Q. 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) $7.7\ \Omega$
Explanation
(a) In figure (b) current through $R_{2} = i - \frac{i}{10} = \frac{9i}{10}$
Potential difference across $R_{2} =$ Potential difference across
$R \Rightarrow R_{2} \times \frac{9}{10}i = R \times \frac{i}{10}\ i.e.\ R_{2} = \frac{R}{9} = \frac{11}{9}\Omega$
$R_{eq} = \frac{R_{2} \times R}{(R_{2} + R)} = \frac{\frac{11}{9} \times \frac{11}{1}}{\frac{11}{9} + \frac{11}{1}} = \frac{11}{10}\Omega$
Total circuit resistance $= \frac{11}{10} + R_{1} = R = 11 \Rightarrow R_{1} = 9.9\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.