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Standard XII

Chemistry

Kohlrausch Law

The maximum p...

Question

The maximum power dissipated in an external resistance R, when connected to a cell of emf E and internal resistance r, will be.

\frac{E^{2}}{r}

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\frac{E^{2}}{2 r}

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\frac{E^{2}}{3 r}

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\frac{E^{2}}{4 r}

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Solution

The correct option is D $\frac{E^{2}}{4 r}$
The current drawn by the resistance is
$i = \frac{E}{r + R}$

The power dissipated in the load resistance is
$P = i^{2} R$

$\Rightarrow P = \frac{E^{2} R}{(r + R)^{2}}$

Power will be maximum when $r = R$
$\Rightarrow P_{m a x} = \frac{E^{2}}{4 r}$ .

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The maximum power dissipated in an external resistance R, when connected to a cell of emf E and internal resistance r, will be.

The correct option is D E24rThe current drawn by the resistance isi=Er+RThe power dissipated in the load resistance isP=i2R⇒P=E2R(r+R)2Power will be maximum when r=R⇒Pmax=E24r.

The correct option is D $\frac{E^{2}}{4 r}$
The current drawn by the resistance is
$i = \frac{E}{r + R}$

The power dissipated in the load resistance is
$P = i^{2} R$

$\Rightarrow P = \frac{E^{2} R}{(r + R)^{2}}$

Power will be maximum when $r = R$
$\Rightarrow P_{m a x} = \frac{E^{2}}{4 r}$ .