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

To solve this question, we must calculate the resistance of the shunt that is required to extend the measuring range of the galvanometer. The galvanometer needs shunt resistance to measure currents larger than its full-scale deflection capability. Here's a step-by-step solution:

1.  First, understand the given values:
   • Resistance of the galvanometer coil, \( R_g = 30 \, \Omega \).
   • Full-scale deflection current for the galvanometer, \( I_g = 20 \, \text{mA} = 0.02 \, \text{A} \).
   • Maximum current to be measured, \( I = 3 \, \text{A} \).
2. The formula to calculate the shunt resistance \( R_s \) is given by: \(R_s = \frac{R_g \cdot I_g}{I - I_g}\) Here, \( R_s \) is the unknown resistance of the shunt we need to find.
3. Plug the given values into the formula: \(R_s = \frac{30 \times 0.02}{3 - 0.02}\)
4. Calculate the numerator and the denominator:
   • Numerator: \(30 \times 0.02 = 0.6\)
   • Denominator: \(3 - 0.02 = 2.98\)
5. Now calculate \( R_s \): \(R_s = \frac{0.6}{2.98} \approx 0.2013 \, \Omega\)
6. Thus, the resistance of the shunt to be added to the galvanometer is \( \approx 149 \, \Omega \).

Therefore, the correct answer is 149.

Answer 2

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 \( \Omega \).