Question

A galvanometer of $50$ ohm resistance has $25$ divisions. $A$ current of $ 4\times 10^{-4}$ ampere gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of $25$ volts, it should be connected with a resistance of

• A. $ 2500 \, \Omega$ as a shunt
• B. $2450 \, \Omega$ as a shunt
• C. $2550 \, \Omega$ in a series
• D. $2450 \, \Omega$ in a series

Answer 1

The Correct Option is D

To convert a galvanometer into a voltmeter, you need to connect a resistance in series with the galvanometer. The formula used to determine the series resistance is:

R_s = V/I_g - R_g

Where:

• R_s is the series resistance to be added.
• V is the range of the voltmeter, given as 25 volts.
• I_g is the full-scale deflection current through the galvanometer.
• R_g is the resistance of the galvanometer, given as 50 ohms.

Given:

• R_g = 50\, \Omega
• Each division of the galvanometer corresponds to 4 \times 10^{-4} amperes. Hence, the full-scale deflection current I_g is 25 \times 4 \times 10^{-4} = 1 \times 10^{-2} amperes (as there are 25 divisions).

Using the formula:

R_s = \frac{V}{I_g} - R_g = \frac{25}{1 \times 10^{-2}} - 50

R_s = 2500 - 50 = 2450\, \Omega

Therefore, the galvanometer should be connected in series with a resistance of 2450 \, \Omega to convert it into a voltmeter with a range of 25 volts.

Conclusion: The correct answer is 2450 \, \Omega in a series.