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Simple Circuit

Two cells, ha...

Question

Two cells, having the same emf are connected in series through an external resistance $R$ . Cells have internal resistance $r_{1}$ and $r_{2} (r_{1} > r_{2})$ respectively. When the circuit is closed, the potential difference across the first cell is zero, The value of $R$ is

A

r_{1} + r_{2}

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B

r_{1} - r_{2}

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C

\frac{r_{1} + r_{2}}{2}

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D

\frac{r_{1} - r_{2}}{2}

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Solution

The correct option is B $r_{1} - r_{2}$
Current inn the circuit is given by Ohm's law.
Net resistance of the circuit= $R_{1} + R_{2} + R$
Net emf in series= $E + E = 2 E$
Therefore, from Ohm's law, current in the circuit,
$I = \frac{N e t e m f}{n e t r e s i s t a n c e}$
$i m p l i e s I = \frac{2 E}{r_{1} + r_{2} + R} \to (1)$
It is given that as circuit is closed, potential difference across the first cell is zero.
$V = E - I r_{1} = 0$
$⟹ I = \frac{E}{r_{1}} \to (2)$
Equating $(1)$ and $(2)$
$\frac{E}{r_{1}} = \frac{2 E}{r_{1} + r_{2} + R}$
$2 r_{1} = r_{1} + r_{2} + R$
$∴$ External Resistance= $R = r_{1} - r_{2}$

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