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

We begin with a battery with EMF $$E$$ and internal resistance $$r$$ connected across an external resistance $$R$$, and we need to find the value of $$R$$ for which the power consumed in $$R$$ is maximum.

Applying Ohm's law to the complete circuit, the current is:

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

The power consumed in the external resistance $$R$$ is:

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

To find its maximum, we differentiate with respect to $$R$$ using the quotient rule, with numerator $$E^2 R$$ and denominator $$(R + r)^2$$:

$$ \frac{dP}{dR} = E^2 \cdot \frac{(R + r)^2 \cdot 1 - R \cdot 2(R + r)}{(R + r)^4} $$

This simplifies to:

$$ = E^2 \cdot \frac{(R + r) - 2R}{(R + r)^3} = E^2 \cdot \frac{r - R}{(R + r)^3} $$

Setting $$\frac{dP}{dR} = 0$$ gives:

$$ r - R = 0 $$

$$ R = r $$

For $$R < r$$, $$\frac{dP}{dR} > 0$$ (power is increasing)

For $$R > r$$, $$\frac{dP}{dR} < 0$$ (power is decreasing)

Since the derivative changes from positive to negative at $$R = r$$, this confirms it is a maximum.

This result is known as the Maximum Power Transfer Theorem: maximum power is delivered to the external load when the load resistance equals the internal resistance of the source.

The correct answer is Option (1): $$R = r$$.