Problem

A galvanometer has a full-scale deflection at $10 \, \mathrm{mA}$ and a resistance of $100 \, \Omega$. How can it be converted into an ammeter with a range of $100 \, \mathrm{A}$?

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Data

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

To find:

• Shunt resistance: $R_s = ?$.

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Prerequisite Concepts

To convert a galvanometer into an ammeter, a shunt resistance $R_s$ is connected in parallel to the galvanometer. The shunt resistance is calculated as:

$$R_s = \frac{I_g R_g}{I - I_g}$$

where:

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

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Solution

1. Substitute the known values into the formula:

$$R_s = \frac{(10 \times 10^{-3}) \cdot 100}{100 - (10 \times 10^{-3})}$$

2. Simplify the numerator:

$$I_g R_g = (10 \times 10^{-3}) \cdot 100 = 1.$$

3. Simplify the denominator:

$$I - I_g = 100 - 0.01 = 99.99.$$

4. Calculate the shunt resistance:

$$R_s = \frac{1}{99.99} \approx 0.01 \, \Omega.$$

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Answer

The shunt resistance required to convert the galvanometer into an ammeter with a range of $100 \, \mathrm{A}$ is:

$$R_s = 0.01 \, \Omega.$$