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Standard XII

Physics

Series Combination of Cells

Two cells of ...

Question

Two cells of equal emf and of internal resistances $r_{1}$ and $r_{2} (r_{1} > r_{2})$ are connected in series. On connecting this combination to an external resistance R, it is observed that the potential difference across the first cell becomes zero. The value of R will be.

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 D $r_{1} - r_{2}$
The required circuit is shown.
$i = \frac{2 E}{R + r_{1} + r_{2}}$
Potential difference across first cell,
$V_{1} = E - i r_{1}$
$0 = E - \frac{2 E}{R + r_{1} + r_{2}} \cdot r_{1}$
$R + r_{1} + r_{2} = 2 r_{1}$
$R = r_{1} - r_{2}$ .

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Two cells of equal emf and of internal resistances r1 and r2(r1>r2) are connected in series. On connecting this combination to an external resistance R, it is observed that the potential difference across the first cell becomes zero. The value of R will be.

The correct option is D r1−r2The required circuit is shown.i=2ER+r1+r2Potential difference across first cell,V1=E−ir10=E−2ER+r1+r2⋅r1R+r1+r2=2r1R=r1−r2.

Two cells of equal emf and of internal resistances $r_{1}$ and $r_{2} (r_{1} > r_{2})$ are connected in series. On connecting this combination to an external resistance R, it is observed that the potential difference across the first cell becomes zero. The value of R will be.

The correct option is D $r_{1} - r_{2}$
The required circuit is shown.
$i = \frac{2 E}{R + r_{1} + r_{2}}$
Potential difference across first cell,
$V_{1} = E - i r_{1}$
$0 = E - \frac{2 E}{R + r_{1} + r_{2}} \cdot r_{1}$
$R + r_{1} + r_{2} = 2 r_{1}$
$R = r_{1} - r_{2}$ .