To obtain 3,mu F capacitance from three capac | vedclass

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Question

(C) Let the three capacitors be $C_1 = C_2 = C_3 = 2\,\mu F$.If we connect two capacitors in series,their equivalent capacitance $C_s$ is given by:$\f

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Two capacitors in series and the third in parallel with the combination of the first two

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(C) Let the three capacitors be $C_1 = C_2 = C_3 = 2\,\mu F$.If we connect two capacitors in series,their equivalent capacitance $C_s$ is given by:$\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{2} + \frac{1}{2} = 1 \implies C_s = 1\,\mu F$.Now,if we connect this combination in parallel with the third capacitor $C_3 = 2\,\mu F$,the total equivalent capacitance $C_{eq}$ is:$C_{eq} = C_s + C_3 = 1\,\mu F + 2\,\mu F = 3\,\mu F$.Thus,the correct arrangement is two capacitors in series and the third in parallel with them.

To obtain $3\,\mu F$ capacitance from three capacitors of $2\,\mu F$ each,they will be arranged as:

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