Chapter 19, Problem 62P

To determine

The internal resistance of the voltmeter.

Answer to Problem 62P

The internal resistance of the voltmeter is $2.4\times {10}^5$ .

Explanation of Solution

Given info:

The voltage across second resistor is, $V_2=4.12\text{V}$ .

The combination voltage of the voltage is, $V_{comb}=3.88\text{V}$ .

The resistance of the resistor is, $R_1=R_2=15000\Omega$ .

Formula Used:

The expression to calculate the current in the voltmeter is,

  $I=\frac{V_2}{R_2}$

The expression to calculate the combined resistance is,

  $R_{comb}=\frac{V_{comb}}{I}$

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

  $r=\frac{R_{comb}{R}_1}{R_1-R_{comb}}$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}I=\frac{4.12\text{V}}{15000\Omega }\\ =2.75\times {10}^{-4}\text{A}\end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}{R}_{comb}=\frac{3.88\text{V}}{2.75\times {10}^{-4}\text{A}}\\ =14109.1\Omega \end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}r=\frac{\left(15000\Omega \right)\left(14109.1\Omega \right)}{15000\Omega -14109.1\Omega }\\ =2.4\times {10}^5\end{array}$

Conclusion:

Thus, the internal resistance of the voltmeter is $2.4\times {10}^5$ .