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Standard XII

Physics

Potentiometer

Two cells, ha...

Question

Two cells, having the same e.m.f., are connected in series through an external resistance R. Cell have internal resistances r1 and r2 $(r 1 > r 2)$ respectively. When the circuit is closed, the potential difference across the first cell is zero. The value of R is:-

r_{1} - r_{2}

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\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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Solution

The correct option is C $r_{1} - r_{2}$
According to question
$E - I r_{1}$ $= 0$ & $I =$ $\frac{E + E}{r_{1} + r_{2} + R}$
$∴ \frac{E}{r_{1}}$ = $\frac{2 E}{r_{1} + r_{2} + R}$

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

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Two cells, having the same e.m.f., are connected in series through an external resistance R. Cell have internal resistances r1 and r2 (r1>r2) respectively. When the circuit is closed, the potential difference across the first cell is zero. The value of R is:-

The correct option is C r1−r2According to questionE−Ir1 =0 & I= E+Er1+r2+R ∴Er1 = 2Er1+r2+R⇒r1 + r2 +R = 2r1 ⇒ R=r1 - r2

Two cells, having the same e.m.f., are connected in series through an external resistance R. Cell have internal resistances r1 and r2 $(r 1 > r 2)$ respectively. When the circuit is closed, the potential difference across the first cell is zero. The value of R is:-

The correct option is C $r_{1} - r_{2}$
According to question
$E - I r_{1}$ $= 0$ & $I =$ $\frac{E + E}{r_{1} + r_{2} + R}$
$∴ \frac{E}{r_{1}}$ = $\frac{2 E}{r_{1} + r_{2} + R}$

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