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
$R_{2}=i-\frac{i}{10}=\frac{9 i}{10}$
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 $=\frac{11}{10}+R_{1}=R=11$
$\Rightarrow R_{1}=9.9\, \Omega$