1. Current in the circuit

For a cell of emf $E$ and internal resistance $r$ connected to an external resistance $R$, the current is

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

2. Power delivered to the external resistance

The power consumed by the external resistance $R$ is

$P = I^2 R$

Substituting $I$:

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

3. Condition for maximum power

We must maximize

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

Since $E^2$ is constant, maximize

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

Differentiate with respect to $R$:

$\frac{df}{dR} = \frac{(R+r)^2 - R\cdot 2(R+r)}{(R+r)^4}$

Simplifying the numerator:

$\frac{df}{dR} = \frac{(R+r)[(R+r)-2R]}{(R+r)^4} = \frac{(R+r)(r-R)}{(R+r)^4}$

For maximum power,

$\frac{df}{dR}=0$

This gives

$r-R=0 \Rightarrow R=r$

4. Check with options

• A: $R=1000r$ ❌
• B: $R=r$ ✅
• C: $R=2r$ ❌
• D: $R=0.001r$ ❌

So, the power delivered to the external resistance is maximum when

$\boxed{R=r}$

This is the maximum power transfer condition.