Question

A galvanometer having a coil of resistance 30 \( \Omega \) needs 20 mA of current for full-scale deflection. If a maximum current of 3 A is to be measured using this galvanometer, the resistance of the shunt to be added to the galvanometer should be \( X \, \Omega \), where \( X \) is:

• A. 447
• B. 298
• C. 149
• D. 596

Hint:

In measuring larger currents with a galvanometer, a shunt resistor is used to bypass excess current, ensuring that the galvanometer only measures the small current for full-scale deflection.

Answer 1

The Correct Option is C

The resistance \( R_g \) of the galvanometer is 30 \( \Omega \), and the current for full-scale deflection is \( I_g = 20 \, \text{mA} = 0.02 \, \text{A} \). To measure a maximum current of 3 A, we use a shunt resistance \( R_s \) in parallel with the galvanometer. The total current passing through the parallel combination is: \[ I = I_g + I_s, \] where \( I_s \) is the current passing through the shunt. The voltage across the galvanometer and the shunt must be the same, so: \[ V = I_g R_g = I_s R_s. \] Since \( I_s = I - I_g = 3 - 0.02 = 2.98 \, \text{A} \), we have: \[ R_s = \frac{I_g R_g}{I_s} = \frac{0.02 \times 30}{2.98} \approx 0.201 \, \Omega. \] Thus, the resistance of the shunt is approximately 149 \( \Omega \).