Question

A galvanometer has resistance \( G = 100 \, \Omega \) and shows full-scale deflection at \( I_q = 1 \, \text{mA} \). To convert it into a voltmeter of range 5 V, what resistance should be connected in series?

• A. 400 \( \Omega \)
• B. 4900 \( \Omega \)
• C. 490 \( \Omega \)
• D. 5000 \( \Omega \)

Hint:

Remember: To convert a galvanometer into a voltmeter, calculate the series resistance using the formula \( R_s = \frac{V}{I_q} - G \).

Answer 1

The Correct Option is B

Given: Resistance of the galvanometer, \( G = 100 \, \Omega \) 
Full-scale deflection current, \( I_q = 1 \, \text{mA} = 1 \times 10^{-3} \, \text{A} \) 
Voltage range of the voltmeter, \( V = 5 \, \text{V} \) 

Step 1: Formula for the Series Resistance The resistance \( R_s \) to be connected in series is given by the formula: \[ R_s = \frac{V}{I_q} - G \] where: - \( V \) is the desired voltage range, - \( I_q \) is the full-scale deflection current, - \( G \) is the resistance of the galvanometer. 

Step 2: Substitute the given values Substitute the given values into the formula: \[ R_s = \frac{5 \, \text{V}}{1 \times 10^{-3} \, \text{A}} - 100 \, \Omega \] \[ R_s = 5000 \, \Omega - 100 \, \Omega \] \[ R_s = 4900 \, \Omega \] 

Step 3: Conclusion Thus, the resistance to be connected in series is \( 4900 \, \Omega \). 

Answer: The correct answer is option (b): 4900 \( \Omega \).