The power delivered by a source of internal resistance $‘ r^{'}$ to a load resistance $R$ is maximum when:

R = r

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R > r

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R < r

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R >> r

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Solution

The correct option is A $R = r$
Effective resistance of circuit $= R + r$

Current in circuit $= I = \frac{V}{R + r}$

Power delivered to $R$ is $P = I^{2} R = \frac{V^{2} R}{(R + r)^{2}}$

Differentiating with respect to $R :$

$\frac{d P}{d R} = V^{2} \frac{(R + r)^{2} - 2 R (R + r)}{(R + r)^{4}}$

$⟹ \frac{d P}{d R} = V^{2} \frac{(R + r) (r - R)}{(R + r)^{4}} = 0$ for maximum

$⟹ r - R = 0$

$⟹ r = R$

Answer-(A)

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The power delivered by a source of internal resistance ‘r′ to a load resistance R is maximum when:

The correct option is A R=rEffective resistance of circuit=R+rCurrent in circuit=I=VR+rPower delivered to R is P=I2R=V2R(R+r)2Differentiating with respect to R:dPdR=V2(R+r)2−2R(R+r)(R+r)4⟹dPdR=V2(R+r)(r−R)(R+r)4=0 for maximum⟹r−R=0⟹r=RAnswer-(A)

The power delivered by a source of internal resistance $‘ r^{'}$ to a load resistance $R$ is maximum when: