Question

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

• A. 4/9
• B. 2
• C. 8/3
• D. 18

Answer 1

The correct option is B 2

The equivalent circuit are as shown in the figure
This circuit represents Wheatstone Bridge. So, this $6R$ resistance is ineffective.
Now the equvalent resistance will be,
$\frac{1}{R_{eq}}=\frac{1}{(2R+4R)}+\frac{1}{(R+2R)}$
$\frac{1}{R_{eq}}=\frac{1}{6R}+\frac{1}{3R}=\frac{1+2}{6R}$
$R_{eq}=2R$
For maximun power,
External resistance = Internal resistance
$2R=4$
$R=2\Omega$