The minimum number of capacitors of 2,mu F ca | vedclass

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(B) To obtain a total capacitance of $5\,\mu F$ using $2\,\mu F$ capacitors,we can use a combination of series and parallel connections.Consider the a

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$4$

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(B) To obtain a total capacitance of $5\,\mu F$ using $2\,\mu F$ capacitors,we can use a combination of series and parallel connections.Consider the arrangement shown in the image:$1$. Two capacitors of $2\,\mu F$ are connected in parallel. Their equivalent capacitance is $C_1 = 2\,\mu F + 2\,\mu F = 4\,\mu F$.$2$. Two capacitors of $2\,\mu F$ are connected in series. Their equivalent capacitance is $C_2 = \frac{2\,\mu F 	imes 2\,\mu F}{2\,\mu F + 2\,\mu F} = 1\,\mu F$.$3$. Now,connecting $C_1$ and $C_2$ in parallel,the total equivalent capacitance is $C_{eq} = C_1 + C_2 = 4\,\mu F + 1\,\mu F = 5\,\mu F$.Thus,the total number of capacitors required is $2 + 2 = 4$.

The minimum number of capacitors of $2\,\mu F$ capacitance each required to obtain a resultant capacitance of $5\,\mu F$ is

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