Consider the circuit shown in the figure. The | vedclass

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(C) For the given circuit,the equivalent resistance $R^{\prime}$ of the parallel combination of $R$ and $X$ is given by:$\frac{1}{R^{\prime}} = \frac{

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$X=\frac{R}{2}$

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(C) For the given circuit,the equivalent resistance $R^{\prime}$ of the parallel combination of $R$ and $X$ is given by:$\frac{1}{R^{\prime}} = \frac{1}{R} + \frac{1}{X} = \frac{R+X}{RX} \implies R^{\prime} = \frac{RX}{R+X}$The total current $i$ in the circuit is:$i = \frac{E}{R + R^{\prime}} = \frac{E}{R + \frac{RX}{R+X}} = \frac{E(R+X)}{R^2 + RX + RX} = \frac{E(R+X)}{R^2 + 2RX}$The voltage $V_X$ across the resistance $X$ is equal to the voltage across $R^{\prime}$:$V_X = i R^{\prime} = \left( \frac{E(R+X)}{R^2 + 2RX} \right) \left( \frac{RX}{R+X} \right) = \frac{ERX}{R^2 + 2RX} = \frac{EX}{R + 2X}$The power $P_X$ dissipated in resistance $X$ is:$P_X = \frac{V_X^2}{X} = \frac{(EX)^2}{X(R+2X)^2} = \frac{E^2 X}{(R+2X)^2}$For $P_X$ to be independent of small variations in $X$,we set $\frac{dP_X}{dX} = 0$:$\frac{dP_X}{dX} = E^2 \left[ \frac{(R+2X)^2(1) - X(2)(R+2X)(2)}{(R+2X)^4} \right] = 0$$(R+2X) - 4X = 0$$R - 2X = 0 \implies X = \frac{R}{2}$

Consider the circuit shown in the figure. The value of the resistance $X$ for which the thermal power generated in it is independent of small variations of its resistance is

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