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In the circuit shown, the cell is ideal ...

In the circuit shown, the cell is ideal , with `emf= 15 V`. Ecah resistance is of `3 Omega.` the potential difference across the capacitor is

A

zero

B

`9V`

C

`12 V`

D

`15 V`

Text Solution

Verified by Experts

The correct Answer is:
C
At steady state capacitor is open circuit
`I = (15)/[(3+3)||3]+3)=3amp`.

`V_(B) +3 xx 3+3xx1 = V_(A)`
`V_(A) - V_(B) = 12 "volt"`
`(2)` Since plate `1` and `4` aer connected by a conducting wire so
`V_(1) = V_(4)`
So `V_(2) -E_(2)d = V_(2) -E_(1) xx2d -E_(1)d`
`E_(2) = 3E_(1)`
`(Q_(2))/(in_(0)A) = (3Q_(1))/(in_(0)A) rArr Q_(2) = 3Q_(1) ...(1)`
Since net charge on `2` is `Q`
`Q_(1) +Q_(2) = Q ...(2)`
From `(1)` and `(2)`
`Q_(1) = (Q)/(4),Q_(2) = (3Q)/(4)`
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