Question

A galvanometer has a coil of resistance $100\, ohm$ and gives a full scale deflection for $30\, mA$ current. If it is to work as a voltmeter of $30\, volt$ range, the resistance required to be added will be

• A. 900 $\Omega$
• B. 1800 $\Omega$
• C. 500 $\Omega$
• D. 1000 $\Omega$

Answer 1

The Correct Option is A

 To convert a galvanometer into a voltmeter, a high resistance is added in series with the galvanometer. The given problem can be approached as follows:

1. Understand the given values:
   • The resistance of the galvanometer coil \(R_g = 100 \, \Omega\).
   • Current for full scale deflection \(I_g = 30 \, \text{mA} = 0.03 \, \text{A}\).
   • Desired voltage range of the voltmeter \(V = 30 \, \text{V}\).
2. Calculate the total resistance required for the desired voltage range:
   • Using Ohm’s law, the voltage across the whole setup is given by \(V = I_g \times (R_g + R_s)\), where \(R_s\) is the series resistance added.
   • Substitute the known values: \(30 = 0.03 \times (100 + R_s)\).
3. Solve for the unknown resistance \(R_s$\)
   • Rearrange the equation: \((100 + R_s) = \frac{30}{0.03}\).
   • Calculate: \((100 + R_s) = 1000\).
   • Thus, \(R_s = 1000 - 100 = 900 \, \Omega\).
4. Conclusion: The resistance required to be added to convert the galvanometer into a voltmeter with a range of \(30 \, \text{V}\) is \(900 \, \Omega\).