Two sources of equal emf are connected to an external resistance R. The internal resistances of two sources are $R_{1}$ and $R_{2}$ ( $R_{2}$ > $R_{1}$ ). If the potential difference across the source having internal resistance $R_{2}$ is zero, then;

R = R_{1} R_{2} / (R_{1} + R_{2})

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R = R_{1} R_{2} / (R - R_{1})

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R = R_{2} \times (R_{1} + R_{2}) / (R_{2} - R_{1})

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R = R_{2} - R_{1}

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Solution

The correct option is D $R = R_{2} - R_{1}$
$I = \frac{2 E}{R_{1} + R_{2} + R}$
According to the question,
$⟹ - (V_{A} - V_{B}) = E - I R_{2} ⟹ 0 = E - I R_{2} ⟹ E = I R_{2}$
$∴ E = I R_{2} = \frac{2 E}{R_{1} + R_{2} + R} R_{2} ⟹ R_{1} + R_{2} + R = 2 R_{2} ⟹ R = R_{2} - R_{1}$

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Two sources of equal emf are connected to an external resistance R. The internal resistances of two sources are R1 and R2(R2>R1). If the potential difference across the source having internal resistance R2 is zero, then;

The correct option is D R=R2−R1I=2ER1+R2+RAccording to the question,⟹−(VA−VB)=E−IR2⟹0=E−IR2⟹E=IR2∴E=IR2=2ER1+R2+RR2⟹R1+R2+R=2R2⟹R=R2−R1

Two sources of equal emf are connected to an external resistance R. The internal resistances of two sources are $R_{1}$ and $R_{2}$ ( $R_{2}$ > $R_{1}$ ). If the potential difference across the source having internal resistance $R_{2}$ is zero, then;

The correct option is D $R = R_{2} - R_{1}$
$I = \frac{2 E}{R_{1} + R_{2} + R}$
According to the question,
$⟹ - (V_{A} - V_{B}) = E - I R_{2} ⟹ 0 = E - I R_{2} ⟹ E = I R_{2}$
$∴ E = I R_{2} = \frac{2 E}{R_{1} + R_{2} + R} R_{2} ⟹ R_{1} + R_{2} + R = 2 R_{2} ⟹ R = R_{2} - R_{1}$