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Q.

Under what condition, current passing through the resistance R can be increased by short circuiting the battery of emf E₂? The internal resistances of the two batteries are r₁ and r₂respectively

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a

$E_{2} r_{1} > E_{1} (R + r_{2})$

b

$E_{2} r_{2} > E_{1} (R + r_{2})$

c

$E_{1} r_{1} > E_{2} (R + r_{1})$

d

$E_{1} r_{2} > E_{2} (R + r_{1})$

answer is B.

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Detailed Solution

Initially $I_{1} = \frac{E_{1} + E_{2}}{R + r_{1} + r_{2}} .$

After short circuiting the second cell, $I_{2} = \frac{E_{1}}{R + r_{1}}$

The condition required is $I_{2} > I_{1} \cdot$

$\Rightarrow \frac{E_{1}}{R + r_{1}} > \frac{E_{1} + E_{2}}{R + r_{1} + r_{2}}$

$E_{1} R + E_{1} r_{1} + E_{1} r_{2} > E_{1} R + E_{1} r_{1} + E_{2} R + E_{2} r_{1}$

$E_{1} r_{2} > E_{2} (R + r_{1})$

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Under what condition, current passing through the resistance R can be increased by short circuiting the battery of emf E2? The internal resistances of the two batteries are r1 and r2 respectively