JEE Advanced | 2014Current Electricity and Electrical InstrumentHard
Question
At time t = 0, terminal A in the circuit shown in the figure is connected to B by a key and an alternating current I(t) = I0cos(ωt), with I0 = 1A and ω = 500 rad s-1 starts flowing in it with the initial direction shown in the figure. At t = 
, the key is switched from B to D. Now onwards only A and D are connected. A total charge Q flows from the battery to charge the capacitor fully. If C = 20 μF, R = 10Ω and the battery is ideal with emf of 50 V, identify the correct statement(s).
, the key is switched from B to D. Now onwards only A and D are connected. A total charge Q flows from the battery to charge the capacitor fully. If C = 20 μF, R = 10Ω and the battery is ideal with emf of 50 V, identify the correct statement(s).Options
A.Magnitude of the maximum charge on the capacitor before t = 
is 1 × 10-3C
is 1 × 10-3CB.The current in the left part of the circuit just before t = is clockwise
is clockwiseC.Immediately after A is connected to D, the current in R is 10 A
D.Q = 2 × 10-3C
Solution
The variation of current is shown below
The variation of current is shown below
Between t = 0 to
, charge will be maximum at 
Q =
I0 cosωtdt = 
(sin ωt)0π/2ω = 
= 2 × 10-3C
At t =
, sense of current will be opposite to initial sense i.e. anticlockwise.
At t =, the charge on upper plate is
I0 cosωtdt = 
[sin ωt]07π/6ω
× sin 
= - 10-3C
Applying KVL immediately after switch is shifted to D,
+ 50 +
- i × 10 = 0
Final charge on C after shifting the switch, Q′= CV = 20 × 10-6 × 50 = 10-3 C.
So, total charged flown from battery = 2 × 10-3C.
The variation of current is shown below
Between t = 0 to
, charge will be maximum at Q =
I0 cosωtdt = (sin ωt)0π/2ω = = 2 × 10-3CAt t =
, sense of current will be opposite to initial sense i.e. anticlockwise. At t =
, the charge on upper plate isI0 cosωtdt = [sin ωt]07π/6ω × sin = - 10-3CApplying KVL immediately after switch is shifted to D,
+ 50 +
- i × 10 = 0Final charge on C after shifting the switch, Q′= CV = 20 × 10-6 × 50 = 10-3 C.
So, total charged flown from battery = 2 × 10-3C.
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