Question

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

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

Hint:

Voltmeter → add high resistance in series.

Answer 1

The Correct Option is D

Step 1: Understanding Voltmeter Conversion:
To convert a galvanometer into a voltmeter, a high resistance (\(R_s\)) is connected in series with it. This series resistor limits the current to the galvanometer's full-scale deflection current (\(I_g\)) when the maximum desired voltage (\(V\)) is applied across the entire combination.
Step 2: Key Formula or Approach:
The total resistance of the voltmeter is \(R_{total} = R_g + R_s\), where \(R_g\) is the galvanometer resistance.
By Ohm's law, when the maximum voltage \(V\) is applied, the current flowing must be \(I_g\).
\[ V = I_g \times (R_g + R_s) \] Step 3: Detailed Calculation:
We are given the following values:
Galvanometer resistance, \(R_g = 100 \, \Omega\).
Full-scale deflection current, \(I_g = 1 \, \text{mA} = 1 \times 10^{-3} \, \text{A}\).
Maximum voltage to be measured, \(V = 2\) V.
We need to find the series resistance \(R_s\).
Rearranging the formula: \(R_g + R_s = \frac{V}{I_g}\).
\[ 100 + R_s = \frac{2}{1 \times 10^{-3}} = 2000 \] Now, solve for \(R_s\):
\[ R_s = 2000 - 100 \] \[ R_s = 1900 \, \Omega \] Step 4: Final Answer:
A resistance of 1900 \(\Omega\) must be connected in series with the galvanometer.