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Initially the capacitor was uncharged Cu...

Initially the capacitor was uncharged Current in the capacitor just after switching on will be ?

A

`(epsilon)/(R)`

B

`(epsilon)/(2R)`

C

`(epsilon)/(5R)`

D

`(epsilon)/(4R)`

Text Solution

Verified by Experts

The correct Answer is:
C

Just after switching the capacitor will act like a conducting wire. So effective circuit will be

By applying KVL at point `P (V - epsilon)/(R)+(V - epsilon)/(R)+(V)/(2R)=0`
`V = (4 epsilon)/(5)`
So, current in the branch containing capacitor `= (epsilon)/(5R)`
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