A battery with EMF E and internal resistance r is connected acros… — Solution | JEE Physics

MCQEasyJEE 2026Cells, EMF & Internal Resistance

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

Answer

Correct answer:A

Standard Method

Given: A battery has EMF $E$ , internal resistance $r$ , and load resistance $R$ .

Find: The condition for maximum power consumption in $R$ .

This question follows the Maximum Power Transfer Theorem. 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 equal to zero:

\frac{dP}{dR} = E^2 \left[ \frac{(R+r)^2(1) - R(2)(R+r)}{(R+r)^4} \right] = 0

So,

(R+r)^2 - 2R(R+r) = 0

(R+r)(R+r-2R) = 0

(R+r)(r-R) = 0

Hence,

r - R = 0 \implies R = r

Therefore, the power consumption is maximum when $R = r$ . The correct option is A.

Using the condition for maximum current instead of maximum power. Maximum current occurs for very small external resistance, but maximum power transfer occurs when $R = r$ . Always write the power expression first.
Forgetting that power is to be calculated in the load resistance $R$ , not in the whole circuit. The correct expression is $P = I^2R$ , not total source power.
Differentiating $\frac{E^2 R}{(R+r)^2}$ incorrectly. The denominator also depends on $R$ , so the quotient or product rule must be applied carefully.

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.