Question #78627

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Question #78627

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Answer 1 · 2017-01-18T04:38:17

(A)

Explanation:

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Two given capacitors are connected as shown in figure above.
The total capacitance $C_{T}$ $C_"T"$ can be found by

$C_{T} = C_{1} + C_{2}$ $C_"T"=C_1+C_2$ .........(1)

For a charge $q$ $q$ given to the assembly

$C_{T} = \frac{q}{V}$ $C_"T"=q/V$
where $V$ $V$ is voltage drop across capacitors and is equal for both capacitors.

If $q_{1} and q_{2}$ $q_1 and q_2$ is the charge distributed on capacitors $C_{1} and C_{2}$ $C_1 and C_2$ respectively.

$C_{1} = \frac{q_{1}}{V}$ $C_1=q_1/V$ ......(2) and
$C_{2} = \frac{q_{2}}{V}$ $C_2=q_2/V$ .....(3)

Dividing equation (2) by (3) we get
$\frac{C_{1}}{C_{2}} = \frac{\frac{q_{1}}{V}}{\frac{q_{2}}{V}}$ $C_1/C_2=(q_1/V)/(q_2/V)$
$\Rightarrow \frac{q_{1}}{q_{2}} = \frac{C_{1}}{C_{2}}$ $=>(q_1)/(q_2)=C_1/C_2$

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Question #78627

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