Q.2. (a) In this circuit, us shown in ligure 1. S is closed at t = 0; and S, is opened at t =
4ms, Determine i(t) for t>0. Assume inductor is initially de-energised
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Answer:
In steady state , charge on C is q
10
=CE and charge on 2C is q
20
=2CE
∴
q
20
q
10
=
2CE
CE
=1/2=1:2
Charge on C at time t is q
1
(t)=q
10
(1−e
−t/C(2R)
)=CE(1−e
−t/2RC
)
Charge on 2C at time t is q
2
(t)=q
20
(1−e
−t/(2C)R)
)=2CE(1−e
−t/2RC
)
Thus,
q
2
q
1
=1/2=1:2 and time constant of both is τ=2RC
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