Question

In the circuit shown, if the current through the resistor $R$ is $\frac{1}{5}A$, the value of $R$ is

• A. $2\Omega$
• B. $3\Omega$
• C. $5\Omega$
• D. $1\Omega$

Answer 1

Answer: (D)

According to the question,

Equivalent circuit of above circuit shown below,

Now, equivalent emf, $=\frac{\frac{V_1}{R_1}-\frac{V_2}{R_2}}{\frac{1}{R_1}+\frac{1}{R_2}}$
$\therefore$ Internal resistance,
$r=\frac{R_1\cdot R_2}{R_1+R_2}$
$r=\frac{2\times 1}{2+1}=\frac{2}{3}\Omega$
$\therefore$Current, $I=\frac{e}{r+R}$ or $\frac{1}{5}=\frac{\frac{1}{3}}{\frac{2}{3}+R}$
$\frac{1}{5}=\frac{\frac{1}{3}}{\frac{2+3R}{3}}$
$\Rightarrow \frac{1}{5}=\frac{3}{3(2+3R)}$
$\Rightarrow 6+9R=15$
$\Rightarrow 9R=15-6$
$\Rightarrow 9R=9$
or$R=1\Omega$