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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))` toward 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 2.
Just after `S_(1)` closed the potential difference across capacitor 1 and 2 are 0 and 2, respectively.
If we take potential of `C` and `D` to be zero than
`V_(A) = epsilon,V_(B) = 2 epsilon`
Since `VB gt VA`, so current will flow
`i= (2 epsilon - epsilon)/(R_(1)) = (epsilon)/(R_(1))` toward left
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