JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Current Electricity

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: (D)

Solution

With the load

R

across a cell of emf

E

and internal resistance

r

, the current is

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

and the power delivered to

R

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

. Setting

\frac{d P}{d R} = 0

gives

(R + r)^{2} - R \cdot 2 (R + r) = 0

, i.e.

(R + r) - 2 R = 0

. The power consumed in

R

is maximum when

R = r

JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Single correct option