1.

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`

Answer» Correct Answer - 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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