Question

A battery with EMF $E$ and internal resistance $r$ is connected across a resistance $R$. The power consumption in $R$ will be maximum when :

• A. $R = r$
• B. $R = r/2$
• C. $R = \sqrt{2} r$
• D. $R = 2r$

Hint:

Maximum Power Transfer Theorem: Power is max when $R_{ext} = R_{int}$. The efficiency at this point is exactly 50%.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This question follows the Maximum Power Transfer Theorem. The power delivered to a load by a source is maximum when the load resistance equals the internal resistance of the source.

Step 2: Detailed Explanation:
The current in the circuit is $I = \frac{E}{R+r}$.
Power dissipated in $R$ is $P = I^2 R = \frac{E^2 R}{(R+r)^2}$.
To find the maximum power, differentiate $P$ with respect to $R$ and set it to zero:
\[ \frac{dP}{dR} = E^2 \left[ \frac{(R+r)^2(1) - R(2)(R+r)}{(R+r)^4} \right] = 0 \]
\[ (R+r)^2 - 2R(R+r) = 0 \]
\[ (R+r) [ R+r - 2R ] = 0 \]
\[ r - R = 0 \implies R = r \]

Step 3: Final Answer:
Power is maximum when $R = r$.