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Two cells of same emf but different internal resistances $\mathrm{r_{1}}$ and $\mathrm{r_{2}}$ are connected in series with a resistance $\mathrm{R}$ . The value of resistance $\mathrm{R}$ .The value of resistance $\mathrm{R}$ ,for which the potential difference across second cell is zero, is :
Option: 1
$\mathrm{r_{2}-r_{1}}$

Option: 2
$\mathrm{r_{1}-r_{2}}$

Option: 3
$\mathrm{r_{1}}$

Option: 4
$\mathrm {r_{2}}$

Answers (1)

best_answer

$\mathrm{I=\frac{E_{1}+E_{2}}{r_{1}+r_{2}+R}=\frac{2 E_{1}}{r_{1}+r_{2}+R}}\\$ ..........(1)

$\mathrm{\left(\because E_{1}=E_{2} \cdots(\text { given })\right)}$

The potential drop across the second cell is zero

$\mathrm{\therefore V=E_{2}-I r_{2}=0 \ldots \text { (given) }}\\$

$\mathrm{E_{2}=E_{1}=I r_{2} }$ .....(2)

From eq (1) and (2)

$\mathrm{\frac{E_{1}}{r_{2}}=\frac{2 E_{1}}{r_{1}+r_{2}+R}}\\$

$\mathrm{r_{1}+r_{2}+R=2 r_{2}}\\$

$\mathrm{r_{1}+R=r_{2}}\\$

$\mathrm{R=r_{2}-r_{1}}$

Hence the correct answer is option 1.

Posted by

Ritika Jonwal

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