Two cells with the same emf $E$ and different internal resistances $r_{1}$ and $r_{2}$ are connected in series to an external resistance $R$ . The value of $R$ so that the potential difference across the first cell be zero, is

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Answer 1

The two cells have potential $E$ each and respective internal resistance $r_{1}$ and $r_{2}$ .
Let the current flowing in the circuit be $i$ .

Using Kirchhoff's voltage law:
$\Rightarrow - E + i r_{1} + i R - E + i r_{2} = 0$
$\Rightarrow i = \frac{2 E}{( R + r _{1} + r _{2} )}$

Potential across first cell, $V_{A B} = i r_{1} - E = \frac{2 E r _{1}}{( R + r _{1} + r _{2} )} - E$

For $V_{A B} = 0$ , we get: $\frac{2 E r _{1}}{( R + r _{1} + r _{2} )} - E = 0$

$\Rightarrow$ $\frac{2 r _{1}}{( R + r _{1} + r _{2} )} = 1$

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

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

Answer 2

The two cells have potential

E

each and respective internal resistance

r_{1}

and

r_{2}

.

Let the current flowing in the circuit be

i

.

Using Kirchhoff's voltage law:

\Rightarrow - E + i r_{1} + i R - E + i r_{2} = 0

\Rightarrow i = \frac{2 E}{( R + r _{1} + r _{2} )}

Potential across first cell,

V_{A B} = i r_{1} - E = \frac{2 E r _{1}}{( R + r _{1} + r _{2} )} - E

For

V_{A B} = 0

, we get:

\frac{2 E r _{1}}{( R + r _{1} + r _{2} )} - E = 0

\Rightarrow

\frac{2 r _{1}}{( R + r _{1} + r _{2} )} = 1

\Rightarrow 2 r_{1} = R + r_{1} + r_{2}

\Rightarrow R = r_{1} - r_{2}