Two cells of same emf are connected in series. Their internal resistances are $r_{1}$ and $r_{2}$ respectively and $r_{1} > r_{2}$ . When this combination is connected to an external resistance R then the potential difference between the terminals of first cell becomes zero. In this condition the value of R will bw

\frac{r_{1} - r_{2}}{2}

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\frac{r_{1} + r_{2}}{2}

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r_{1} - r_{2}

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r_{1} + r_{2}

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Solution

The correct option is C $r_{1} - r_{2}$
Let emf of each cell is $E$ .
As they are connected in series so current in circuit is $I = \frac{E + E}{r_{1} + r_{2} + R} = \frac{2 E}{r_{1} + r_{2} + R}$
Potential across terminal of first cell is $V_{1} = E - I r_{1} = E - \frac{2 E r_{1}}{r_{1} + r_{2} + R}$
As $V_{1} = 0 \Rightarrow E - \frac{2 E r_{1}}{r_{1} + r_{2} + R} = 0$
$r_{1} + r_{2} + R - 2 r_{1} = 0$
$R = r_{1} - r_{2}$

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Two cells of same emf are connected in series. Their internal resistances are r1 and r2 respectively and r1>r2. When this combination is connected to an external resistance R then the potential difference between the terminals of first cell becomes zero. In this condition the value of R will bw

The correct option is C r1−r2Let emf of each cell is E.As they are connected in series so current in circuit is I=E+Er1+r2+R=2Er1+r2+R Potential across terminal of first cell is V1=E−Ir1=E−2Er1r1+r2+R As V1=0⇒E−2Er1r1+r2+R=0 r1+r2+R−2r1=0 R=r1−r2