Question

A galvanometer has a coil of resistance $100\, ohm$ and gives a full scale deflection for $30\, mA$ current. If it is to work as a voltmeter of $30$ volt range, the resistance required to be added will be

• A. $900 \,\Omega$
• B. $1800 \,\Omega$
• C. $500 \,\Omega$
• D. $1000 \,\Omega$

Answer 1

Answer: (A)

Here,
Resistance of galvanometer, $G = 100\, \Omega$
Current for full scale deflection, $I_g=30 \,mA$
$ =30 \times10^{-3}A$
Range of voltmeter, $V =30 \,V $
To convert the galvanometer into an voltmeter of a given range, a resistance $R$ is connected in series with it as shown in the figure.

From figure.
$ V=I_g(G+R)$
or $R=\frac{V}{I_g}-G=\frac{30}{30 \times 10^{-3}}-100 \Omega$
$=1000-100=900\, \Omega$