Q63 / 75 · JEE Main 2026 (21 January Shift 2) · Physics
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$R=2 r$
- B$R=r / 2$
- Correct answerC$R=r$
- D$R=\sqrt{2} r$
Solution
Power consumed in external resistance R is $P = \frac{E^2 R}{(R + r)^2}$.
To find maximum, differentiate with respect to R:
$\frac{dP}{dR} = E^2 \frac{(R + r)^2 - 2R(R + r)}{(R + r)^4} = E^2 \frac{r - R}{(R + r)^3}$.
$\frac{dP}{dR} = 0$ gives $R = r$.
This is the maximum power transfer theorem: maximum power is delivered when the external load resistance equals the source internal resistance.
To find maximum, differentiate with respect to R:
$\frac{dP}{dR} = E^2 \frac{(R + r)^2 - 2R(R + r)}{(R + r)^4} = E^2 \frac{r - R}{(R + r)^3}$.
$\frac{dP}{dR} = 0$ gives $R = r$.
This is the maximum power transfer theorem: maximum power is delivered when the external load resistance equals the source internal resistance.
Hence, the correct answer is option: 'C'.