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Physics-

General

Easy

Question

In the circuit shown here $C subscript 1 end subscript equals 6 mu F comma C subscript 2 end subscript equals 3 mu F$ and battery $B equals 20 V$ . The switch $S subscript 1 end subscript$ is first closed. It is then opened and afterwards $S subscript 2 end subscript$ is closed. What is the charge finally on $C subscript 2 end subscript$

$120 μ C$
$80 μ C$
$40 μ C$
$20 μ C$

The correct answer is: $40 mu C$

Common potential $V equals fraction numerator 6 cross times 20 plus 3 cross times 0 over denominator left parenthesis 6 plus 3 right parenthesis end fraction equals fraction numerator 120 over denominator 9 end fraction V o l t$
So, charge on $3 mu F$ capacitor
$Q subscript 2 end subscript equals 3 cross times 1 0 to the power of negative 6 end exponent cross times fraction numerator 120 over denominator 9 end fraction equals 40 mu C$

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