Question

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

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

Answer 1

The correct option is B

$0.10\mathrm{W}$

Step 1: Given data

Terminal voltage of a battery, $\mathrm{V}=2.5\mathrm{volt}$

Voltage of a battery, $\mathrm{E}=3.0\mathrm{volt}$

Power dissipated through resistor $\text{'}\mathrm{R}\text{'}={\mathrm{P}}_{\mathrm{R}}=0.5\mathrm{W}$

Step 2: Calculating the power dissipated within the internal resistance

Let current through the given circuit$=\mathrm{i}$

And let the internal resistance$=\mathrm{r}$

Let power dissipated across $\text{'}\mathrm{r}\text{'}={\mathrm{P}}_{\mathrm{r}}$

We all know that:

$$\begin{gathered} {\mathrm{P}}_{\mathrm{R}}=0.5\left(1\right) \\ \Rightarrow {\mathrm{i}}^2\mathrm{R}=0.5 \end{gathered}$$

Also we know that:

$$\begin{gathered} \mathrm{V}=\mathrm{E}-\mathrm{ir} \\ \Rightarrow 2.5=3-\mathrm{ir} \\ \Rightarrow \mathrm{ir}=0.5 \end{gathered}$$

The power dissipated across $\text{'}\mathrm{r}\text{'}={\mathrm{P}}_{\mathrm{r}}={\mathrm{i}}^2\mathrm{r}\left(2\right)$

We know that $\mathrm{V}=\mathrm{iR}=2.5$, therefore

$$\begin{gathered} \frac{\mathrm{iR}}{\mathrm{ir}}=\frac{2.5}{0.5} \\ \Rightarrow \frac{\mathrm{R}}{\mathrm{r}}=5 \end{gathered}$$

Now dividing the first equation by second, we get

$$\begin{gathered} \frac{{\mathrm{P}}_{\mathrm{R}}}{{\mathrm{P}}_{\mathrm{r}}}=\frac{{\mathrm{i}}^2\mathrm{R}}{{\mathrm{i}}^2\mathrm{r}} \\ \Rightarrow \frac{{\mathrm{P}}_{\mathrm{R}}}{\mathrm{Pr}}=\frac{\mathrm{R}}{\mathrm{r}} \\ \Rightarrow \frac{{\mathrm{P}}_{\mathrm{R}}}{{\mathrm{P}}_{\mathrm{r}}}=5 \\ \Rightarrow {\mathrm{P}}_{\mathrm{r}}=\frac{{\mathrm{P}}_{\mathrm{R}}}{5} \\ \Rightarrow {\mathrm{P}}_{\mathrm{r}}=\frac{0.5}{5} \\ \Rightarrow {\mathrm{P}}_{\mathrm{r}}=0.10\mathrm{W} \end{gathered}$$

Therefore power dissipated within internal resistance is equal to $0.10\mathrm{W}$.

Hence, option(B) is correct answer.