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+3 R}{3}} $
$\Rightarrow \frac{1}{5}=\frac{3}{3(2+3 R)} $
$ \Rightarrow 6+9\, R=15$
$ \Rightarrow 9 \,R=15-6 $
$\Rightarrow 9 \,R=9 $
or$ R=1 \,\Omega $