Assertion : If three capacitors of capacitanc | vedclass

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(C) For capacitors connected in parallel,the equivalent capacitance is $C_P = C_1 + C_2 + C_3$.Since $C_1, C_2, C_3 > 0$,it follows that $C_P > C_1$,$

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If the Assertion is correct but Reason is incorrect.

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(C) For capacitors connected in parallel,the equivalent capacitance is $C_P = C_1 + C_2 + C_3$.Since $C_1, C_2, C_3 > 0$,it follows that $C_P > C_1$,$C_P > C_2$,and $C_P > C_3$.For capacitors connected in series,the equivalent capacitance $C_S$ is given by $\frac{1}{C_S} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$.It is a known property that for any set of capacitors,$C_P > C_S$.Thus,the Assertion is correct.The Reason provided is $\frac{1}{C_P} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$,which is the formula for series combination,not parallel. Therefore,the Reason is incorrect.

Assertion : If three capacitors of capacitances $C_1 < C_2 < C_3$ are connected in parallel,then their equivalent capacitance $C_P > C_S$,where $C_S$ is the equivalent capacitance in series.
Reason : $\frac{1}{C_P} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$

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Assertion : If three capacitors of capacitances $C_1 < C_2 < C_3$ are connected in parallel,then their equivalent capacitance $C_P > C_S$,where $C_S$ is the equivalent capacitance in series.Reason : $\frac{1}{C_P} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$