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 $=i_1=\frac{E}{R_{eq}}$
$\Rightarrow i_1=\frac{nE}{n{R}_0+R}$
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,
$R_{eq}=R_0+nR$
Current in circuit is
$i_2=\frac{E}{R_{eq}}=\frac{E}{R_0+nR}$
Power dissipated in $n$ resistors,
$P_2=\left(i_2^2\right)(nR)=\frac{n{E}^{2}R}{{\left(R_0+nR\right)}^2}$
As, $P_1=P_2$
$\Rightarrow \frac{n{E}^2}{{\left(n{R}_0+R\right)}^2}=\frac{n{E}^{2}R}{{\left(R_0+nR\right)}^2}$
$\Rightarrow n{R}_0+R=R_0+nR$
$\Rightarrow (n-1)R_0=(n-1)R$
$\Rightarrow R_0=R$
or $\frac{R_0}{R}=1$