Question

A galvanometer has a coil of resistance \(200 \, \Omega\) with a full scale deflection at \(20 \, \mu A\). The value of resistance to be added to use it as an ammeter of range (0–20) mA is:

• A. \(0.40 \, \Omega\)
• B. \(0.20 \, \Omega\)
• C. \(0.50 \, \Omega\)
• D. \(0.10 \, \Omega\)

Answer 1

The Correct Option is B

To transform a galvanometer into an ammeter, a low-resistance component known as a "shunt" is connected in parallel. This shunt resistor redirects the majority of the current, enabling the ammeter to measure significantly larger current values than the galvanometer alone could handle.

Provided Information:

• Galvanometer coil resistance: \(R_g = 200 \, \Omega\)
• Galvanometer full-scale deflection current: \(I_g = 20 \, \mu A = 20 \times 10^{-6} \, A\)
• Target ammeter range: \(I = 20 \, mA = 20 \times 10^{-3} \, A\)

The current passing through the shunt resistor, \(I_s\), is calculated as:

\(I_s = I - I_g\)

Substituting the given values:

\(I_s = 20 \times 10^{-3} - 20 \times 10^{-6} = 19.98 \times 10^{-3} \, A\)

Since the voltage across the galvanometer and the shunt is identical, the following relationship holds:

\(I_g \times R_g = I_s \times R_s\)

Rearranging this equation to determine the shunt resistance, \(R_s\):

\(R_s = \frac{I_g \times R_g}{I_s}\)

Substituting the known values into the equation:

\(R_s = \frac{20 \times 10^{-6} \times 200}{19.98 \times 10^{-3}}\)

The calculated shunt resistance is:

\(R_s = \frac{4 \times 10^{-3}}{19.98 \times 10^{-3}} \approx 0.20 \, \Omega\)

Consequently, a resistance of \(0.20 \, \Omega\) must be added for the galvanometer to function as an ammeter with a range of 0-20 mA.

The determined value is: \(0.20 \, \Omega\).