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In the given circuit, a charge of +80muC...

In the given circuit, a charge of `+80muC` is given to the upper plate of the `4muF` capacitor. Then in the steady state, the charge on the upper plate of the `3muF` capacitor is

A

(a) `+32muC`

B

(b) `+40muC`

C

(c) `48muC`

D

(d) `+80muC`

Text Solution

Verified by Experts

The correct Answer is:
C
The total charge on plate A will be `-80muC`. If `q_B` and `q_C` be the charges on plate B and C then
`q_B+q_C=80muC`
Also `2muF` and `3muF` capacitors are in parallel. Therefore,
`(q_B)/(2)-(q_C)/(3)`
`:.` `(80-q_C)/(2)=(q_C)/(3)`
`:.` `240-3q_C=2q_C`
This charge will obviously be poistive.
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