In the circuit shown,n-identical resistors R | vedclass

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(A) In case $I$ (parallel combination):Total circuit resistance,$R_{	ext{eq}, 1} = R_0 + \frac{R}{n} = \frac{n R_0 + R}{n}$.Circuit current,$i_1 = \f

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(A) In case $I$ (parallel combination):Total circuit resistance,$R_{	ext{eq}, 1} = R_0 + \frac{R}{n} = \frac{n R_0 + R}{n}$.Circuit current,$i_1 = \frac{E}{R_{	ext{eq}, 1}} = \frac{n E}{n R_0 + R}$.Power dissipated in the $n$ resistors,$P_1 = i_1^2 \cdot \left(\frac{R}{n}ight) = \left(\frac{n E}{n R_0 + R}ight)^2 \cdot \frac{R}{n} = \frac{n E^2 R}{(n R_0 + R)^2}$.In case $II$ (series combination):Total circuit resistance,$R_{	ext{eq}, 2} = R_0 + n R$.Circuit current,$i_2 = \frac{E}{R_{	ext{eq}, 2}} = \frac{E}{R_0 + n R}$.Power dissipated in the $n$ resistors,$P_2 = i_2^2 \cdot (n R) = \left(\frac{E}{R_0 + n R}ight)^2 \cdot n R = \frac{n E^2 R}{(R_0 + n R)^2}$.Given $P_1 = P_2$:$\frac{n E^2 R}{(n R_0 + R)^2} = \frac{n E^2 R}{(R_0 + n R)^2}$.This implies $(n R_0 + R)^2 = (R_0 + n R)^2$.Taking the square root,$n R_0 + R = R_0 + n R$ (since $R_0, R, n > 0$).$(n - 1) R_0 = (n - 1) R$.Since $n > 1$,we get $R_0 = R$,so $\frac{R_0}{R} = 1$.

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