Question

A galvanometer of resistance $20\Omega$ is to be converted into an ammeter of range $1\mathrm{A}$. If a current of $1\mathrm{mA}$ produces full scale deflection, the shunt required for the purpose is}

• A. $0.01\Omega$
• B. $0.05\Omega$
• C. $0.02\Omega$
• D. $0.04\Omega$

Hint:

For ammeter conversion, $S = \frac{I_g G}{I - I_g}$.

Answer 1

The Correct Option is C

To convert a galvanometer into an ammeter, a shunt resistor is connected in parallel with the galvanometer. The shunt resistor allows the ammeter to measure larger currents by bypassing most of the current through the shunt, leaving only a small fraction to pass through the galvanometer.

Let's first understand the problem:

1. The galvanometer has a coil resistance \(R_g = 20 \, \Omega\).
2. The full scale deflection current of the galvanometer is \(I_g = 1 \, \text{mA} = 0.001 \, \text{A}\).
3. We want to convert it to measure up to \(I = 1 \, \text{A}\).

The shunt resistance \(R_s\) can be calculated using the formula:

\(R_s = \frac{R_g \cdot I_g}{I - I_g}\)

Substituting the given values:

• \(R_s = \frac{20 \, \Omega \times 0.001 \, \text{A}}{1 \, \text{A} - 0.001 \, \text{A}}\)
• \(R_s = \frac{0.02 \, \Omega}{0.999 \, \text{A}}\)
• \(R_s \approx 0.020 \, \Omega\)

Therefore, the value of the shunt resistance required is approximately \(0.02 \, \Omega\).

Hence, the correct answer is $0.02\Omega$.