Three capacitors C_1, C_2 and C_3 are connect | vedclass

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(A) The equivalent capacitance of the circuit is given by:$\frac{1}{C_{eq}} = \frac{1}{C_3} + \frac{1}{C_1 + C_2}$(since $C_1$ and $C_2$ are in parall

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$(C_1 + C_2) V [1 - C_3/(C_1 + C_2 + C_3)]$

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(A) The equivalent capacitance of the circuit is given by:$\frac{1}{C_{eq}} = \frac{1}{C_3} + \frac{1}{C_1 + C_2}$(since $C_1$ and $C_2$ are in parallel,and this combination is in series with $C_3$).Thus,$\frac{1}{C_{eq}} = \frac{C_1 + C_2 + C_3}{C_3(C_1 + C_2)}$$	herefore C_{eq} = \frac{C_3(C_1 + C_2)}{C_1 + C_2 + C_3}$Since $V$ is the voltage of the battery,the initial total charge is $q = C_{eq} V = \frac{C_3(C_1 + C_2) V}{C_1 + C_2 + C_3}$.If the capacitor $C_3$ breaks down (acts as a short circuit),the effective capacitance becomes $C_{eq}' = C_1 + C_2$.The new charge is $q' = C_{eq}' V = (C_1 + C_2) V$.The change in total charge is $\Delta q = q' - q = (C_1 + C_2) V - \frac{C_3(C_1 + C_2) V}{C_1 + C_2 + C_3}$.Factoring out $(C_1 + C_2) V$,we get:$\Delta q = (C_1 + C_2) V \left[ 1 - \frac{C_3}{C_1 + C_2 + C_3} ight]$.

Three capacitors $C_1, C_2$ and $C_3$ are connected as shown in the figure to a battery of $V$ volt. If the capacitor $C_3$ breaks down electrically,the change in total charge on the combination of capacitors is

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