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In the circuit shown in, C(1)=6 muF, C(2...

In the circuit shown in, `C_(1)=6 muF, C_(2)=3 muF`, and battery `B=20 V`. The switch `S_(1)` is first closed. It is then opened, and `S _(2)` is closed. What is the final charge on `C_(2)`
.

A

`120 muC`

B

`80 muC`

C

`40 muC`

D

`20 muC`

Text Solution

Verified by Experts

The correct Answer is:
C
Common potential `V = (6xx20+3xx0)/((6+3)) = (120)/(9)`volt
lt So, charge on `3muF` capacitor
`Q_(2)= 3 xx 10^(-6) xx (120)/(9) = 40muC`
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