In the circuit shown, $n$-identical resistors $R$ are connected in parallel $(n > 1)$ and the combination is connected in series to another resistor $R_0$. In the adjoining circuit $n$ resistors of resistance $R$ are all connected in series alongwith $R_0$. The batteries in both circuits are identical and net power dissipated in the $n$ resistors in both circuits is same. The ratio $R_0 / R$ is
KVPY 2012, Diffcult
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(a)

In case $I$,

Total circuit resistance,

$R_{\text {eq }}=R_0+\frac{R}{n}=\frac{n R_0+R}{n}$

Circuit current $=i_1=\frac{E}{R_{ eq }} \Rightarrow i_1=\frac{n E}{n R_0+R}$

Power dissipated in $n$ resistors,

$P_1=i_1^2 \cdot\left(\frac{R}{n}\right)=\frac{n E^2 R}{\left(n R_0+R\right)^2}$

Total circuit resistance, $R_{ eq }=R_0+n R$ Current in circuit is

$i_2=\frac{E}{R_{ eq }}=\frac{E}{R_0+n R}$

Power dissipated in $n$ resistors,

$P_2=\left(i_2^2\right)(n R)=\frac{n E^2 R}{\left(R_0+n R\right)^2}$

$\text { As, } P_1=P_2$

$\Rightarrow \frac{n E^2}{\left(n R_0+R\right)^2}=\frac{n E^2 R}{\left(R_0+n R\right)^2}$

$\Rightarrow n R_0+R=R_0+n R$

$\Rightarrow (n-1) R_0=(n-1) R$

$\Rightarrow R_0=R \text { or } \frac{R_0}{R}=1$

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