Question

A galvanometer having a coil resistance 100 Ω gives a full scale deflection when a current of 1mA is passed through it. What is the value of the resistance which can convert this galvanometer into a voltmeter given full scale deflection for a potential difference of 10V?

Answer 1

9900 Ω

Answer 2

To convert a galvanometer into a voltmeter, a high resistance (R) is connected in series with the galvanometer.

Given data:

• Coil resistance of the galvanometer, $G = 100 \, \Omega$
• Full scale deflection current of the galvanometer, $I_g = 1 \, \text{mA} = 1 \times 10^{-3} \, \text{A}$
• Desired full scale deflection voltage for the voltmeter, $V = 10 \, \text{V}$

Formula: When a resistance R is connected in series with the galvanometer, the total resistance of the voltmeter circuit is $R_{total} = G + R$. For full scale deflection, the voltage across the voltmeter is given by Ohm's law: $V = I_g \times (G + R)$

Rearranging the formula to solve for R: $G + R = \frac{V}{I_g}$ $R = \frac{V}{I_g} - G$

Calculation: Substitute the given values into the formula: $R = \frac{10 \, \text{V}}{1 \times 10^{-3} \, \text{A}} - 100 \, \Omega$ $R = \frac{10}{0.001} - 100$ $R = 10000 - 100$ $R = 9900 \, \Omega$

Therefore, a resistance of 9900 Ω must be connected in series with the galvanometer to convert it into a voltmeter with a 10V range.