Question

A battery of $3.0V$ is connected to a resistor dissipating $0.5W$ of power. If the terminal voltage of the battery is $2.5V,$ the power dissipated within the internal resistance is :

• A. $0.50W$
• B. $0.125W$
• C. $0.072W$
• D. $0.10W$

Answer 1

Answer: (D)

$P_R=0.5W$
$\Rightarrow i^{2}R=0.5W$
Also, $V=E-$ ir
$2.5=3-$ ir
$\Rightarrow$ ir $=0.5$
Power dissipated across 'r' $:P_r=i^{2}r$
Now $iR=2.5$
ir $=0.5$
On dividing : $\frac{R}{r}=5$
Now $\frac{P_R}{P_r}=\frac{i^{2}R}{i^{2}r}$
$\Rightarrow \frac{P_R}{P_r}=\frac{R}{r}$
$\Rightarrow \frac{P_R}{P_r}=5$
$\Rightarrow P_r=\frac{P_R}{5}$
$\Rightarrow P_r=\frac{0.50}{5}$
$\Rightarrow P_r=0.10W$