Question

A galvanometer of resistance 100 \(\Omega\) gives a full scale deflection for a current of 1 mA through it. The resistance required to convert it into a voltmeter which can read up to 2 V is:

• A. 1175 \(\Omega\)
• B. 1200 \(\Omega\)
• C. 1525 \(\Omega\)
• D. 1900 \(\Omega\)
• E. 2025 \(\Omega\)

Hint:

To convert a galvanometer into a voltmeter, add a series resistance so that the voltage drop across the combination is equal to the maximum voltage that the voltmeter can measure.

Answer 1

The Correct Option is D

Step 1: Understanding the concept of converting a galvanometer to a voltmeter.}
To convert a galvanometer into a voltmeter, we need to add a series resistance, \( R_s \), so that the voltmeter can measure a voltage up to the required value. The total resistance of the voltmeter is the sum of the galvanometer resistance (\( R_g \)) and the series resistance (\( R_s \)). The voltage \( V \) across the galvanometer is given by: \[ V = I \cdot (R_g + R_s) \] where:
- \( V \) is the maximum voltage (2 V),
- \( I \) is the full-scale current (1 mA = 0.001 A),
- \( R_g \) is the resistance of the galvanometer (100 \(\Omega\)).

Step 2: Apply the formula to find the series resistance.}
We are given: - \( V = 2 \, \text{V} \),
- \( I = 0.001 \, \text{A} \),
- \( R_g = 100 \, \Omega \).
Using the formula: \[ V = I \cdot (R_g + R_s) \] Substitute the given values: \[ 2 = 0.001 \cdot (100 + R_s) \] Solve for \( R_s \): \[ 2 = 0.001 \cdot (100 + R_s) \] \[ \frac{2}{0.001} = 100 + R_s \] \[ 2000 = 100 + R_s \] \[ R_s = 2000 - 100 = 1900 \, \Omega \]
Step 3: Final Answer.}
The required resistance to convert the galvanometer into a voltmeter is \( 1900 \, \Omega \).