Question:medium

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?

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Remember: To convert a galvanometer into a voltmeter, calculate the series resistance using the formula \( R_s = \frac{V}{I_q} - G \).
Updated On: Nov 26, 2025
  • 400 \( \Omega \)
  • 4900 \( \Omega \)
  • 490 \( \Omega \)
  • 5000 \( \Omega \)
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The Correct Option is B

Solution and Explanation

Given: Galvanometer resistance, \( G = 100 \, \Omega \); Full-scale deflection current, \( I_q = 1 \, \text{mA} = 1 \times 10^{-3} \, \text{A} \); Voltmeter voltage range, \( V = 5 \, \text{V} \).

Step 1: Series Resistance Formula The series resistance \( R_s \) is calculated using: \[ R_s = \frac{V}{I_q} - G \] where \( V \) is the voltage range, \( I_q \) is the full-scale deflection current, and \( G \) is the galvanometer resistance.

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

Step 3: Final Result The required series resistance is \( 4900 \, \Omega \).

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

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