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Simple Circuit

If a battery ...

Question

If a battery of emf E and internal resistance $r$ is connected across a load of resistance R. Show that the rate at which energy is dissipated in R is maximum when R = r and this maximum power is $P = E^{2} / 4 r$ .

P_{m a x} = \frac{E^{2}}{4 r}

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P_{m a x} = \frac{E^{4}}{4 r}

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P_{m a x} = \frac{E^{2}}{8 r}

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P_{m a x} = \frac{E^{2}}{4 r^{2}}

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Solution

The correct option is A $P_{m a x} = \frac{E^{2}}{4 r}$
Total resistance of the circuit $R_{e q} = R + r$
Thus current in the circuit $I = \frac{E}{R_{e q}} = \frac{E}{R + r}$
Power dissipated $P = I^{2} R = \frac{E^{2} R}{(R + r)^{2}}$
For maxima, $\frac{d P}{d R} = 0$
$∴$ $E^{2} [\frac{(R + r)^{2} \times 1 - R (2) (R + r) \times 1}{(R + r)^{4}}] = 0$
OR $\frac{E^{2}}{(R + r)^{3}} \times (r - R) = 0$
$∴$ $r - R = 0$ $⟹ R = r$
Maximum power dissipated $P_{m a x} = \frac{E^{2} r}{(r + r)^{2}} = \frac{E^{2}}{4 r}$

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If a battery of emf E and internal resistance r is connected across a load of resistance R. Show that the rate at which energy is dissipated in R is maximum when R = r and this maximum power is P=E2/4r.

The correct option is A Pmax=E24rTotal resistance of the circuit Req=R+rThus current in the circuit I=EReq=ER+rPower dissipated P=I2R=E2R(R+r)2For maxima, dPdR=0∴ E2[(R+r)2×1−R(2)(R+r)×1(R+r)4]=0OR E2(R+r)3×(r−R)=0∴ r−R=0 ⟹R=rMaximum power dissipated Pmax=E2r(r+r)2=E24r

If a battery of emf E and internal resistance $r$ is connected across a load of resistance R. Show that the rate at which energy is dissipated in R is maximum when R = r and this maximum power is $P = E^{2} / 4 r$ .