NCERT Class XII Chapter
Alternating Current
Questions With Solutions

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Question : 12
Total: 26
An LC circuit contains a 20 mH inductor and a 50 μF capacitor with initial charge of 10 mC. The resistance of the circuit in negligible. Let the instant the circuit is closed be t = 0.
(a) What is the total energy stored initially? Is it conserved during LC oscillations?
(b) What is the natural frequency of the circuit?
(c) At what time is the energy stored
(i) completely electrical (i.e., stored in the capacitor)?
(ii) completely magnetic (i.e., stored in the inductor)?
(d) At what times is the total energy shared equally between the inductor and capacitor?
(e) If a resistor is inserted in the circuit, how much energy is eventually dissipated as heat?
Solution:  
(a) Total energy is initially in the form of electric field within the plates of charged capacitor.

Ue =
Q2
2C
=
(10×103)2
2×50×106
= 1 J
If we neglect the losses due to resistance of connecting wires, the total energy remain consumed during LC oscillations.
(b) Natural frequency of the circuit
f =
1
2πLC
=
1
2π×20×103×50×106
=
500
π
Hz = 159 Hz
(c) Instantaneous electrical energy
Ue =
q02cos2ωt
2C

At ωt = 0, π, 2π, 3π… the energy is completely electrical.
t =
nπ
2πf
=
n
2f
=
nπ
1000
s , n = 0 , 1 , 2 , 3 , 4 or t = T/2 , T , 3T/2 ...
Instantaneous magnetic energy
UB =
1
2
L
q02
ω2
s
i
n2
ω
t
or UB =
q02
2C
s
i
n2
ω
t

so at ωt = π/2, 3π/2, 5π/2…
The energy is completely magnetic
t =
(2n+1)π
2(2πf)
=
(2n+1)
4f
=
(2n+1)π
2000
s
where n = 0, 1, 2, 3, 4 … or t = T/4, 3T/4, 5T/4…
(d) timings for energy shared equally between inductor and capacitor.
UB = UE
q02
2C
s
i
n2
ω
t
=
q02
2C
c
o
s2
ω
t

tan2ωt = 1 or tan ωt = tan π/4
t =
π
4ω
,
3π
4ω
,
5π
4ω
... or t =
T
8
,
3T
8
,
5T
8
...
(e) When a resistor is inserted in the circuit, eventually all the energy will be lost as heat across resistance. The oscillation will be damped.
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