Two cells with the same e.m.f. E and different internal resistances are connected in series to an external resistances R. The value of R so that the potential difference across the first cell to be zero is:

\sqrt{r_{1} r_{2}}

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r_{1} + r_{2}

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

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r_{1} - r_{2}

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Solution

The correct option is C $r_{1} - r_{2}$
$\Rightarrow I [R + r_{1} + r_{2}] = E_{1} + E_{2}$
$I = \frac{E_{1} + E_{2}}{R + r_{1} + r_{2}} [G i v e n E_{1} = E_{2}]$
$I = \frac{2 E_{1}}{R + r_{1} + r_{2}}$
P.D across 1st cell $= E_{1} - I r_{1} = 0$
$\Rightarrow \frac{E_{1}}{r_{1}} = I$
$\Rightarrow \frac{E_{1}}{r_{1}} = \frac{2 E_{1} 2}{R + r_{1} + r_{2}}$
$2 r_{1} = R + r_{1} + r_{2}$
$\Rightarrow R = r_{1} - r_{2}$

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Two cells with the same e.m.f. E and different internal resistances are connected in series to an external resistances R. The value of R so that the potential difference across the first cell to be zero is:

The correct option is C r1−r2⇒I[R+r1+r2]=E1+E2 I=E1+E2R+r1+r2[GivenE1=E2] I=2E1R+r1+r2P.D across 1st cell =E1−Ir1=0 ⇒E1r1=I ⇒E1r1=2E12R+r1+r2 2r1=R+r1+r2 ⇒R=r1−r2