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Standard X

Physics

Combination of Resistances

Two cells hav...

Question

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

(r_{1} + r_{2})

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(r_{1} - r_{2})

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

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

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Solution

The correct option is B $(r_{1} - r_{2})$
Current in the circuit
$I = \frac{2 E}{R + r_{1} + r_{2}}$
$∴$ Potential difference across first cell
$V = E - I_{r_{1}} = E - \frac{2 E r_{1}}{(R + r_{1} + r_{2})} = 0$
On solving we get, $R = (r_{1} - r_{2})$

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Two cells having the same emf are connected in series through an external resistance R. Cells have internal resistances r1 and r2 such that r1>r2 respectively. When the circuit is closed, the potential difference across the first cell is zero, then the value of R is

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

The correct option is B $(r_{1} - r_{2})$
Current in the circuit
$I = \frac{2 E}{R + r_{1} + r_{2}}$
$∴$ Potential difference across first cell
$V = E - I_{r_{1}} = E - \frac{2 E r_{1}}{(R + r_{1} + r_{2})} = 0$
On solving we get, $R = (r_{1} - r_{2})$