Chapter 24, Problem 74A

a.

To determine

The total resistance of the given circuit.

Answer to Problem 74A

The total resistance of the circuit is $2\times {10}^5\Omega$.

Explanation of Solution

Given:

A galvanometer with a resistor connected in series to it. Its full scale deflection corresponds to a current of 50.0-µA.

Formula used:

Consider the circuit shown in Figure 1. The circuit consists of a galvanometer and a resistor called multiplier connected in series with it to form a voltmeter circuit.

  

Figure 1

In this circuit, the galvanometer measures the current through the resistor and this can be used to obtain the voltage across it using Ohm’s law as,

  $V=IR$

Here, R represents the total resistance of the circuit consisting of the galvanometer resistance and the multiplier given by,

  $R=R_g+R_{multiplier}$

It can be calculated for a maximum current measured by galvanometer and the full-scale deflection voltage as,

  $R=\frac{V_{full-scale}}{I_g}$

Calculation:

Substituting the full scale deflection voltage of $V_{full-scale}=10\text{ V}$ and the current of the galvanometer $I_g=50\mu \text{A}$ , the total resistance of the circuit in Figure 1 is obtained as,

  $\begin{array}{c}R=\frac{10}{50\times {10}^{-6}}\\ =2\times {10}^5\Omega \end{array}$

Conclusion The total resistance of the series resistance along with the galvanometer resistance is $2\times {10}^5\Omega$ .

b.

To determine

The resistance of the multiplier resistor.

Answer to Problem 74A

The resistance of the multiplier resistor is $\text{199 k}\Omega$.

Explanation of Solution

Given:

A galvanometer with a resistor connected in series to it. Its full scale deflection corresponds to a current of 50.0-µA.

Formula used:

Consider the circuit shown in Figure 2. The circuit consists of a galvanometer and a resistor called multiplier connected in series with it to form a voltmeter circuit.

  

Figure 2

In this circuit, the galvanometer measures the current through the resistor and this can be used to obtain the voltage across it using Ohm’s law as,

  $V=IR$

Here, R represents the total resistance of the circuit consisting of the galvanometer resistance and the multiplier (shown in Figure 2) given by,

  $R=R_g+R_{multiplier}$ …… (1)

It can be calculated for a maximum current measured by galvanometer and the full-scale deflection voltage as,

  $R=\frac{V_{full-scale}}{I_g}$

Given the galvanometer resistance $R_g$ , the resistance of the multiplier can be found using (1).

Calculation:

Substituting the full scale deflection voltage of $V_{full-scale}=10\text{ V}$ and the current of the galvanometer $I_g=50\mu \text{A}$ , the total resistance of the circuit in Figure 1 is obtained as,

  $\begin{array}{c}R=\frac{10}{50\times {10}^{-6}}\\ =2\times {10}^5\Omega \end{array}$

Now, the resistance of the multiplier for a given galvanometer resistance $R_g=1.0\text{}\text{ k}\Omega$ ,can be calculated using (1) as,

  $\begin{array}{c}{R}_{multiplier}=R-R_g\\ =2\times {10}^5-{10}^3\\ =199\text{ k}\Omega \end{array}$

Conclusion:

The resistance of the multiplier resistor is $199\text{ k}\Omega$ .