Question

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

• A. $1$
• B. $n$
• C. $n^{2}$
• D. $1 / n$

Answer 1

Answer: (A)

In case I,

Total circuit resistance,
$R_{ eq }=R_{0}+\frac{R}{n}=\frac{n R_{0}+R}{n}$
Circuit current $=i_{1}=\frac{E}{R_{ eq }} $
$\Rightarrow i_{1}=\frac{n E}{n R_{0}+R}$
Power dissipated in $n$ resistors,
$P_{1}=i_{1}^{2} \cdot\left(\frac{R}{n}\right)=\frac{n E^{2} R}{\left(n R_{0}+R\right)^{2}}$
In case II,

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,
$P_{2}=\left(i_{2}^{2}\right)(n R)=\frac{n E^{2} R}{\left(R_{0}+n R\right)^{2}}$
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 $
or $ \frac{R_{0}}{R}=1$