Question

A galvanometer has a resistance of 50 \(\Omega\) and a current of 2 mA is needed for its full scale deflection. Calculate resistance required to convert it. (i) into an ammeter of 0.5 A range. (ii) into a voltmeter of 10 V range.

Hint:

{Ammeter}: Add a tiny resistor in {parallel} (a "shunt") to bypass most of the current. {Voltmeter}: Add a huge resistor in {series} to limit the current and drop most of the voltage.

Answer 1

Given:
Galvanometer resistance, \(R_g = 50 \, \Omega\)
Full scale deflection current, \(I_g = 2 \, mA = 2 \times 10^{-3} \, A\)
(i) Conversion into an Ammeter:
To convert a galvanometer into an ammeter of range \(I\), a small resistance called a shunt (\(S\)) is connected in parallel with it.
The required range is \(I = 0.5 \, A\).
The formula for the shunt resistance is: \[ S = \frac{I_g R_g}{I - I_g} \] \[ S = \frac{(2 \times 10^{-3}) \times 50}{0.5 - (2 \times 10^{-3})} = \frac{0.1}{0.5 - 0.002} = \frac{0.1}{0.498} \] \[ S \approx 0.2008 \, \Omega \] A shunt resistance of approximately 0.201 \(\Omega\) is required.
(ii) Conversion into a Voltmeter:
To convert a galvanometer into a voltmeter of range \(V\), a large resistance (\(R_s\)) is connected in series with it.
The required range is \(V = 10 \, V\).
The formula for the series resistance is: \[ R_s = \frac{V}{I_g} - R_g \] \[ R_s = \frac{10}{2 \times 10^{-3}} - 50 = \frac{10000}{2} - 50 \] \[ R_s = 5000 - 50 = 4950 \, \Omega \] A series resistance of 4950 \(\Omega\) is required.