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. $\frac{4}{9}$
• B. $\frac{24}{3}$
• C. $\frac{8}{3}$
• D. $\frac{1}{8}$

Answer 1

The correct option is B $\frac{24}{3}$

In order to get maximum power

Internal resistance (all)=External resistance$\to \left(1\right)$

Resolving network resistance

In loop $1=\frac{6R\times 6R}{6R+6R}=\frac{36\times R^2}{I^{2}R}=3R$

This loop $\left(3R\right)$ is in series with $2R$

In loop $2=\frac{5R\times R}{5R+R}=\frac{5R^2}{6R}=\frac{5}{6}R$

From $\left(1\right)\to \frac{5}{6}R=4\Omega R=\frac{24}{5}\Omega$