
? ? ?
2
t
T 4
? ? ? ?
?
? ?
? ?
?
T
t
8
?
1
2.5
8 50
? ?
?
ms
Example
If  E = 20 sin (100 ? ?t) volt then calculate value of E at t =
1
600
s
Solution
At t =
1
600
s  E = 20 Sin
1
100
600
? ?
? ?
? ?
? ?
= 20 sin
6
? ? ?
? ?
? ?
= 20 ×
1
2
= 10V
Example
A periodic voltage wave form has been shown in fig.
Determine. (a) Frequency of the wave form.
(b) Average value.
Solution
(a) After 100 ms wave is repeated so time period is
T = 100 ms. ?   f =
1
T
= 10 Hz
(b) Average value = Area/time period  =
( / )
( )
1 2 100 10
100
? ?
= 5 volt
Example
Explain why A.C. is more dangerous than D.C. ?
Solution
There are two reasons for it :
? ? A.C. attracts while D.C. repels.
? A.C. gives a huge and sudden shock.
O
-E
E
t
+E = 311.08
0
-E = -311.08
0
0.01 sec
for 220 V ac  V
rms
= 220 V
Hence, V
0
=
rms
2.V = 12.414 × 220 = 311.08 V
Voltage change from +V
0
(positive peak) to –V
0
(negative peak) = 311.08 – (–311.08) = 622.16 V
This change takes place in half cycle i.e., in
1
100
s (for a 50 Hz A.C.)
A shock of 622.16 within 0.01 s is huge and sudden, hence fatal.
JEEMAIN.GURU
J E E - P h y s i c s
E
3 5
Example
If a direct current of value a ampere is superimposed on
an alternating current I = b sin ?t flowing through a wire,
what is the
effective value of the resulting current in the circuit ?
I
t
AC
b
=? I
t
DC a
+
Solution
As current at any instant in the circuit will be,
I = I
DC
+ I
AC
= a + b sin ?t
?
T T
2 2
eff
0 0
1 1
I I dt (a b sin t) dt
T T
? ? ? ?
? ?
T
2 2 2
0
1
(a 2ab sin t b sin t)dt
T
? ? ? ? ?
?
but as
T
0
1
sin tdt 0
T
? ?
?
and
T
2
0
1 1
sin tdt
T 2
? ?
?
?
2 2
eff
1
I a b
2
? ?
SOME IMPORTANT WAVE FORMS AND THEIR RMS AND AVER AGE VALUE
Nature of Wave–for m RMS Value Average or mean
wave form Val ue
Sinusoidal
0
+
–
?
2 ?
0
I
2
0
2I
?
= 0.707 I
0
= 0.637 I
0
Half wave
0
?
2 ?
0
I
2
0
I
?
rectifired = 0.5 I
0
= 0.318 I
0
Full wave
0
?
2 ?
0
I
2
0
2I
?
rectifired = 0.707 I
0
0.637 I
0
Square or
+
–
I
0
I
0
Rectangular
Saw Tooth wave
2 ? ?
0
0
I
3
0
I
2
JEEMAIN.GURU
J E E - P h y s i c s
3 6
E
MEASUREMENT OF A.C.
Alternating current and voltages are measured by a.c. ammeter and a.c. voltmeter respectively. Working
of these instruments is based on heating effect of current, hence they are also called hot wire instruments.
Te r m s D.C. meter A.C. meter
Name moving coil hot wire
Based on magnetic effect of current heating effect of current
If used in A.C. circuit then they reads zero A.C. or D.C. then meter works
? average value of A.C. = zero properly as it measures rms value
Deflection deflection ? current deflection ? heat
? ? ? (linear) ? ? ?
?
rms
(non linear)
Scale Uniform Seperation Non uniform sepration
? = Number ? -  1  2  3  4  5 ? -  1  2  3   4   5
of divisions ? - 1  2  3  4  5 ? - 1  4  9  16  25
• D.C meter in AC circuit reads zero because < AC > = 0 ( for complete cycle)
• AC meter works in both AC and DC
PHASE AND PHASE DIFFERENCE
( a ) P h a s e
I  = I
0
sin ( ?t + ?)
Initial phase = ? (it does not change with time)
Instantaneous phase = ?t + ? (it changes with time)
• Phase decides both value and sign. ? UNIT: radian
(b ) Phase difference
Voltage V = V
0
sin ( ?t + ?
1
) Current I ? ? ? ? I
0
sin  ( ?t + ?
2
)
• Phase difference of I w.r.t. V ? ?= ?
2
–  ?
1
• Phase difference of V w.r.t. I ? = ? ?
1
– ?
2
( a ) V leads I or I lags V ? It means, V reach maximum before I
I=I
0
? sin ( t ) ? ? ?
?t
V ,I
V=Vsin t
0
?
Let if V = V
0
sin ?t then I = I
0
sin ( ?t – ? ?
and if V = V
0
sin ( ?t+ ? ) then I = I
0
sin ?t
(b ) V lags I or I leads V ? ?It means V reach maximum after I
I=I
0
? sin ( t + ) ? ?
?t
V ,I
V=V sin t
0
?
Let if V = V
0
sin ?t then I = I
0
sin ( ?t + ? ?
and if V = V
0
sin ( ?t – ? ) then I = I
0
sin ?t
PHASOR AND PHASOR DIAGRAM
A diagram representing alternating current and voltage (of same frequency) as vectors (phasor) with the phase
angle between them is called phasor diagram.
?
?t
V
V
0 I
0
I
?
V
0
I
0
fig (a) fig (b)
Y
X
Y
X
Let V = V
0
sin ?t and I  = I
0
sin ( ?t + ?)
In figure (a) two arrows represents phasors. The length of phasors
represents the maximum value of quantity. The projection of a phasor
on y-axis represents the instantaneous value of quantity
• A.C. is cheaper than D.C
• It can be easily converted into D.C. (by rectifier)
• It can be controlled easily (choke coil)
• It can be transmitted over long distance at negligible power loss.
• It can be stepped up or stepped down with the help of transformer.
JEEMAIN.GURU
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