Question

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

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

Answer 1

The correct option is B $2\Omega$

Consider the problem,

The given circuit is a balanced Wheatstone bridge, so the wire containing $6ohm$ will be eliminated.

$R,R,R$ are in series while $R,R$ and $4R$ are in series.

So,

$\begin{array}{c}{R}^{\prime }=R+R+R=3R\\ R^{\prime \prime }=R+R+4R=6R\end{array}$

$R^{\prime }$ and $R^{\prime \prime }$ are in parallel

Effective resistance $=\frac{1}{3}R+\frac{1}{6}R=\frac{3}{6}R=\frac{6R}{3}=2R$

For delivering maximum power, internal resistance $=$ external resistance

Given, internal resistance $=4\Omega$

So,

$\begin{array}{c}2R=4\\ R=2\Omega \end{array}$

Hence, Option $B$ is the correct answer.