Question

Consider the circuits shown in the figure. Both the circuits are taking 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 $R_1$

• A. $9.9\Omega$
• B. $11\Omega$
• C. $8.8\Omega$
• D. $7.7\Omega$

Answer 1

Answer: (A)

In Figure (b) current through
$R_2=i-\frac{i}{10}=\frac{9i}{10}$
Potential difference across $R_2=$
Potential difference across $R$
$\Rightarrow R_2\times \frac{9}{10}i=R\times \frac{i}{10}$
i .e $R_2=\frac{R}{9}=\frac{11}{9}\Omega$
$R_{eq}=\frac{R_2\times R}{\left(R_2+R\right)}=\frac{\frac{11}{9}\times \frac{11}{1}}{\frac{11}{9}+\frac{11}{1}}$
$=\frac{11}{10}\Omega$
Total circuit resistance $=\frac{11}{10}+R_1=R=11$
$\Rightarrow R_1=9.9\Omega$