Question

Two sources of equal emf are connected to a resistance $R$. The internal resistance of these sources are $r_1$ and $r_2 (r_1 > r_2 )$. If the potential difference across the source having internal resistance $r_2$ is zero, then

• A. $R = \frac{r_1 r_2}{r_2 - r_1}$
• B. $R = r_2 \left( \frac{r_1 + r_2}{r_2 - r_1} \right)$
• C. $R = \left( \frac{r_1 r_2}{r_2 - r_1} \right)$
• D. $R = r_2 - r_1$

Answer 1

Answer: (D)

$R = r_2 - r_1$

Answer 2

Let E be the emf of each source. When they are connected in series, then the current in the circuit is given by
$I = \frac{E_{tot}}{R_{tot}} = \frac{E +E}{r_{1 } + r_{2} + R}$
$= \frac{2E}{r_{1} + r_{2} + R}$
So, potential drop across the cell of internal resistance
$r_{2} , \left(\frac{2E}{r_{1} + r_{2} + R}\right) r_{2}$
Hence, $E - \frac{2E}{ \left(r_{1} + r_{2} + R\right)} r_{2} = 0$
$r_{1} + r_{2} + R = 2r_{2}$
So $R = r_{2} - r_{1}$