Question

A battery of internal resistance $4\Omega$ is connected to the network of resistances as shown in the figure. In order that the maximum power can be delivered to the network, the value of 𝑅 in $\Omega$ should be

• A. $\frac{4}{9}$
• B. $2$
• C. $\frac{8}{3}$
• D. $18$

Answer 1

The correct option is B $2$

We know that
$P=\frac{E^2}{R_{eq}+r}$
For maximum power $r=R_{eq}$
$r=4\Omega$
Let us now try to find $R_{eq}$ in this circuit


This circuit can be further represented as

This is a classic case of wheatstone's bridge. It means that no current will flow through the branch containing $6R$ resistance.
Now, the equivalent resistance will be
$R_{eq}=\frac{3R\times 6R}{3R+6R}$
$⟹R_{eq}=2R$
Now, for maximum power to be delivered
$r=2R$
$4\Omega =2R$
$⟹R=2\Omega$