Two cells of same emf E but different internal resistances $r_{1}$ and $r_{2}$ are connected in series with an external resistance $R$ . The potential drop across the first cell is found to be zero. The external resistance $R$ is

r_{1} + r_{2}

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r_{1} - r_{2}

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r_{2} - r_{1}

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r_{1} r_{2}

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Solution

The correct option is A $r_{1} - r_{2}$
Answer is B.

Batteries and cells have an internal resistance (r) which is measures in ohms. When electricity flows round a circuit the internal resistance of the cell itself resists the flow of current and so thermal (heat) energy is wasted in the cell itself.
$ε = I (R + r)$
where,
$ε$ = electromotive force in volts, V
I = current in amperes, A
R = resistance of the load in the circuit in ohms,
r = internal resistance of the cell in ohms.
The drop across the internal resistance of the first cell r1 is E. So the total drop across r2 and R must also be E. In the serially connected circuit, the current through the circuit is always the same.
So, $r_{1} = I (r_{2} + R) t h a t i s R = r_{1} - r_{2}$ .
Hence, the external resistance R is $R = r_{1} - r_{2}$ .

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Two cells of same emf E but different internal resistances r1 and r2 are connected in series with an external resistance R. The potential drop across the first cell is found to be zero. The external resistance R is

Two cells of same emf E but different internal resistances $r_{1}$ and $r_{2}$ are connected in series with an external resistance $R$ . The potential drop across the first cell is found to be zero. The external resistance $R$ is