Chapter 19, Problem 12P

To determine

The internal resistance of the battery .

Answer to Problem 12P

The internal resistance of the battery is $0.4\Omega$ .

Explanation of Solution

Given info:

emf of the battery, $\epsilon =12\text{V}$

Terminal voltage of the battery, $V=11.8\text{V}$

Power of the light bulb, $P=3\text{W}$

Formula Used:

The expression to calculate the current through the battery is,

  $I=\frac{2P}{V}$

The expression to calculate the internal resistance of the battery is,

  $r=\frac{\epsilon -V}{I}$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}I=\frac{2\left(3\text{W}\right)}{12\text{V}}\\ =0.5\text{A}\end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}r=\frac{12\text{V}-11.8\text{V}}{0.5\text{A}}\\ =0.4\Omega \end{array}$

Conclusion:

Therefore, the internal resistance of the battery is $0.4\Omega$ .