Question

A battery of $3.0\text{V}$ is connected to a resistor dissipating $0.5\text{W}$ of power. If the terminal voltage of the battery is $2.5\text{V}.$ The power dissipated within the internal resistance is-

• A. $\text{0.50 W}$
• B. $\text{0.072 W}$
• C. $\text{0.10 W}$
• D. $\text{0.125 W}$

Answer 1

The correct option is C $\text{0.10 W}$

Given,

$E=3\text{V};V=E-ir=2.5\text{V}{P}_R=0.5\text{W}$

Where, $R=\text{External resistance}r=\text{internal resistance}$


$\because P_R=i^{2}R=0.5\text{W and}$

$\because V=E-ir\Rightarrow 2.5=3-ir\Rightarrow V_r=ir=0.5$

Now, $V_R=V=iR=2.5$

$\therefore \frac{R}{r}=5$

$\therefore \frac{P_R}{P_r}=\frac{i^{2}R}{i^{2}r}=\frac{R}{r}=5$

$\Rightarrow P_r=\frac{P_R}{5}=\frac{0.50}{5}=0.10\text{W}$
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.