Chapter 19, Problem 63P

To determine

The emf and internal resistance of the battery.

Answer to Problem 63P

The emf of the battery is $10.41\text{V}$ and the internal resistance of the battery is $2.57\Omega$ .

Explanation of Solution

Given info:

The voltage across first resistor is, $V_1=9.7\text{V}$ .

The voltage across second resistor is, $V_2=8.1\text{V}$ .

The resistance of the first resistor is, $R_1=35\Omega$ .

The resistance of the second resistor is, $R_1=9\Omega$ .

Formula Used:

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

  $r=\left(\frac{V_1-V_2}{V_2{R}_1-V_1{R}_2}\right)R_1{R}_2$

The expression to calculate the emf of the battery is,

  $E=\frac{V_1}{R_1}\left(r+R_1\right)$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}r=\left(\frac{9.7\text{V}-8.1\text{V}}{\left(35\Omega \right)\left(8.1\text{V}\right)-\left(9\Omega \right)\left(9.7\text{V}\right)}\right)\left(35\Omega \right)\left(9\Omega \right)\\ =2.57\Omega \end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}E=\left(\frac{9.7\text{V}}{35\Omega }\right)\left(2.57\Omega +35\Omega \right)\\ =10.41\text{V}\end{array}$

Conclusion:

Thus, the emf of the battery is $10.41\text{V}$ and the internal resistance of the battery is $2.57\Omega$ .