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In the circuit shown, S(2) is closed fir...

In the circuit shown, `S_(2)` is closed first and is kept closed for a long time. Now `S_(1)` is closed. Just after that instant the current through `S_(1)`is

A

`epsilon/R_1` toward right

B

`epsilon/R_1` towards left

C

zero

D

`(2epsilon)/(R_1)`

Text Solution

Verified by Experts

The correct Answer is:
B
b. Just before `S_1` is closed, the potential difference across capacitor
2 is `2epsilon`.
Just after `S_1` is closed, the potential differences across
capacitors 1 and 2 are 0 and `2epsilon`, respectively. Applying KVL to
loop ABCD immediately after
`S_1` is closed,
`epsilon = -iR_1 + 0+ 2epsilon, i = E/R_1` towards left .
.
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