Two cells E 1 and E 2 having equal e.m.f ' E ' and internal resistances r 1 and r 2 ( r 1 r 2 ) respectively are connected in series. This combination is connected to an extern

Question

Two cells E 1 ​ and E 2 ​ having equal e.m.f ' E ' and internal resistances r 1 ​ and r 2 ​ ( r 1 ​ > r 2 ​ ) respectively are connected in series. This combination is connected to an external resistance ' R '. It is observed that the potential difference across the cell E 1 ​ becomes zero. The value of R will be

TestHub · Accepted Answer

The total e.m.f. of the series combination is $E_{e q} = E + E = 2 E$ .

The total resistance of the circuit is $R_{e q} = R + r_{1} + r_{2}$ .

The current flowing through the circuit is given by Ohm's law:

$I = \frac{2 E}{R + r _{1} + r _{2}}$

The potential difference across the first cell $E_{1}$ is $V_{1} = E - I r_{1}$ .

It is given that the potential difference across $E_{1}$ is zero, so $V_{1} = 0$ .

$E - I r_{1} = 0 \Rightarrow I = \frac{E}{r _{1}}$

Equating the two expressions for the current $I$ :

$\frac{2 E}{R + r _{1} + r _{2}} = \frac{E}{r _{1}}$

Dividing both sides by $E$ and cross-multiplying:

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

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

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

Answer: $r_{1} - r_{2}$

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