A galvanometer coil has a resistance of 12 Ω and the metre shows full scale deflection for a current of 3 mA.
Question:
A galvanometer coil has a resistance of 12 Ω and the metre shows full scale deflection for a current of 3 mA. How will you convert the metre into a voltmeter of range 0 to 18 V?
Solution:
Resistance of the galvanometer coil, G = 12 Ω
Current for which there is full scale deflection, $I_{g}=3 \mathrm{~mA}=3 \times 10^{-3} \mathrm{~A}$
Range of the voltmeter is 0, which needs to be converted to 18 V.
$\therefore V=18 \mathrm{~V}$
Let a resistor of resistance R be connected in series with the galvanometer to convert it into a voltmeter. This resistance is given as:
$R=\frac{V}{I_{\mathrm{g}}}-\mathrm{G}$
$=\frac{18}{3 \times 10^{-3}}-12=6000-12=5988 \Omega$
Hence, a resistor of resistance $5988 \Omega$ is to be connected in series with the galvanometer.