Question

Consider the circuits shown in the figures. Both the circuits are taking the same current from battery but current through R in the second circuit is $\frac{1}{10}th$ of current through R in the first circuit. If R is $11\Omega$ the value of $R_1\left(in\Omega \right)$ is

Answer 1

In first figure, from ohm's law
$i=E/R$

In second figure by KCL at junction A, current through

$R_2=i-\frac{i}{10}=\frac{9i}{10}$

Potential difference across $R_2$ = Potential difference across R (parallel combination)

$\Rightarrow R_2\times \frac{9}{10}i=R\times \frac{i}{10}$

$R_2=\frac{R}{9}=\frac{11}{9}\Omega$
for $R_2$ and R connected in parallel, equivalent resistance

$R_{eq}=\frac{R_2\times R}{(R_2+R)}=\frac{\frac{11}{9}\times \frac{11}{1}}{\frac{11}{9}+\frac{11}{1}}=\frac{11}{10}\Omega$

$R_1$ is in series with above combination
for second figure, total circuit resistance

$=\frac{11}{10}\Omega +R_1=R=11\Omega$

$\Rightarrow R_1=9.9\Omega$

Hence $R_1=9.9\Omega$