JEE-Physics
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 = I0 sin  t
or
I = I0 cos  t
where I = Instantaneous value of current at time t,
I 0 = Amplitude or peak value
 = Angular frequency  =
I0
2
= 2f
T
T = time period
I
I0
T 3T
2 4
I
T
2
T
T
4
f = frequency
T
3T
4
T
4
t
t
–I0
–I0
I as a sine function of t
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.
triangular AC
I
+
+
t
–
I
square wave AC
saw tooth wave
I
t
I0
–
t
I0
I0
t
t
t
mixture of AC and DC
Not AC (direction not change)
Not AC (not periodic)
32
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sinosudial AC
I
E
JEE-Physics
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.
T /2
 Idt
average value of current for half cycle < I > =
0
T /2
 dt
0
Average value of I = I 0 sin t over the positive half cycle :
T
2
I av 
 I sin t dt
0
0
T
2
=
 dt
< sin  > = < sin 2 >=0
< cos  >= < cos 2 >= 0
< sin  cos  > = 0
2
2
< sin  > = < cos  >= 1
2
T
2 I0
2I
 cos t 02  0

T

0
•
For symmetric AC, average value over full cycle = 0,
Average value of sinusoidal AC
Full cycle
(+ve) half cycle
(–ve) half cycle
0
2I0

–2I0

MAXIMUM VALUE
•
I = a sin
•
I Max. = a

I = a sin + b cos  
I Max. =
a 2  b2
As the average value of AC over a
complete cycle is zero, it is always
defined over a half cycle which must
be either positive or negative
•
I = a + b sin 
I Max. = a + b ( if a and b > 0)
•
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
I rms
2
 I dt

 dt
0
rms value = Virtual value = Apparent value
T
0
rms value of
I = I 0 sin t :
I rms 
 (I sin t) dt
=
 dt
T
0
2
0
T
0
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•
E
I 20 T 2
sin t dt = I
T 0
0
If nothing is mentioned then values printed in a.c circuit on electrical appliances, any given
values, reading of AC meters are assumed to be RMS.
Current
Average
Peak
I 1 = I 0 sin t
0
I0
I2 = I 0sin t cos t =
I0
sin 2 t
2
I 3 = I 0sint + I 0cost
For above varieties of current rms =
RMS
I0
I0
I0
2
2 2
0
2 I0
I0
2
33
or
Angular fequency

2
0
Peak value
•
T
1 T  1  cos 2 t 
1  t sin 2 t 
I

dt  I 0
= 0





0
T 
2
T  2 2  2 0

2
2

unknown
JEE-Physics
Example
t ampere then calculate average and rms values over t = 2 to 4 s
If I = 2
Solution
4
< I > =
 2 t.dt
2
4
 dt
4
3
4 (t 2 )24 2 

 8  2 2 
3 (t)24
3
and I rms =
4
2
2
4
 (2 t ) dt   4t dt = 2  t   2 3 A
2
2 
 dt
2
2
4
2
2
2
Example
Find the time required for 50Hz alternating current to change its value from zero to rms value.
Solution
 I = I0 sin  t 
I0
2
 I 0 sin t  sin t 
1
t 
2

 2 

t 
 

4
4
 T 
 t
T
1

 2.5 ms
8 8  50
Example
If E = 20 sin (100 t) volt then calculate value of E at t =
1
s
600
Solution
At t =
1 


1
s E = 20 Sin 100  
= 20 sin  

600 
600

6 
= 20 ×
1
= 10V
2
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
1
T = 100 ms.  f =
= 10 Hz
T
(b)
Average value = Area/time period =
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.
for 220 V ac V rms = 220 V
Hence, V 0 =
2.Vrms
E
O
-E
+E0= 311.08
t
0.01 sec
-E0= -311.08
= 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
s (for a 50 Hz A.C.)
100
A shock of 622.16 within 0.01 s is huge and sudden, hence fatal.
34
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(1 / 2)  100  10
= 5 volt
(100 )
E
JEE-Physics
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
a
I
effective value of the resulting current in the circuit ?
AC
DC
t
b
+I
t
Solution
As current at any instant in the circuit will be,
I = I DC + I AC = a + b sint
T
1 2
I dt 
T 0

I eff 
but as
1
sin tdt  0
T 0
T
1
(a  b sin t)2 dt 
T 0
T
1
(a 2  2ab sin t  b 2 sin 2 t)dt
T 0
T
T
1
1
sin 2 tdt 

T0
2
and

I eff  a 2 
1 2
b
2
SOME IMPORTANT WAVE FORMS AND THEIR RMS AND AVER AGE VALUE
Nature of
Wa ve – f o r m
RMS Value
wave form
Va l u e
I0
Sinusoidal
0
+
2

–
= 0.707 I 0
= 0.637 I 0
I0
2
I0

rectifired
= 0.5 I 0
= 0.318 I 0
Full wave
I0
2I 0

0

2
2
0

rectifired
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2I 0

2
Half wave
E
Average or mean
Square or
+
Rectangular
Saw Tooth wave
2
= 0.707 I 0
0.637 I 0
I0
I0
I0
I0
2
–
0
 2
3
35
=?
JEE-Physics
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
Name
Based on
Reads
If used in
Deflection
Scale
 = Number
of divisions
•
•
D.C. meter
moving coil
magnetic effect of current
average value
A.C. circuit then they reads zero
 average value of A.C. = zero
deflection  current
   (linear)
Uniform Seperation
 - 1 2 3 4 5
 - 1 2 3 4 5
A.C. meter
hot wire
heating effect of current
r.m.s. value
A.C. or D.C. then meter works
properly as it measures rms value
deflection  heat
   rms (non linear)
Non uniform sepration
 - 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)
Phase
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  I0 sin (t +2)
•
Phase difference of I w.r.t. V
 =  2 –  1
•
Phase difference of V w.r.t. I
 =  1 – 2
LAGGING AND LEADING CONCEPT
(a)
(b)
I=I0 sin (t )
V leads I or I lags V  It means, V reach maximum before I
Let if
V = V 0 sin t
then I = I0 sin (t – 
and if
V = V 0 sin (t+ )
then I = I 0 sin t
t
V,I
V=V0sint
I=I0 sin (t +)
V lags I or I leads V  It means V reach maximum after I
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
t
V,I
V=V0sint
angle between them is called phasor diagram.
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
Y
V
I
Y
I0

t
fig (a)
ADVANTAGES OF AC
•
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.
36
I0
V0

X
fig (b)
V0
X
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PHASOR AND PHASOR DIAGRAM
A diagram representing alternating current and voltage (of same frequency) as vectors (phasor) with the phase
E
JEE-Physics
•
GOLDEN KEY POINTS
AC can't be used in
(a) Charging of battery or capacitor (as its average value = 0)
(b) Electrolysis and electroplating (Due to large inertia, ions can not follow frequency of A.C)
Minimum, at that instant when they are near their peak values
•
The rate of change of A.C.
•
•
For alternating current I0 > I rms > I av.
•
•
•
Maximum, at that instant when they change their direction.
+
Average value over half cycle is zero if one quarter is positive
–
and the other quarter is negative.
Average value of symmetrical AC for a cycle is zero thats why average potential difference on any
element in A.C circuit is zero.
The instrument based on heating effect of current are works on both A.C and D.C supply and also provides
same heating for same value of A.C (rms) and D.C. that's why a bulb bright equally in D.C. and A.C. of same
value.
If the frequency of AC is f then it becomes zero 2f times in one second and the direction of current changes
2f times in one second. Also it become maximum 2f times in one second.
Example


The Equation of current in AC circuit is I = 4sin 100  t   A. Calculate.
3


(i) RMS Value
(ii) Peak Value
(iii) Frequency
(iv) Initial phase
Solution
(i)
I rms =
I0
2
4
=
= 2 2A
(ii)
2
(iii)

(iv)
Initial phase =
(v) Current at t = 0
Peak value I 0 = 4A
100 
= 50 Hz
2
 = 100  rad/s

frequency f =

3
(v)


At t = 0, I = 4sin 100   0   = 4 ×
3

3
=2 3 A
2
Example


E = E 0 cos t   . Calculate phase difference between E and I
3

If I = I0 sin t,
Solution


E = E 0sin   t  
3
2
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I = I 0 sin t and
E

phase difference =


5
+
=
2
3
6
Example
If E = 500 sin (100 t) volt then calculate time taken to reach from zero to maximum.
Solution
 = 100 T =
2
1
T
1
=
s, time taken to reach from zero to maximum =
=
s
100 
50
4
200
Example
If Phase Difference between E and I is

and f = 50 Hz then calculate time difference.
4
Solution

2  T
Phase difference time difference

 Time difference =
2
T
37
T

T
1
×
=
=
= 2.5ms
2
4
8
50  8
JEE-Physics
Example
Show that average heat produced during a cycle of AC is same as produced by DC with I = I rms.
Solution
For AC, I = I 0 sint, the instantaneous value of heat produced (per second) in a resistance R is,
H = I 2R = I 02sin 2t × R
H av
T
T
0
0
2
0
the average value of heat produced during a cycle is :
2
 H dt   (I sin t  R )dt  1 I R

2
 dt
 dt
T
T
0
0
2
0
However, in case of DC, H DC = I 2 R...(ii)
2
T
1 2 

 I0 
2
2
R  I 2rms R .....(i)
 0 I 0 sin t dt  2 I 0 T   H av  


 2 
 I = I rms so from equation (i) and (ii) H DC = H av
AC produces same heating effects as DC of value I = Irms. This is also why AC instruments which are based
on heating effect of current give rms value.
DIFFERENT TYPES OF AC CIRCUITS
In order to study the behaviour of A.C. circuits we classify them into two categories :
(a) Simple circuits containing only one basic element i.e. resistor (R) or inductor (L) or capacitor (C) only.
(b) Complicated circuit containing any two of the three circuit elements R, L and C or all of three elements.
AC CIRCUIT CONTAINING PURE RESISTANCE
R
Let at any instant t the current in the circuit = 
V =IR
Potential difference across the resistance =  R.
s
with the help of kirchoff’s circuital law E –  R = 0
E = E0sint
E 0 sin t =  R   
E0
E
sin t   0 sin t (  0  0
R
R
= peak or maximum value of current)
Alternating current developed in a pure resistance is also of sinusoidal
E = E0 sint
E and I
I=I0 sint
nature. In an a.c. circuits containing pure resistance, the voltage and
current are in the same phase. The vector or phasor diagram which
O
represents the phase relationship between alternating current and al-
t
3/2

2
/2
X
ternating e.m.f. as shown in figure.
In the a.c. circuit having R only, as current and voltage are in the same
phase, hence in fig. both phasors E0 and 0 are in the same direction,
E
instantaneous values of alternating current and voltage.
i.e.
 =  0 sint and
Since
0
E
E
 0
 0  0 , hence
2 R 2
R
I
E = E 0 sint.

t
P
E0
I0
t
E
 rms  rms
R
X
O
AC CIRCUIT CONTAINING PURE INDUCTANCE
L
A circuit containing a pure inductance L (having zero ohmic resistance)
connected with a source of alternating emf. Let the alternating e.m.f. E = E0 sin t
38
d
dt
s
When a.c. flows through the circuit, emf induced across inductance  L
E = E0sin t
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Y
making an angle t with OX. Their projections on Y-axis represent the
E
JEE-Physics
Negative sign indicates that induced emf acts in opposite direction to that of applied emf.
Because there is no other circuit element present in the circuit other then inductance so with the help of
Maximum current
0 
d 

E   L   0

dt 
 EL
E0
E
 1  0 , Hence,
L
L
E
d

so we get   0 sin  t  

L
2
dt


  0 sin  t  

2
In a pure inductive circuit current always lags behind the emf by
or alternating emf leads the a. c. by a phase angle of

.
2
E and I
Kirchoff’s circuital law
O

.
2
E = E0 sint
/2
3/2

t
2
I = I0 sin (t -/2)
E
E0
resembles the expression
R.

L
This non-resistive opposition to the flow of A.C. in a circuit is called the
inductive reactance (XL) of the circiut.
Expression  0 
Y
PE
E
0
X L = L = 2  f L where f = frequency of A.C.
t
O
Unit of X L : ohm
L) = Unit of L × Unit of  = henry × sec –1

0
Q
E = E0 sint
I=I0 cos t
Volt
Volt

 sec 1 
 ohm
Ampere / sec
Ampere
Inductive reactance X L  f
Higher the frequency of A.C. higher is the inductive reactance offered by an inductor
in an A.C. circuit.
For d.c. circuit, f = 0  X L = L = 2  f L = 0
Hence, inductor offers no opposition to the flow of d.c. whereas
X
/2 – t
XL
a resistive path to a.c.
f
AC CIRCUIT CONTAINING PURE CAPACITANCE
C
A circuit containing an ideal capacitor of capacitance C connected with a source of
alternating emf as shown in fig. The alternating e.m.f. in the circuit E = E 0 sin t
q = C E
q
C
E and 
E = E0 sint
E = E0sint
/2
3/2 t
2

O
 q = CE 0 sin t
The instantaneous value of current  
dq
d
  CE 0 sin t   CE 0  cos t
dt dt
I= I0sin(t + /2)
Y
P
E
E0


sin  t    0 sin  t    where I 0 = CV 0

2
1 / C  
2 

In a pure capacitive circuit, the current always leads the e.m.f. by a phase
angle of . The alternating emf lags behinds the alternating current by a
phase angle of .
39
Q
I0
E0
I
°
 
90
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E
Instantaneous potential difference across the capacitor E =
s
When alternating e.m.f. is applied across the capacitor a similarly varying alternating current
flows in the circuit.
The two plates of the capacitor become alternately positively and negatively charged
and the magnitude of the charge on the plates of the capacitor varies sinusoidally
with time. Also the electric field between the plates of the capacitor varies sinusoidally
with time.Let at any instant t charge on the capacitor = q
O
t
X
JEE-Physics
IMPORTANT POINTS
f
E
is the resistance R when both E and  are in phase, in present case they

XC

1
, hence
is not the resistance of the capacitor,,
2
C
the capacitor offer opposition to the flow of A.C. This non-resistive opposition
differ in phase by
1
1

C 2 fC
to the flow of A.C. in a pure capacitive circuit is known as capacitive reactance X C. X C 
Unit of X C : ohm
Capacitive reactance XC is inversely proportional to frequence of A.C. XC decreases as the frequency increases.
This is because with an increase in frequency, the capacitor charges and discharges rapidly following the flow
of current.
1
  but has a very small value for a.c.
For d.c. circuit f = 0
 X C 
2 fC
This shows that capacitor blocks the flow of d.c. but provides an easy path for a.c.
INDIVIDUAL COMPONENTS (R or L or C)
R
L
TERM
C
L
R
C
Circuit
Supply Voltage
V = V 0sin t
V = V 0 sin t
V = V 0 sin t
Current
I = I 0 sin t
Peak Current
I0 
Impedance ( )
V0
R
I0

I = I0 sin (t – )
2
V0
I0 
L
V0
 L  X L
I0
R = Resistance
X L=Inductive reactance.
zero (in same phase)

Z
V0 Vrms

I0
I rms
Phase difference
V0
R

(V leads I)
2
I = I 0 sin (t+

)
2
V0
=V 0 C
1 C
V0
1

 XC
I0
C
I0 
X C =Capacitive reactance.


(V lags I)
2
V
I
I
V
I
R
R does not depend on 
G=1/R = conductance.
Same in
A C and D C
V R = IR
XC  1
f
f
f
Ohm's law
I
XL f
Variation of Z with f
G,S L ,S C
(mho, seiman)
Behaviour of device
in D.C. and A.C
V
XL
V
S L = 1/X L
Inductive susceptance
L passes DC easily
(because X L = 0) while
gives a high impedance
for the A.C. of high
S C = 1/X C
Capacitive susceptance
C - blocks DC
(because X C =) while
provides an easy path
for the A.C. of high
frequency (X L f)
1

frequency  X C  
f

V C = IX C
V L = IX L
40
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Phasor diagram
E
JEE-Physics
•
•
•
GOLDEN KEY POINTS
Phase diference between capacitive and inductive reactance is 
Inductor called Low pass filter because it allow to pass low frequency signal.
Capacitor is called high pass filter because it allow to pass high frequency signal.
Example
What is the inductive reactance of a coil if the current through it is 20 mA and voltage across it is 100 V.
Solution
  VL = IXL
X L =
VL
100

= 5 k
I
20  10 3
Example
The reactance of capacitor is 20 ohm. What does it mean?
What will be its reactance if frequency of AC is doubled?
What will be its, reactance when connected in DC circuit? What is its consequence?
Solution
The reactance of capacitor is 20 ohm. It means that the hindrance offered by it to the flow of AC at a specific
1
1

C 2 fC
Therefore by doubling frequency, the reactance is halved i.e., it becomes 10 ohm. In DC circuit f = 0. Therefore
reactance of capacitor =  (infinite). Hence the capacitor can not be used to control DC.
Example
A capacitor of 50 pF is connected to an a.c. source of frequency 1kHz Calculate its reactance.
Solution
frequency is equivalent to a resistance of 20 ohm. The reactance of capacitance X C 
XC =
1
1
10 7
=
=

C
2   10 3  50  10 12

L
Example
A
In given circuit applied voltage V = 50 2 sin 100t volt and
ammeter reading is 2A then calculate value of L
Solution
V rms = I rms X L
 Reading of ammeter = I rms
XL =
Vrms
=
I rms
V0
2 I rms
=
50 2
2 2
= 25  L =
XL
25
=

100
V
=
1
H
4
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Example
A 50 W, 100 V lamp is to be connected to an AC mains of 200 V, 50 Hz. What capacitance is essential to
be put in series with the lamp ?
Solution
E
 resistance of the lamp R 
V 100 1
Vs2 (100)2
= 200  and the maximum curent I  
 A

R 200 2
W
50
 when the lamp is put in series with a capacitance and run at 200 V AC, from V = IZ
Z
V 200

 400 
1
I
2
Now as in case of C–R circuit Z  R 2 
1
,
( C)2
1
1
1
 (400) 2 
 16  10 4  (200)2  12  10 4 
 12  10 2
2
2
( C )
( C )
C
1
100
F 
F  9.2 F
C 
2
100   12  10
 12
2
R 
41
JEE-Physics
RESISTANCE AND INDUCTANCE IN SERIES (R-L CIRCUIT)
A circuit containing a series combination of a resistance R and an inductance L, connected with a source
of alternating e.m.f. E as shown in figure.
L
R
s
E = E0 sint
PHASOR DIAGRAM FOR L-R CIRCUIT
Let in a L-R series circuit, applied alternating emf is E = E 0 sint. As R
and L are joined in series, hence current flowing through both will be same
at each instant. Let  be the current in the circuit at any instant and V L and
V R the potential differences across L and R respectively at that instant.
Then V L = X L
and
V R = R
Now, V R is in phase with the current while V L leads the current by
Y
R
Q
E
VL

.
2
X
P
VR
So V R and V L are mutually perpendicular (Note : E  VR + V L )
The vector OP represents V R (which is in phase with ), while OQ represents V L (which leads  by 90°).
The resultant of V R and V L = the magnitude of vector OR E  VR2  VL2
Thus
E² = V R² + V L² = ² (R² + X L ²)   
E
2
R  X 2L
XL
ZL
The phasor diagram shown in fig. also shows that in L-R circuit the applied
emf E leads the current  or conversely the current  lags behind the e.m.f.
E. by a phase angle  tan  
VL X L X L L
 L 
      tan 1 



 R 
VR
R
R
R

R
Inductive Impedance Z L :
In L-R circuit the maximum value of current  0 
E0
2
Here
2
2
R 2  2 L2 represents the effective
R  L
opposition offered by L-R circuit to the flow of a.c. through it. It is known as impedance of L-R circuit and
is represented by Z L. Z L  R 2   2 L2
1

ZL
1
C
2
R  2 L2
R
PHASOR DIAGRAM FOR R-C CIRCUIT
Current through both the resistance and capacitor will be same at every instant
and the instantaneous potential differences across C and R are
O
V C =  X C and V R =  R
where X C = capacitive reactance and  = instantaneous current.
VC
Now, V R is in phase with , while V C lags behind  by 90°.
The phasor diagram is shown in fig.
Q
The vector OP represents V R (which is in phase with )

and the vector OQ represents V C (which lags behind  by
).
2
The vector OS represents the resultant of V R and
V C = the applied e.m.f. E.
42
s
RESISTANCE AND CAPACITOR IN SERIES (R-C CIRCUIT)
A circuit containing a series combination of a resistance R and a capacitor
C, connected with a source of e.m.f. of peak value E 0 as shown in fig.
E = E0 sint
P
VR

E=
VR 2
+V
C
2
S
E (applied emf)
X
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YL 
 R 2  (2 fL ) 2 The reciprocal of impedance is called admittance
E
JEE-Physics
 E  VR2  VC2
V R2 + V C2 = E 2
Hence
P

Z=
E
 E² = ² (R² + X C²)   
The term
O
2
R X
R
R2
+X
2
C
C
(R 2  X 2C ) represents the effective resistance of the R-C
Q
X
2
XC
S
circuit and called the capacitive impedance Z C of the circuit.
 1 
Z C  R 2  X 2C  R 2  
 C 
Hence, in C-R circuit
2
Capacitive Impedance Z C :
R 2  X 2C effective opposition offered by R-C circuit to the flow of a.c. through
In R-C circuit the term
it. It is known as impedance of R-C circuit and is represented by Z C
The phasor diagram also shows that in R-C circuit the applied e.m.f. lags behind the current 
(or the current  leads the emf E) by a phase angle  given by
tan  
VC X C 1 / C
1
X
1
1 
1 



tan   C 
   tan 
VR
R
R
CR ,
 CR 
R
CR
COMBINATION OF COMPONENTS (R-L or R-C or L-C)
TERM
R-L
R-C
L
R
L-C
R
L
C
C
Circuit
I is same in R & L
I is same in R & C
VL
VR
I is same in L & C
VL
I
I
VR
V 2 = VR2  VL2
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Phase difference
E
V leads I ( = 0 to
R 2  X C 
Variation of Z
with f
2

)
2
VC
VC
V 2 = VR2  VC2
V = V L – V C (V L >V C )
V = V C – V L (V C >V L )
V lags I ( = –

to 0 )
2
V lags I ( = 

,if X L>X C )
2
Impedance
V leads I ( = 
in between V & I
Z =
I
V
Phasor diagram

,if X C>X L )
2
Z  R 2  X 2L
Z = X L  Xc
as f,Z 
as f,
as f, Z first  then
Z
Z
Z
R
R
f
f
f
At very low f
Z ~ R (XL  0)
Z ~ XC
Z ~ XC
At very high f
Z ~ XL
Z ~ R (XC  0)
Z ~ XL
43
JEE-Physics
Example
30
Calculate the impedance of the circuit shown in the figure.
40
Solution
R 2  (X c ) 2 =
Z =
(30)2  (40)2 =
2500 = 50 
Example
If XL = 50  and X C = 40  Calculate effective value of current in given circuit.
Solution
XL=50
Z = X L – X C = 10 
 =
V0
40
 =
= 4 A
Z
10

I rms =
4
2
XC=40
= 2 2 A
V=40sin 100 volt
Example
VR=60
In given circuit calculate, voltage across inductor
VL=?
Solution
 V 2  VR2  VL2

V L2 = V 2 – V R2
2
2
V=100 2 sint volt
V 2  VR2 =
VL
(100)  (60)
Example
In given circuit find out
Solution
=
6400 = 80 V
(i) impedance of circuit
(ii) current in circuit
8
6
R 2  X 2C  (6)2  (8)2 = 10 
(i)
Z =
(ii)
V = IZ

I
V0 20

 2 A so I rms =
Z
10
2
=
2
V=20sint volt
2 A
Example
When 10V, DC is applied across a coil current through it is 2.5 A, if 10V, 50 Hz A.C. is applied current reduces
to 2 A. Calculate reactance of the coil.
Solution
10
V 20
 4  For 10 V A.C.  V = I Z  Z  
 5
I 10
2.5
For 10 V D.C. V = IR  Resistance of coil R =
 Z 
R 2  X 2L  5  R 2  X 2L  25  X 2L = 52 – 42 XL= 3 
Example
circuit and is in phase with the applied voltage. When the same voltage is applied across another device
Y, the same current again flows through the circuit but it leads the applied voltage by /2 radians.
(a)
Name the devices X and Y.
(b)
Calculate the current flowing in the circuit when same voltage is applied across the series
combination of X and Y.
Solution
(a)
(b)
X is resistor and Y is a capacitor
Since the current in the two devices is the same (0.5A at 220 volt)
When R and C are in series across the same voltage then
R = XC =
220
= 440   I rms =
0.5
Vrms
R
2
 X 2C
44
=
220
2
(440 )  (440 )
2
=
220
440 2
= 0.35A
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When an alternating voltage of 220V is applied across a device X, a current of 0.5 A flows through the
E
JEE-Physics
INDUCTANCE, CAPACITANCE AND RESISTANCE IN SERIES
L
(L-C-R SERIES CIRCUIT)
C
R
A circuit containing a series combination of an resistance R, a coil of inductance
L and a capacitor of capacitance C, connected with a source of alternating e.m.f.
of peak value of E 0, as shown in fig.
A.C. source
E = E0 sint
PHASOR DIAGRAM FOR SERIES L-C-R CIRCUIT
Let in series LCR circuit applied alternating emf is E = E 0 sin t.
As L,C and R are joined in series, therefore, current at any instant
through the three elements has the same amplitude and phase.
However voltage across each element bears a different
Y
VL Q
phase relationship with the current.
Let at any instant of time t the current in the circuit is 
Let at this time t the potential differences across L, C, and R
O
V L =  X L, V C =  X C and V R =  R
X
VR P
VC
Now, V R is in phase with current  but V L leads  by 90°
While V C legs behind  by 90°.
The vector OP represents V R (which is in phase with ) the vector
OQ represent VL (which leads  by 90°)
Y
VL Q
T
and the vector OS represents V C (which legs behind  by 90°)
If VL > VC (as shown in figure) the their resultant will be (VL – VC) which
is represented by OT.
Z  R 2  (X L  X C )2 =
1 

R   L 

C 
P
X
VC
E
R  (X L  X C ) 2
Y
XL Q
T
2
The phasor diagram also shown that in LCR circuit the applied e.m.f.
45
VR
2
2
XL  XC
leads the current by a phase angle tan =
R
ed
pli
ap
f)
em

K
(XL-XC )
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E
Impedance
 
E(
O
Finally, the vector OK represents the resultant of VR and (VL – VC), that
is, the resultant of all the three = applied e.m.f.
Thus E  VR2  (VL  VC ) 2 = I R 2  (X L  X C )2
K
(VL-VC)
V L and V C are opposite to each other.
O
XC
Z

R
P
X
JEE-Physics
SERIES LCR AND PAR ALLEL LCR COMBINATION
SERIES L-C-R CIRCUIT
1.
Circuit
PARALLEL L-C-R CIRCUIT
diagram
R
R
L
L
C
C
V same for R,L and C
V same for R, L & C
VL
IC
I
V
VR
IR
VC
(i)
IL
If V L > V C then
VL–VC
(i)
if I C > I L then
IC-IL
I
V
VR
(ii)
VR
VC–VL
(iii)
IR
If V C > V L then
(ii)
I
VR2  (VL  VC ) 2
V =
Impedance Z =
(iii)
R 2  (X L  X C )2
IL-IC
IR
I =
I2R  (I L  I C ) 2
XL  XC
V  VC
= L
R
VR
Impedance triangle

G 2  (S L  S C ) 2
IL  IC
SL  SC
=
IR
G
Admittance triangle
tan =
(iv)
G

Z
V
Admittance Y =
tan =
(iv)
if I L > I C then
X=XL–XC
SL-SC
Y
R
GOLDEN KEY POINTS
•
(a)
(b)
Series
if X L > X C then V leads I, (positive)
circuit nature inductive
if X C> XL then V lags I, (negative)
circuit nature capacitive
(a)
(b)
Parallel
if S L > S C (X L < X C ) then V leads I, (positive)
circuit nature inductive
if S C> S L ( X C < X L) then V lags I,  (negative)
circuit nature capacitive
•
In A.C. circuit voltage for L or C may be greater than source voltage or current but it happens
only when circuit contains L and C both and on R it never greater than source voltage or current.
•
In parallel A.C.circuit phase difference between I L and I C is 
46
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2.
I same for R, L & C
Phasor diagram
E
JEE-Physics
Example
Find out the impedance of given circuit.
9
4
6
Solution
Z  R 2  (X L  X C ) 2
 4 2  (9  6)2 =
4 2  3 2  25  5 
( X L > XC  Inductive)
Example
Find out impedance of given circuit.
3
Solution
1 1 1 


Y = G + (S L – S C )
36  6 3 
2
2
Y =
6
2
2
2
 
6
6
6
Z =
2
  (capacitive, because X L > X C)
Example
Find out reading of A C ammeter and also calculate the potential difference across, resistance and capacitor.
Solution
Z  R 2  (X L  X C ) 2  10 2 

I0 

ammeter reads RMS value, so its reading =
so
V R = 5 × 10 = 50 V
V0
100
10


A
Z
10 2
2
10
A
10
2
20
10
2
= 5A
E=100sint100t volt
and
V C = 5 × 10 = 50 V
Example
In LCR circuit with an AC source R = 300  , C = 20 F, L = 1.0 H, Erms = 50V and f = 50/ Hz. Find RMS
current in the circuit.
Solution
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I rms 
E

E rms

Z
E rms
1 

R 2  L 
C 

2
50

50
1


300 2  2  
1 
50 

6
20  10  2  


 

50
I rms 
3

10 
(300) 2  100 
2 

2

50
100 9  16

2
1
 0.1A
10
Example
Calculate impedance of the given circuit :
5V
12V 4V
0.2F
240
Z
120
(i)
I=2A
(ii)
Z=?
1H
3
(iii)
V=100sint volt
47
JEE-Physics
Solution
(i)
It is parallel circuit so Y is evaluated
Y  SL  SC 
1
1
1


120 240
240
V s = 13 volt therefore Z 
(ii)
V s 2 = 5 2 + 12 2 = 169
(iii)
R = 3, X L = L = 1 as ( = 1)

1
1
=
= 5
C
(0.2).1
XC =
Z = 240  (inductive)

so
Vs 13

 6.5 
I
2
Z 2 = R 2 + (X L – X C) 2 = 3 2 + (1 – 5) 2 = 25  Z = 5 
RESONANCE
A circuit is said to be resonant when the natural frequency of circuit is equal to frequency of the applied voltage.
For resonance both L and C must be present in circuit.
There are two types of resonance :
SERIES RESONANCE
(a)
At Resonance
(i)
XL = XC
(iv)
(b)
(ii) Parallel Resonance
VL = VC
(iii)
 = 0 (V and I in same phase)
(impedance minimum)
(v)
I max =
V
(current maximum)
R
Resonance frequency

(c)
Z min = R
(ii)
(i) Series Resonance
1
1
2
 rL =  C   r 

LC
r
XL = XC
r 
1
LC
 fr 
1
2  LC
Variat ion of Z w it h f
(i)
If f< f r then X L < X C
circuit nature capacitive,  (negative)
(ii)
At f = f r then X L = X C
circuit nature, Resistive,  = zero
(iii)
If f > f r then X L > X C
circuit nature is inductive, (positive)
Variation of I with f as f increase, Z first decreases then increase
Z
R
fr
f
Imax
Imax
2
I
f1
•
as f increase, I first increase then decreases
f
fr
f2
f
At resonance impedance of the series resonant circuit is minimum so it is called 'acceptor circuit' as it most readily
accepts that current out of many currents whose frequency is equal to its natural frequency. In radio or TV tuning
we receive the desired station by making the frequency of the circuit equal to that of the desired station.
Half power frequencies
The frequencies at which, power become half of its maximum value called half power frequencies
Band width  = f = f 2 – f 1
Quality factor Q    Q-factor of AC circuit basically gives an idea about stored energy & lost energy.
Q  2
maximum energy stored per cycle
max imum energy loss per cycle
(i) It represents the sharpness of resonance. (ii) It is unit less and dimension less quantity
(iii) Q =
(X L ) r
R
=
(X C ) r
2 fr L
1
=
=
R
R
R
fr
fr
L
=
=
C
f
band width
48
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(d)
I max= V
R
E
JEE-Physics
Magnification
At resonance
V L or V C = QE (where E = supplied voltage)
So at resonance
Magnification factor = Q-factor
I
R1
R1 < R2 < R3
Sharpness
R2
Sharpness  Quality factor  Magnification factor
R decrease   Q increases   Sharpness increases
R3
fr
PARALLEL RESONANCE
(a)
At resonance
(i) S L = S C (ii) IL = I C
(iii)   = 0
L
(iv) Z max = R (impedance maximum)
R
V
(current minimum)
R
(v) I min =
(b)
Resonant frequency f r =
(c)
Z
1
2  LC
Variation of Z with f as f increases , Z first increases then decreases

If f < f r thenS L > S C,  (positive), circuit nature is inductive

If f > f r then S C > S L,  (negative), circuit nature capacitive.
fr
(d)
f
C
I
Imin= V
R
Variat ion of I w it h f
as f increases , I first decreases then increases
Note : For this circuit fr 
fr
L
1
R2
1
1
R2
 2
 2  Z max 
For resonance
RC
LC L
2  LC L
f
GOLDEN KEY POINTS
• Series resonance circuit gives voltage amplificaltion while parallel resonance circuit gives current amplification.
• At resonance current does not depend on L and C, it depends only on R and V.
• At half power frequencies :
• As R increases ,
net reactance = net resistance.
bandwidth increases
• To obtain resonance in a circuit following parameter can be altered :
(i) L
(ii) C
(iii) frequency of source.
• Two series LCR circuit of same resonance frequency f are joined in series then resonance frequency of series
combination is also f
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• The series resonance circuit called acceptor whereas parallel resonance circuit called rejector circuit.
E
• Unit of
LC is second
Example
For what frequency the voltage across the resistance R will be maximum.
Solution
R
It happens at resonance
f
1
2  LC

1
1
1
2
 10 6 


 500 Hz
49
1
 F
1
H
JEE-Physics
Example
R=220
A capacitor, a resistor and a 40 mH inductor are connected in series to an
AC source of frequency 60Hz, calculate the capacitance of the capacitor, if
the current is in phase with the voltage. Also calculate the value of X and I.
V1
V2
V3
300V
300V
XV
I
110V, 60Hz
Solution
At resonance
L 
1
1
1
1

 176 F
, C 2 
2 2
2
2
4   (60)  40  10 3
 L 4 f L
C
V = VR

and I 
X = 110 V
V 110

 0.5 A
R 220
Example
A coil, a capacitor and an A.C. source of rms voltage 24 V are connected in series, By varying the frequency
of the source, a maximum rms current 6 A is observed, If this coil is connected to a bettery of emf 12 V, and
internal resistance 4 , then calculate the current through the coil.
Solution
At resonance current is maximum. I =
V
V
24
   Resistance of coil R =
=
= 4 
R
I
6
When coil is connected to battery, suppose I current flow through it then I =
12
E
=
44
Rr
= 1.5 A
Example
Radio receiver recives a message at 300m band, If the available inductance is 1 mH, then calculate required
capacitance
Solution
Radio recives EM waves. ( velocity of EM waves c = 3 x 10 8 m/s)

c = f  f =
3  10 8
= 10 6 Hz
300
Now f =
1
2  LC
= 1 × 10 6 C =
1
= 25 pF
4  L  10 12
2
Example
In a L–C circuit parallel combination of inductance of 0.01 H and a
capacitor of 1 F is connected to a variable frequency alternating current
source as shown in figure. Draw a rough sketch of the current variation
as the frequency is changed from 1kHz to 3kHz.
A
L
Solution
L and C are connected in parallel to the AC source,
10 4
so resonance frequency f 


 1.6kHz
2
2  LC 2  0.01  10 6
1
I
1
In case of parallel resonance, current in L–C circuit at resonance is zero,
so the I-f curve will be as shown in figure.
50
1.6kHz
f
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C
E
JEE-Physics
POWER IN AC CIRCUIT
The average power dissipation in LCR AC circuit
Let V = V 0 sint
and
I = I0 sin (t – )
Instantaneous power P = (V 0 sint)(I 0 sin(t – ) = V 0I 0 sint (sintcos – sincost)
T
Average power <P> =
1
(V0 I 0 sin 2 t cos   V0 I 0 sin t cos t sin )dt
T 0
T
 T

= V0 I 0  1 sin 2 t cos dt  1 sin t cos t sin dt   V0 I 0  1 cos   0  sin  


2

T0


T 0

V0 I 0 cos 
 = V rms I rm,s cos
2
Average power/actual power/
Virtual power/ apparent
dissipated power/power loss
Power/rms Power
P = V rms I rms cos 
P = V rms I rms

Instantaneous
power
P = VI
•
•
<P> =
Peak power
P = V0 I0
Irms cos is known as active part of current or wattfull current, workfull current. It is in phase with voltage.
Irms sinis known as inactive part of current, wattless current, workless current. It is in quadrature (90) with voltage.
Power factor :
Average power P  E rms I rms cos   r m s power  cos 
Power factor (cos ) 
Power factor : (i)
Average power
r m s Power
is leading if I leads V
and
cos =
R
Z
(ii) is lagging if I lags V
GOLDEN KEY POINTS
• P av < P rms .
• Power factor varies from 0 to 1
• Pure/Ideal

V
Power factor = cos
R
0
V, I same Phase
1 (maximum)
L


2
V leads I
0
0
C


2
V lags I
0
0
0
0

V leads I
2
• At resonance power factor is maximum

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Choke coil
E
Average power
V rms. I rms
( = 0 so cos = 1) and P av = V rms I rms
Example
1
F capacitor in series with a resistor of 500. Find

the power factor of the circuit and the power dissipated.
Solution
A voltage of 10 V and frequency 103 Hz is applied to

XC 
1

2 f C
Power factor cos=
1
10 6
2   10 

 500   Z =
R 2  X 2C  (500)2  (500) 2  500 2 
3
R
(10) 2
1
1
V2
500
1


=
=
  Power
=V rms I rms cos  rms cos =
W
Z 500 2
Z
500 2
2 10
2
dissipated
51
JEE-Physics
Example

) mA for an A.C. circuit then find out
3
total impedance, reactance, resistance
components contains by circuits
If V = 100 sin 100 t volt and I = 100 sin (100 t +
(a)
(c)
Solution
phase difference between V and I
power factor and power dissipated
(b)
(d)

(I leads V)
3
(a)
Phase difference

(b)
Total impedance
Z =
(c)
reactance X  Z sin 60   1000  3  500 
2
3
 = – 60° 
Power factor = cos = cos (–60°) = 0.5 (leading)
V0
100

 1k  Now resistance R  Z cos 60   1000  1  500 
I0
100  10 3
2
Power dissipated P  Vrms I rms cos  
(d)
Circuit must contains R as  
100
2

0.1

2
R
60°
X
Z
1
 2.5 W
2

and as  is negative so C must be their, (L may exist but X C > X L)
2
Example
1
when applied voltage is V = 100 sin 100t volt and resistance of
2
circuit is 200 then calculate the inductance of the circuit.
Solution
If power factor of a R-L series circuit is
cos  =
R
1 R
 =  Z = 2R R 2  X 2L = 2R
Z
2 Z
L =
3 R
3R
=

L =
XL 3 R

3  200
2 3
H
=
100 

Example
A circuit consisting of an inductance and a resistacne joined to a 200 volt supply (A.C.). It draws a current
of 10 ampere. If the power used in the circuit is 1500 watt. Calculate the wattless current.
Solution
Apparent power = 200 × 10 = 2000 W
Power factor cos =
True power
1500
3
=
=
4
Apparent power 2000
2
Example
A coil has a power factor of 0.866 at 60 Hz. What will be power factor at 180 Hz.
Solution
Given that cos  = 0.866,  = 2f = 2 × 60 = 120 rad/s, ’ = 2f’ = 2 × 180 = 360 rad/s
Now,
cos  = R/Z

R = Z cos  = 0.866 Z
But
Z =
R 2  ( L ) 2 
L =
Z2  R 2 =
Z 2  (0.866 Z)2 = 0.5 Z  L =
0.5Z
0.5Z
=

120 
When the frequency is changed to ’ = 2 × 180 = 3 × 120 = 300 rad/s, then
inductive reactance ’ L = 3 L = 3 × 0.5 Z = 1.5 Z
 New impedence Z’ =
[R ' (  'L )2 ] =
 New power factor
R
0.866 Z
=
= 0.5
Z'
1.732 Z
=
(0.866 Z)2  (1.5 Z)2 = Z
52
[(0.866 )2  (1.5) 2 ] = 1.732Z
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Wattless current =  rms sin  = 10
10 7
 3
1  
A
 4
4
E
JEE-Physics
CHOKE COIL
In a direct current circuit, current is reduced with the help of a resistance.
tube light rod
Hence there is a loss of electrical energy I 2 R per sec in the form of
heat in the resistance. But in an AC circuit the current can be reduced
by choke coil which involves very small amount of loss of energy. Choke
starter
coil is a copper coil wound over a soft iron laminated core. This coil is
put in series with the circuit in which current is to be reduced. It also
known as ballast.
choke coil
Circuit with a choke coil is a series L-R circuit. If resistance of choke coil = r (very small)
The current in the circuit I 
E
with
Z
Z  (R  r )2  ( L )2 So due to large inductance L of the coil, the
current in the circuit is decreased appreciably. However, due to small resistance of the coil r,
The power loss in the choke
cos  
P av = V rms I rms cos   0 
r

Z
r
2
2
2
r  L

r
0
L
GOLDEN KEY POINT
•
Choke coil is a high inductance and negligible resistance coil.
•
Choke coil is used to control current in A.C. circuit at negligible power loss
•
Choke coil used only in A.C. and not in D.C. circuit
•
Choke coil is based on the principle of wattless current.
•
Iron cored choke coil is used generally at low frequency and air cored at high frequency.
•
Resistance of ideal choke coil is zero
Example
A choke coil and a resistance are connected in series in an a.c circuit and a potential of 130 volt is applied
to the circuit. If the potential across the resistance is 50 V. What would be the potential difference across the
choke coil.
Solution
V =
VR2  VL2

VL  V 2  VR2  (130)2  (50)2 = 120 V
Node-6\E:\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No.-10\EMI & AC\English\03 AC.p65
Example
An electric lamp which runs at 80V DC consumes 10 A current. The lamp is connected to
100 V – 50 Hz ac source compute the inductance of the choke required.
Solution
E
Resistance of lamp R  V  80  8 
I 10
Let Z be the impedance which would maintain a current of 10 A through the Lamp when it is run on
100 Volt a.c. then. Z =

V
100
=
= 10  but
I
10
Z =
R 2  ( L ) 2
(L) 2 = Z 2 – R 2 = (10) 2 – (8) 2 = 36L = 6 L =
6
6
=
= 0.02H

2   50
Example
Calculate the re sistance or i nductance required to operate a lamp (60V, 10W) from a source of
(100 V, 50 Hz)
53
JEE-Physics
Solution
R
(a)
Maximum voltage across lamp = 60V

V Lamp + V R = 100

10
1
Wattage
=
=
A
60
6
voltage
Now current througth Lamp is =
But
(b)
V R= IR

40 =
V R = 40V
1
(R)
6

100V, 50Hz
L
R = 240 
Now in this case (V Lamp) 2 + (V L) 2 = (V) 2
(60) 2 + (V L ) 2 = (100) 2
Also
V L = IX L =
1
X
6 L
so
 V L = 80 V
100V, 50Hz
X L = 80 × 6 = 480  = L (2f)  L = 1.5 H
A capacitor of suitable capacitance replace a choke coil in an AC circuit, the average power consumed in
a capacitor is also zero. Hence, like a choke coil, a capacitor can reduce current in AC circuit without power
dissipation.
Cost of capacitor is much more than the cost of inductance of same reactance that's why choke coil is used.
Example
L, R
A choke coil of resistance R and inductance L is connected in series with
C
a capacitor C and complete combination is connected to a.c.
voltage, Circuit resonates when angular frequency of supply is  = 0 .
(a)
Find out relation betwen 0, L and C
(b)
What is phase difference between V and I at resonance,
~
V=V0 sinwt(volt)
is it changes when resistance of choke coil is zero.
Solution
1
(a) At resonance condition X L = X C  0L =  C     0 =
0
(b)
  cos  =
R
R
=
= 1
Z
R
  = 0°
1
LC
No, It is always zero.
LC OSCILL ATION
UNDA MPED OSCILL ATION
When the circuit has no resistance, the energy taken once from the source and given to capacitor keeps on
oscillating between C and L then the oscillation produced will be of constant amplitude. These are called undamped
oscillation.
I
C
L
t
54
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The oscillation of energy between capacitor (electric field energy) and inductor (magnetic field energy) is called
LC Oscillation.
E
JEE-Physics
After switch is closed
Q
di
L
0
C
dt
Q
d2 Q
L 2 0
C
dt

d2 Q
1

Q 0
2
LC
dt

 d2 x

2
By comparing with standard equation of free oscillation  2   x  0 
 dt

2 
1
LC
Frequency of oscillation f 
1
2  LC
Charge varies sinusoidally with time q = q m cos t
current also varies periodically with t
I =
dq

= q m  cos ( t  )
dt
2
If initial charge on capacitor is q m then electrical energy strored in capacitor is U E =
1 q 2m
2 C
At t = 0 switch is closed, capacitor is starts to discharge.
As the capacitor is fully discharged, the total electrical energy is stored in the inductor in the form of magnetic
energy.
UB =
1 2
LI m
2
where I m = max. current
(U max ) EPE = (U max ) MPE
1 q2
1 2
 m  LI m
2 C
2
DA MPED OSCILL ATION
Practically, a circuit can not be entirely resistanceless, so some part of energy is lost in resistance and amplitude
of oscillation goes on decreasing. These are called damped oscillation.
R
L
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C
E
Angular frequency of oscillation  
frequency of oscillation f 
oscillation to be real if
1
R2
 2
LC 4L
I
1
1
R2
 2
2  LC 4L
t
1
R2
 2 0
LC 4L
Hence for oscilation to be real
1
R2

LC 4L2
55
JEE-Physics
GOLDEN KEY POINTS
• In damped oscillation amplitude of oscillation decreases exponentially with time.
• At t 
T 3T 5 T
,
,
..... energy stored is completely magnetic.
4 4 4
• At t 
T 3T 5 T
,
,
..... energy is shared equally between L and C
8 8 8
• Phase difference between charge and current is
 when charge is maximum, current minimum
2 when charge is minimum,current maximum
Example
An LC circuit contains a 20mH inductor and a 50F capacitor with an initial charge of 10mC. The resistance
of the circuit is negligible. Let the instant the circuit is closed to be t = 0.
(a)
What is the total energy stored initially.
(b)
What is the natural frequency of the circuit.
(c)
At what time is the energy stored is completely magnetic.
(d)
At what times is the total energy shared equally between inductor and the capacitor.
Solution
(a)
UE =
(b)
 =
(c)

1 q2
2 C
1
LC
=
1 (10  10 3 ) 2

= 1.0J
2
50  10 6
1
=
20  10
3
 50  10
6
= 10 3 rad/sec

f = 159 Hz
q = q 0 cos t
Energy stored is completely magnetic (i.e. electrical energy is zero, q = 0)
(d)
t
T 3T 5 T
,
,
......... where T  1 = 6.3 ms
4 4 4
f
Energy is shared equally between L and C when charge on capacitor become
so,
at t 
T 3T 5 T
,
,
...... energy is shared equally between L and C
8 8 8
56
q0
2
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at
E
JEE-Physics
SOME WORKED OUT EXAMPLES
Example#1
For the given circuit
(A) The phase difference between I L & I R 1 is 0°
(B) The phase difference between V C & VR 2 is 90°
(C) The phase difference between I L & I R 1 is 180°
(D) The phase difference between V C & VR 2 is 180°
Solution
Ans. (B)
The phase difference between V C and VR
2
is  rad or 90°
Example#2
A periodic voltage V varies with time t as shown in the figure. T is the time period. The r.m.s. value of the
voltage is :V
V0
T
T/4
(A)
V0
8
(B)
V0
2
t
(C) V 0
(D)
V0
4
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Solution
E
Ans. (B)
T/4
2
0
 V dt
0
Root mean square value <V> =
T
 dt

T
V02  
 4  
T
T
V02  
4 
T
V02
V
 0
4
2
0
Example#3
The potential difference V and current I flowing through the AC circuit is given by V = 5 cos(t – /6) volt and
I = 10sint ampere. The average power dissipated in the circuit is
(A)
25 3
W
2
(B) 12.5 W
(C) 25 W
57
(D) 50 W
JEE-Physics
Solution
Ans. (B)
V = 5 cos (t - /6); i = 10 sin t = 10 cos (t - /2)

  
VI
5 10 1
  ; P
cos  
  12.5 W
2 6 3
2
2
2
Example#4
The radius of a coil decreases steadily at the rate of 10 –2 m/s. A constant and uniform magnetic field of induction 10–3 Wb/m2 acts perpendicular to the plane of the coil. The radius of the coil when the induced e.m.f.
in the coil is 1V, is :(A)
2
cm

(B)
3
cm

(C)
4
cm

(D)
5
cm

Solution
e
Ans. (D)
dr
e
10 6
5
d d

r 
r 
 cm
(  r 2 B) = 2  rB
dt dt
dr 
dt
2   10 3  10 2 

 2 B dt 


Example#5
A time varying voltage V = 2t volt is applied across an ideal inductor of inductance
L = 2H as shown in figure. Then select incorrect statement
(A) current versus time graph is a parabola
.
.
2H
V=2t
(B) energy stored in magnetic field at t = 2 s is 4J
(C) potential energy at time t = 1 s in magnetic field is increasing at a rate of 1 J/s
(D) energy stored in magnetic field is zero all the time
Solution
Ans. (D)
V = 2t  L
U =
di
di
di
t2
 i – t graph parabola
= 2t  2 ×
= 2t   i =
dt
dt
dt
2
dU
di
1
1
t2
Li2 =
× 2 × 4 = 4J and
=Li
= 2 ×
× t = t3 = 1 J/s
dt
dt
2
2
2
A circular coil of 500 turns encloses an area of 0.04 m 2 . A uniform magnetic field of induction
0.25 Wb/m 2 is applied perpendicular to the plane of the coil. The coil is rotated by 90° in 0.1 second at a
constant angular velocity about one of its diameters. A galvanometer of resistance 25 was connected in series
with the the coil. The total charge that will pass through the galvanometer is (A) 0.4 C
(B) 1 C
(C) 0.2 C
Solution
Induced current I 
Total charge
(D) Zero
Ans. (C)
e
dq e  d   1
 NBA

   q 

R
dt R  dt  R
R
R
q =
500  0.25  0.04
= 0.2 C
25
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Example#6
E
JEE-Physics
Example#7
A condenser of capacity 6 µF is fully charged using a 6-volt battery. The battery is removed and a
resistanceless 0.2 mH inductor is connected across the condenser. The current which is flowing through the
inductor when one-third of the total energy is in the magnetic field of the inductor is :(A) 0.1 A
(B) 0.2 A
(C) 0.4 A
(D) 0.6 A
Solution
Ans. (D)
Total energy = Intial energy on capacitor =

1
1 2
CV 2 , Magnetic field energy =
LI
2
2
1 1
1 2
LI = 
CV 2  I =
3 2
2
CV 2

3L
6  10 6  6  6
= 0.6 A
3  2.0  10 3
Example#8
For the circuit shown, which of the following statement(s) is(are) correct?
(A) Its time constants is 2 second.
(B) In steady state, current through inductance will be 1A.
(C) In steady state, current through 4  resistance will be 2/3 A.
(D) In steady state, current throgh 8  resistance will be zero.
Solution
Node-6\E:\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No.-10\EMI & AC\English\03 AC.p65
Time constant  
E
Ans. (B)
L 12

 1s
R 12
In steady state
current through 4  resistance =
6
1
1  A
6  12
3
59
JEE-Physics
Example#9
In the circuit shown the capacitor has charge Q. At t = 0 sec the key is closed. The charge on the capacitor at
the instant potential difference across the inductor L1 is zero, is
(A) Q
(B)
Q
3
(C)
2Q
3
(D) 0
Ans. (D)
Solution
When Vacross L = 0  i = imax  qc = 0
Example#10
The figure shows a rod of length  with points A and B on it. The rod is moved in a uniform magnetic field
(B 0 ) i n di fferent ways as show n. In wh i ch case potent i al di fference (V A –V B ) b et ween
A & B is minimum?
B0
B
B0

(A)
A

B0
B

v=
2

(B)
A
(C)
B0
/2
/2 B

A
/2
(D)
2 B
A /

Ans. (C)
Solution
1
1
B 2  VA  VB   B 2
2
2
1
1
  
2
2
For (B) : VB  VA  Bv  B     B   VA  VB   B 
2
2
2
For (C) : VA _ VB = 0
For (A): VB  VA 
2
Example#11
A bent rod PQR (PQ = QR = ) shown here is rotating about its end P with angular speed  in a region of
transverse magnetic field of strength B.
Q
(A) e.m.f. induced across the rod is B2
60°
 B
(B) e.m.f. induced across the rod is B2/2

(C) Potential difference between points Q and R on the rod is B2/2
P
R
(D) Potential difference between points Q and R on the rod is zero
Solution
Ans. (B,D)
The rod in equivalent to a rod joining the ends P and R of the rod rotating is the same sense.
B
60°
P
VR  V P 

R

R
B 2
B  2
B 2
; VQ  VP 
; VR  V P 
; VQ – VR = 0
2
2
2
60
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1
1
3
3
 
2
2
2
For (D) : VB  VA  B   B     B   VA  VB   B 
2
2
2
8
8
E
JEE-Physics
y
Example#12
B
A conducting loop is kept so that its center lies at the origin of the coordinate
x
system. A magnetic field has the induction B pointing along Z-axis as shown
in the figure
(A) No emf and current will be induced in the loop if it rotates about Z-axis
(B) emf is induced but no current flows if the loop is a fiber when it rotates about y-axis.
(C) emf is induced and induced current flows in the loop if the loop is made of copper & is rotated about y-axis.
(D) If the loop moves along Z-axis with constant velocity, no current flows in it.
Solution
A n s. ( A , C, D )
If the loop rotates about Z axis, the variation of flux linkage will be zero. Therefore no emf is induced.
Consequently no current flows in the loop. When it rotates about y axis, its flux linkage changes. However,
in insulators there can not be motional emf. If the loop is made of copper, it is conductive therefore
induced current is set up. If the loop moves along the Z axis variation of flux linkage is zero. Therefore the
emf and current will be equal to zero.n
Example#13
Initially key was placed on (1) till the capacitor got fully charged. Key is placed on (2) at t=0. The time
when the energy in both capacitor and inductor will be same(1)
(2)
L
E
C
(A)
 LC
4
(B)
 LC
2
(C)
5  LC
4
(D)
5  LC
2
Solution.
Ans. (A,C)
For given situation
q
di
d2 q
q
d2 q
L
0 2 
 0  2  2 q  0  q = q0 cos t & i = –q0sint
C
dt
LC
dt
dt
q 0 cos 2 t 1 2 2 2
q2 1 2

Li

 Lq 0  sin t
According to given conditions
2C 2
2C
2
 3  5  7
 LC 3  LC 5  LC 7  LC
, ,
,
,
,
.........  t 
.....
4 4 4 4
4
4
4
4
 cot2t = 1  t = ,
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Example#14
E
For an LCR series circuit, phasors of current i and applied voltage V = V 0 sin t are shown in diagram
at t =0. Which of the following is/are CORRECT?
(A) At t =

, instantaneous power supplied by source is negative.
2
(B) From 0 < t <
(C) At t =
2
, average power supplied by source is positive.
3
5
, instantaneous power supplied by source is negative.
6
(D) If  is increased slightly, angle between the two phasors decreases.
61
i0
/3
V0
JEE-Physics
Solution
Ans. (BCD)
i/v
The graph shows V & I as function of time.
Current leads the voltage by /3
current
t
Power is positive if V & I are of same sign.
Voltage
Power is negative if V & I are of opposite sign
1
 thus angle decreases.
If  
C
Example#15 to 17
Consider a conducting circular loop placed in a magnetic field as shown in
figure. When magnetic field changes with time, magnetic flux also changes
and emf
e
d
is induced.
dt
If resistance of loop is R then induced current is i 
e
. For current, charges must have non–zero average
R
velocity. Magnetic force cannot make the stationary charges to move. Actually there is an induced electric field
 
d
in the conductor caused by changing magnetic flux, which makes the charges to move,  E  d   
.This
dt
induced electric field is non conservative by nature.
15.
A cylindrical space of radius R is filled with a uniform magnetic induction B
parallel to the axis of the cylinder. If
dB
= constant, the graph, showing thee
dt
variation of induced electric field with distance r from the axis of cylinder, is
16.
(B)
(C)
A square conducting loop is placed in the time varying magnetic field
(D)
 dB

  ve constant  . The centre of

dt


square coincides with axis of cylindrical region of magnetic field. The directions of induced electric field at point
a, b and c.
62
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(A)
E
JEE-Physics
(A)
17.
(B)
(C)
(D)
A line charge  per unit length is pasted uniformly onto the rim of a wheel of mass m and radius R. The wheel
has light non–conducting spokes and is free to rotate about a vertical axis as shown in figure.A uniform magnetic
field B exist as shown in figure. What is the angular velocity of the wheel when the field is suddenly switched off?
(A)
2 a 2 B
mR
(B)
a 2 B
mR
(C)
3 a 2 B
mR
(D)
a 2 B
2mR
Solution
1 5 . Ans. (B)
For r  R;
E  2 r   r 2
dB
E r
dt
16.
Ans. (A)
Induced electric field is circular.
17.
Ans. (B)

For r  R;
E  2 r   R 2
dB
1
E
dt
r
a 2 BR
 2 B 
L I mR 2 
mR 2  a 2 BR
a 2 B





but     2 R  ER    a
R 

t 
t
t t
t
t
t
mR
Node-6\E:\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No.-10\EMI & AC\English\03 AC.p65
Example#18 to 20
E
A conducting square wire frame ABCD of side  is pulled by horizontal force so that it moves with constant
velocity v. A uniform magnetic field of strength B is existing perpendicular to the plane of wire. The resistance
per unit length of wire is  and negligible self inductance. If at t= 0, frame is just at the boundary of magnetic
field. Then
63
JEE-Physics
18.
The emf across AB at t =
(A)
19.
(C) Bv
(B) zero
Potential difference across BC at time t =
(A)
20.
Bv
4

is
2v
Bv
4
(D) None of these

is
2v
(B) Bv
(C)
3Bv
4
(D) None of these
Find the applied horizontal force on BC, (F) as a function of time 't' (t <
(A)
B 2 v
t

(B)
B 2 v
2
(C)

)
v
B 2 v
4
(D) None of these
Solution
18.
Ans. (B)
  
 
e  B . (   v ) as e| | v   = 0°  e = 0
19.
Ans. (C)
A
B
B v
3B v
[ ] 
4 (  )
4
V = E – ir = Bv –
D
20.
C
Ans. (C)
F  Bi 
B (B v) B 2 2 v B 2 v


4 ( )
4 ( )
4
Example#21
(Wb)
10
8
2
×
×
×
×
×
×
×
×
×
×
×
×
10 14 16
6
t(s)
–10
Column–I
C ol u m n– I I
(A)
At 1 second induced current is
(P)
Clockwise
(B)
At 5 second induced current is
(Q)
Anticlockwise
(C)
At 9 second induced current is
(R)
0.5 A
(D)
At 15 second induced current is
(S)
5A
(T)
None of these
64
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Magnetic flux in a circular coil of resistance 10 changes with time as shown in figure.  direction indicates a
direction perpendicular to paper inwards.
E
JEE-Physics
Solution
Ans. (A) Q,R (B) T (C) P,R (D) Q,R
For (A)
d
5
dt
5
1

A, Anticlockwise
10 2
For (B)
d
 0  i=0 = zero
dt
For (C)
d
5
1
  A, Clockwise
= –5  i  
dt
10
2
For (D)
d
5
1
 A , Anticlockwise
=5  i 
dt
10 2
 i
Example#22
In the circuit shown in figure E=25V, L=2H, C=60 F, R1 = 5 and R2 = 10. Switch S is closed at t = 0.
S
C
L
E
R1
R2
Column-I
Co lu mn -I I
(A)
Current through R 1 at t = 0
(P)
0
(B)
Current through R 2 at t = 0
(Q)
5A
(C)
Current through R 1 at t = 
(R)
2.5 A
(D)
Current through R 2 at t = 
(S)
7.5 A
(T)
None of these
Ans. (A)  (Q), (B)  (P), (C)  (P), (D)  (R)
Solution
At t = 
At t = 0
E
R1
R2
E
E
25
I1 = R =
= 5A
5
1
R1
R2
I1 = 0
E
25
I2 = R =
= 2.5 A
10
2
I2 = 0
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Example#23
E
A uniform but time-varying magnetic field B(t) exists in a circular region of radius a and is directed into
the plane of the paper, as shown. The magnitude of the induced electric field at point P at a distance
r from the centre of the circular region is proportional to 1/r n. Find the value of n.
r
x x
x ax
x
P
Solution
Ans. 1
d
1
 E  2 r   E   n  1
dt
r
65
JEE-Physics
Example#24
An inductor (XL = 2) a capacitor (XC = 8) and a resistance (8) is connected in series with an ac source. The
voltage output of A.C source is given by V = 10 cos 2 50t. Find the instantaneous p.d. between A and B when
the voltage output from source is half of its maximum.
A
X L=2  XC= 8  8
B
~
Ans. 3
Solution
VAB 
6  10 
   3
10  2 
 R AB  X C  X C and Z 
2
XC  XL   R2
Example#25
Figure shows a uniform circular loop of radius ‘a’ having specific resistance  placed in a uniform magnetic field
B perpendicular to plane of figure. A uniform rod of length 2a & resistance R moves with a velocity v as shown.
Find the current in the rod when it has moved a distance
[Given B = 3, a =2, v =3,  
a
from the centre of circular loop.
2
9
, R = 2/3 all in SI units]
4
x
x
x
a
x
x
x
B
x
x
x
a/2
x
x
x
x
x
x
v
6
3
1
3
18
a 3 × B × v = 18 V
66
 i 
18
 6A
3
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Ans. 6
Solution
E