Question

A galvanometer of resistance 20 Ω gives a full scale deflection when a current of 0.04 A is passed through it. To convert it into an ammeter of range 20 A, the resistance that must be connected in series with the coil of the galvanometer is (Galvanometer is shunted by 0.05 Ω)

• A. 5.94 Ω
• B. 4.95 Ω
• C. 12.62 Ω
• D. 9.45 Ω

Hint:

To convert a galvanometer into an ammeter, the series resistance must be chosen so that the full scale deflection corresponds to the required current range.

Answer 1

The Correct Option is B

Step 1: Formula for converting a galvanometer into an ammeter.
To convert a galvanometer into an ammeter, the resistance \( R_s \) that should be connected in series with the galvanometer is given by the formula: \[ R_s = \frac{V_g}{I_{\text{max}}} - R_g \] where \( V_g = I_g \cdot R_g \) is the potential drop across the galvanometer, \( I_{\text{max}} \) is the full scale current, and \( R_g \) is the resistance of the galvanometer.
Step 2: Calculating the series resistance.
Given \( I_g = 0.04 \, \text{A} \), \( R_g = 20 \, \Omega \), and \( I_{\text{max}} = 20 \, \text{A} \), we can find the value of \( R_s \). First, we calculate the potential drop across the galvanometer: \[ V_g = I_g \cdot R_g = 0.04 \times 20 = 0.8 \, \text{V} \] Now, using the formula for \( R_s \): \[ R_s = \frac{0.8}{20} - 0.05 = 4.95 \, \Omega \] Step 3: Conclusion.
Thus, the resistance to be connected in series with the galvanometer is 4.95 Ω, corresponding to option (B).