Question

Two sources of equal emf are connected to an external resistance $R$. The internal resistances of the two sources are $R_1$ and $R_2\left(R_2>R_1\right)$. If the potential difference across the source having internal resistance $R_2$ is zero, then

• A. $R=\frac{R_2 \times\left(R_1+R_2\right)}{\left(R_2-R_1\right)}$
• B. $R=R_2-R_1$
• C. $R=\frac{R_1 R_2}{\left(R_1+R_2\right)}$
• D. $R=\frac{R_1 R_2}{\left(R_2-R_1\right)}$

Answer 1

Answer: (B)

$R_{e q}=R_1+R_2+R$
$\therefore I \equiv \frac{2 E}{R_1+R_2+R}$

According to the question,
$ -\left(V_A-V_B\right)=E-I R_2$
$ 0=E-I R_2 $
$E=I R_2$
$ E=\frac{2 E}{R_1+R_2+R} R_2 $
$R_1+R_2+R=2 R_2$
$\therefore R=R_2-R_1$