Three capacitors each of capacity 4,mu F are | vedclass

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(B) Let the three capacitors be $C_1 = C_2 = C_3 = 4\,\mu F$.If we connect two capacitors in series,their equivalent capacitance $C_s$ is given by:$\f

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Connecting two in series and one in parallel

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(B) Let the three capacitors be $C_1 = C_2 = C_3 = 4\,\mu F$.If we connect two capacitors in series,their equivalent capacitance $C_s$ is given by:$\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$So,$C_s = 2\,\mu F$.Now,if we connect this combination in parallel with the third capacitor $C_3$,the total effective capacitance $C_{eq}$ is:$C_{eq} = C_s + C_3 = 2\,\mu F + 4\,\mu F = 6\,\mu F$.Thus,connecting two capacitors in series and one in parallel gives an effective capacitance of $6\,\mu F$.

Three capacitors each of capacity $4\,\mu F$ are to be connected in such a way that the effective capacitance is $6\,\mu F$. This can be done by:

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