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 convert a galvanometer into an ammeter, a low resistance called a "shunt" is connected in parallel with the galvanometer. The shunt resistance allows the ammeter to measure larger currents by diverting most of the current around the galvanometer.

Given:

• Resistance of the galvanometer coil, \(R_g = 200 \, \Omega\)
• Full scale deflection current of the galvanometer, \(I_g = 20 \, \mu A = 20 \times 10^{-6} \, A\)
• Desired range of the ammeter, \(I = 20 \, mA = 20 \times 10^{-3} \, A\)

The current flowing through the shunt resistor, \(I_s\), is given by:

\(I_s = I - I_g\)

Substitute the values:

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

The voltage across the galvanometer and the shunt is the same. Therefore, we have:

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

We can rearrange to find the shunt resistance, \(R_s\):

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

Substitute the values:

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

Calculate the value:

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

Therefore, the value of the resistance to be added so the galvanometer can be used as an ammeter with a range of 0-20 mA is \(0.20 \, \Omega\).

The correct answer is: \(0.20 \, \Omega\).

Answer 2

Step 1: Formula for shunt resistance The shunt resistance \( R_s \) is given by:

\[ R_s = \frac{I_g R_g}{I - I_g}, \]

where:

• \( I_g = 20 \, \mu A = 20 \times 10^{-6} \, \text{A} \) (full-scale deflection current),
• \( R_g = 200 \, \Omega \) (resistance of the galvanometer),
• \( I = 20 \, mA = 20 \times 10^{-3} \, \text{A} \) (ammeter range).

Step 2: Substitute the values

\[ R_s = \frac{(20 \times 10^{-6}) \cdot 200}{(20 \times 10^{-3}) - (20 \times 10^{-6})}. \]

Simplify the numerator:

\[ (20 \times 10^{-6}) \cdot 200 = 4 \times 10^{-3}. \]

Simplify the denominator:

\[ (20 \times 10^{-3}) - (20 \times 10^{-6}) = 19.98 \times 10^{-3}. \]

Thus:

\[ R_s = \frac{4 \times 10^{-3}}{19.98 \times 10^{-3}} \approx 0.20 \, \Omega. \]

Final Answer: \( 0.20 \, \Omega. \)