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Question
The switches in figure (a) and (b) are closed at t = 0 and reopened after a long time at t = t0.
(a) The charge on C just after t = 0 is εC.
(b) The charge on C long after t = 0 is εC.
(c) The current in L just before t = t0 is ε/R.
(d) The current in L long after t = t0 is ε/R.
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Solution
(b) The charge on C long after t = 0 is εC.
(d) The current in L long after t = t0 is ε/R.
The charge on the capacitor at time ''t'' after connecting it with a battery is given by,
`Q=C epsilon[1-e^(-t"/RC")]`
Just after t = 0, the charge on the capacitor will be
`Q=C epsilon[1-e^0]=0`
For a long after time, t → ∞
Thus, the charge on the capacitor will be
`Q = Cepsilon[1-e^(-infty)]`
`rArr Q=Cepsilon[1-0]=Cepsilon`
The current in the inductor at time ''t'' after closing the switch is given by
`I=V_b/R(1-e^(-tR"/L"))`
Just before the time t0, current through the inductor is given by
`I=V_b/R(1-e^(-t_0R"/L"))`
It is given that the time t0 is very long.
∴ t0 → ∞
`I=epsilon/R(1-e^-infty)=epsilon/R`
When the switch is opened, the current through the inductor after a long time will become zero.
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