Question

A battery has an open circuit potential difference of $6\text{V}$ between its terminals. When a load resistance of $60\Omega$ is connected across the battery, the total power dissipated by the battery is $0.4\text{W}$. What should be the load resistance $R$, so that maximum power will be dissipated in $R$ ?

• A. $20\Omega$
• B. $30\Omega$
• C. $40\Omega$
• D. $60\Omega$

Answer 1

The correct option is B $30\Omega$

Figure shows the situation described in question. Here $r$ is the internal resistance of the battery and $E$ its $\text{EMF}$.


Power supplied by the battery in this case is given as

$P=\frac{E^2}{R+r}$

Substituting the values gives,

$0.4=\frac{(6{)}^2}{60+r}$

$\Rightarrow r=30\Omega$

Maximum power is dissipated in the circuit when external resistance is equal to net internal resistance, which gives
$R=r$

$\Rightarrow R=30\Omega$

Hence, option $\left(B\right)$ is correct answer.