Explanation (Complete Step-by-Step Derivation)

Current in the circuit:

\[ I = \frac{E}{R + r} \]

Power consumed in external resistance R:

\[ P = I^2 R \]

Substitute value of current:

\[ P = \left(\frac{E}{R+r}\right)^2 R \]

\[ P = \frac{E^2 R}{(R+r)^2} \]

To find maximum power, differentiate P with respect to R and equate to zero:

\[ \frac{d}{dR}\left(\frac{E^2 R}{(R+r)^2}\right)=0 \]

Since E² is constant, consider:

\[ f(R)=\frac{R}{(R+r)^2} \]

Using quotient rule:

\[ \frac{d}{dR}\left(\frac{R}{(R+r)^2}\right) = \frac{(R+r)^2(1)-R(2)(R+r)}{(R+r)^4} \]

Set numerator equal to zero:

\[ (R+r)^2 - 2R(R+r)=0 \]

\[ (R+r)\left[(R+r)-2R\right]=0 \]

\[ (R+r)(r-R)=0 \]

Ignoring trivial solution, we get:

\[ r-R=0 \]

\[ R=r \]

Therefore, power is maximum when:

\[ \boxed{R=r} \]

This is known as Maximum Power Transfer Theorem.

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