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Standard XII

Physics

Series Combination of Cells

Two cells of ...

Question

Two cells of same emf are in series with an external resistance R. The internal resistance $r_{1}$ > $r_{2}$ . The p.d across the $1^{s t}$ cell is found to be zero 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 $r_{1} - r_{2}$

$V_{1} = 0; I = \frac{2 E}{R + r_{1} + r_{2}} V_{2} = E - I r_{2} I r_{2} = E - V_{2} = E - I R E = I [R + r_{2}] \to (2)$
Sub equ (2) in equ (1)
$I = \frac{2 I [R + r_{2}]}{R + r_{1} + r_{2}} r + r_{1} + r_{2} = 2 R + 2 r_{2} r_{1} - r_{2} = R$

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Two cells of same emf are in series with an external resistance R. The internal resistance r1 > r2. The p.d across the 1st cell is found to be zero the value of R is

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Two cells of same emf are in series with an external resistance R. The internal resistance $r_{1}$ > $r_{2}$ . The p.d across the $1^{s t}$ cell is found to be zero the value of R is

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

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Sub equ (2) in equ (1)
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