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.

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