Chapter 19, Problem 56P

(a)

To determine

The value of the resistance to use the galvanometer as the ammeter.

Answer to Problem 56P

The resistance in the parallel for using the galvanometer as ammeter is $2.9\times {10}^{-4}\Omega$ .

Explanation of Solution

Given info:

The sensitivity of the galvanometer is, $s=35k\Omega /\text{V}=35\times {10}^3\Omega /\text{V}$ .

The internal resistance of the galvanometer is, $r=20\Omega$ .

The reading of the ammeter is, $I_a=2\text{A}$ .

Formula Used:

The expression to calculate the current in the galvanometer is,

  $I=\frac{1}{s}$

The expression to calculate the resistance for using the galvanometer as ammeter is,

  $R=\frac{Ir}{I_a-I}$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}I=\frac{1}{35\times {10}^3\Omega /\text{V}}\\ =2.85\times {10}^{-5}\text{A}\end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}R=\frac{\left(2.85\times {10}^{-5}\text{A}\right)\left(20\Omega \right)}{2\text{A}-2.85\times {10}^{-5}\text{A}}\\ =2.9\times {10}^{-4}\Omega \end{array}$

Conclusion:

Thus, the resistance in the parallel for using the galvanometer as ammeter is $2.9\times {10}^{-4}\Omega$ .

(b)

To determine

The value of the resistance to use the galvanometer as the voltmeter.

Answer to Problem 56P

The resistance for using the galvanometer as voltmeter is $35\times {10}^3\Omega$ .

Explanation of Solution

Given info:

The voltage at full scale deflection of the voltmeter is, $V=1\text{V}$ .

Formula Used:

The expression to calculate the resistance for using the galvanometer as voltmeter is,

  $R_1=\frac{V}{I}-r$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}{R}_1=\frac{1\text{V}}{2.9\times {10}^{-4}\text{A}}-20\Omega \\ \approx 35\times {10}^3\Omega \end{array}$

Conclusion:

Thus, the resistance for using the galvanometer as voltmeter is $35\times {10}^3\Omega$ .