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Potentiometer

Two cells wit...

Question

Two cells with the same emf E and different internal resistance $r_{1} a n d r_{2}$ , are connected in series to an external resistance R. The value of R so that the pd across the ﬁrst cell is zero, is:

A

\sqrt{r_{1} r_{2}}

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B

r_{1} + r_{2}

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C

r_{1} - r_{2}

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D

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

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Solution

The correct option is C $r_{1} - r_{2}$

Current through the circuit
$I = \frac{E + E}{r_{1} + r_{2} + R} = \frac{2 E}{r_{1} + r_{2} + R} . . . f r o m (1)$
$V_{1} = 0$ (given) voltage across $1^{s t}$ cell is zero.
$E - I r_{1} = 0 E = I r_{1} E = \frac{2 E}{r_{1} + r_{2} + R} r_{1} f r o m (1) 2 r 1 = r_{1} + r_{2} = R R = r_{1} - r_{2}$

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