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Two sources of equal emfs are connected in series. This combination is connected to an external resistance $\mathrm{\mathrm{R}. }$ The internal resistances of the two sources are $\mathrm{\mathrm{r}_{1} }$ and $\mathrm{r_{2} \left(r_{1}>r_{2}\right).}$ If the potential difference across the source of internal resistance $\mathrm{r_{1}}$ is zero, then the value of $\mathrm{\mathrm{R}}$ will be :
Option: 1
$\mathrm{r_{1}-r_{2}}$

Option: 2
$\mathrm{\frac{r_{1} r_{2}}{r_{1}+r_{2}}}$

Option: 3
$\mathrm{\frac{r_{1}+r_{2}}{2}}$

Option: 4
$\mathrm{r_{2}-r_{1}}$

Answers (1)

best_answer

Let the source of emf be e

$\mathrm{I=\frac{2 e}{\left(R+r_1+r_2\right)} }$

$\mathrm{V=e-I r_1=0 \text { (Given) } }$

$\mathrm{O=e-\frac{2 e r_1}{\left(R+r_1+r_2\right)} }$

$\mathrm{1=\frac{2 r_1}{(\left.R+r_1+r_2\right)} }$

$\mathrm{(\left.R+r_1+r_2\right)=2 r_1 }$

$\mathrm{R+r_2=r_1 }$

$\mathrm{R=r_1-r_2}$

Hence 1 is correct option.

Posted by

Ajit Kumar Dubey

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