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(A) The circuit consists of two parts in series. The first part has three resistors of resistance $ma$ connected in parallel. The equivalent resistanc

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$3$

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(A) The circuit consists of two parts in series. The first part has three resistors of resistance $ma$ connected in parallel. The equivalent resistance of this part is $R_1 = \frac{ma}{3}$.The second part has two resistors of resistance $\frac{a}{m}$ connected in parallel. The equivalent resistance of this part is $R_2 = \frac{a/m}{2} = \frac{a}{2m}$.The total equivalent resistance $R$ of the circuit is $R = R_1 + R_2 = \frac{ma}{3} + \frac{a}{2m}$.To find the value of $m$ for which $R$ is minimum,we differentiate $R$ with respect to $m$ and set it to zero:$\frac{dR}{dm} = \frac{a}{3} - \frac{a}{2m^2} = 0$.Solving for $m$:$\frac{a}{3} = \frac{a}{2m^2}$$2m^2 = 3$$m^2 = \frac{3}{2}$$m = \sqrt{\frac{3}{2}}$.Comparing this with the given form $m = \sqrt{\frac{x}{2}}$,we get $x = 3$.

In the given circuit,$a$ is an arbitrary constant. The value of $m$ for which the equivalent circuit resistance is minimum is given by $\sqrt{\frac{x}{2}}$. The value of $x$ is:

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