Question

A battery of internal resistance $4\Omega$ is connected to a network of resistances as shown in the figure. In order that the maximum power can be delivered to the network, the value of $R\left(\text{in}\Omega \right)$ should be-

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

Answer 1

The correct option is B $2$

From maximum power transfer theorem,

External resistance = Internal resistance

Given, $R_{in}=4\Omega$

$\therefore$ External resistance should also be $4\Omega \dots \left(1\right)$



The network of resistances, connected across the battery, is actually a Wheatstone bridge in balanced condition.

$\therefore R_{eq}=\frac{3R\times 6R}{3R+6R}=\frac{3R\times 6R}{9R}=2R......\left(2\right)$

For maximum power, $2R=4\Omega$

$\Rightarrow R=2\Omega$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.