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Physics-

General

Easy

Question

The charge deposited on $4 mu F$ capacitor the circuit is

$6 \times 10^{- 6} C$
$12 \times 10^{- 6} C$
$24 \times 10^{- 6} C$
$36 \times 10^{- 6} C$

The correct answer is: $24 cross times 10 to the power of negative 6 end exponent C$

As the capacitors 4 $mu F$ and 2 $mu F$ are connected in parallel and are in series with 6 $blank mu F$ capacitor, their equivalent capacitance is
$fraction numerator left parenthesis 2 plus 4 right parenthesis cross times 6 over denominator 2 plus 4 plus 6 end fraction equals 3 blank mu F$
Charge in the circuit
$Q equals blank 3 blank mu F cross times 12 V equals 36 mu C$

Since, the capacitors $4 mu F$ and 2 $mu F$ are connected in parallel, therefore potential difference across them is same.
$rightwards double arrow fraction numerator Q subscript 1 end subscript over denominator Q subscript 2 end subscript end fraction equals fraction numerator C subscript 1 end subscript over denominator C subscript 2 end subscript end fraction equals fraction numerator 4 over denominator 2 end fraction o r Q subscript 1 end subscript equals 2 Q subscript 2 end subscript$
Also $Q equals Q subscript 1 end subscript plus Q subscript 2 end subscript$
$therefore blank 36 mu C equals 2 Q subscript 2 end subscript plus Q subscript 2 end subscript$ or $Q subscript 2 end subscript equals fraction numerator 36 mu C over denominator 3 end fraction equals 12 mu C$
$Q subscript 1 end subscript equals Q minus Q subscript 2 end subscript equals 36 mu C minus 12 mu C$
$equals 24 blank mu C equals 24 cross times 10 to the power of negative 6 end exponent C$

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