Question

A galvanometer of resistance 99.9 \( \Omega \) gives a full scale deflection when 5 mA current is passed through it. The resistance to be connected to the galvanometer such that it can be converted into an ammeter of range 0-5 A is:

• A. \( 0.01 \, \Omega \)
• B. \( 10 \, \Omega \)
• C. \( 1 \, \Omega \)
• D. \( 0.1 \, \Omega \)

Hint:

To convert a galvanometer into an ammeter, use the formula for the equivalent resistance of the parallel combination of the galvanometer and the shunt resistor.

Answer 1

The Correct Option is D

Let the resistance \( R \) be connected in parallel with the galvanometer. The total resistance in the circuit becomes the resistance of the galvanometer \( G \) in parallel with \( R \).
We are given:
- The full scale deflection current of the galvanometer is 5 mA \( (I_g = 5 \, \text{mA} = 0.005 \, \text{A}) \),
- The range of the ammeter is 0-5 A,
- The resistance of the galvanometer \( G = 99.9 \, \Omega \).
The ammeter is designed for a range of 0-5 A, so the total resistance required in the circuit is: \[ R_{total} = \frac{V}{I_{max}} = \frac{I_g \times G}{I_g - I_{max}} = \frac{(0.005 \, \text{A}) \times (99.9 \, \Omega)}{0.005 - 5} \] Calculating this value gives \( R \approx 0.1 \, \Omega \).
Thus, the required resistance to convert the galvanometer into an ammeter is \( 0.1 \, \Omega \).