Chapter 19, Problem 75GP

To determine

The emf and internal resistance of the battery.

Answer to Problem 75GP

The emf of the battery is $31.55\text{V}$ and the internal resistance of the battery is $1.25\Omega$ .

Explanation of Solution

Given info:

The first voltage rating of battery is, $V_1=40.8\text{V}$ .

The first current ratting of battery is, $I_1=7.40\text{}\text{A}$ .

The second voltage rating of battery is, $V_2=47.3\text{V}$ .

The second current ratting of battery is, $I_2=2.20\text{}\text{A}$ .

Formula Used:

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

  $r=\frac{V_2-V_1}{I_1-I_2}$

The expression to calculate the emf of the battery is,

  $\epsilon =V_1+I_{1}r$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}r=\frac{47.3\text{V}-40.8\text{V}}{7.40\text{A}-2.20\text{A}}\\ =1.25\Omega \end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}\epsilon =40.8\text{V}-\left(7.40\text{}\text{A}\right)\left(1.25\Omega \right)\\ =31.55\text{V}\end{array}$

Conclusion:

Thus, the emf of the battery is $31.55\text{V}$ and the internal resistance of the battery is $1.25\Omega$ .