Problem

How can a $5 \, \mathrm{mA}, 100 \, \Omega$ galvanometer be converted into a voltmeter with a range of $20 \, \mathrm{V}$?

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Data

• Current through the galvanometer: $I_g = 5 \, \mathrm{mA} = 5 \times 10^{-3} \, \mathrm{A}$,
• Total voltage range: $V = 20 \, \mathrm{V}$,
• Resistance of the galvanometer: $R_g = 100 \, \Omega$.

To find:

• High resistance to be connected in series: $R_h = ?$.

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Prerequisite Concepts

To convert a galvanometer into a voltmeter, a high resistance $R_h$ is connected in series with the galvanometer. The high resistance is calculated as:

$$R_h = \frac{V}{I_g} - R_g$$

where:

• $V$ is the total voltage range,
• $I_g$ is the galvanometer current,
• $R_g$ is the galvanometer resistance.

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Solution

1. Substitute the given values into the formula:

$$R_h = \frac{20}{5 \times 10^{-3}} - 100$$

2. Simplify the denominator:

$$\frac{20}{5 \times 10^{-3}} = \frac{20}{0.005} = 4000.$$

3. Subtract the galvanometer resistance:

$$R_h = 4000 - 100 = 3900 \, \Omega.$$

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Answer

The high resistance required to convert the galvanometer into a voltmeter with a range of $20 \, \mathrm{V}$ is:

$$R_h = 3900 \, \Omega.$$