Chapter 19, Problem 86GP

(a)

To determine

To prove: The expression $\frac{1}{R}=\frac{1}{V}-\frac{1}{R_V}$

Explanation of Solution

Given information:

Consider the expression. $V=I{R}^1$

Formula Used:

The equivalent resistance is given by,

  $R^1=\frac{R_{V}R}{R_V+R}$

Here, $R_V$ is the voltmeter resistance.

Calculation:

Use both the equations,

  $\begin{array}{l}V=I\left(\frac{R_{V}R}{R_V+R}\right)\\ \frac{V}{I}=\frac{R_{V}R}{R_V+R}\\ \frac{V}{I}\left(R_V+R\right)=R_{V}R\\ \frac{V}{I}{R}_V=R\left(R_V-\frac{V}{I}\right)\end{array}$

Solve further as,

  $\begin{array}{l}R=\frac{\frac{V}{I}{R}_V}{R_V-\frac{V}{I}}\\ \frac{1}{R}=\frac{1}{V}-\frac{1}{R_V}\end{array}$

Conclusion:

Thus, the expression $\frac{1}{R}=\frac{1}{V}-\frac{1}{R_V}$ is proved.

(b)

To determine

The expression $R=\frac{V}{I}-R_A$ .

Explanation of Solution

Given information:

Consider the expression. $V=I{R}^1$

Formula Used:

The equivalent resistance is given by,

  $R^1=R+R_A$

Here, $R_A$ is the ammeter resistance.

Calculation:

Use both the equations,

  $\begin{array}{l}V=I\left(R+R_A\right)\\ \frac{V}{I}=R+R_A\\ R=\frac{V}{I}-R_A\end{array}$

Conclusion:

Thus, the expression $R=\frac{V}{I}-R_A$ is proved.