Question

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

• A. $900$
• B. $1800$
• C. $500$
• D. $1000$

Answer 1

Answer: (A)

Given data:-

Resistance of Galvanometer$=R_G=100\Omega$

Current through galvanometer for full scale deflection$=I_g=30mA$

Now if we want to convert galvanometer into voltmeter, then we have to connect a large resistance R in series with galvanometer.

Now equivalent Resistance of galvanometer and large resistance R$=R_G+R$

Current through galvanometer and large resistance$=I_g=\frac{V}{R_G+R}$ V=voltage

$30\ast {10}^{-3}=\frac{30}{100+R}$

$\frac{30}{1000}=\frac{30}{100+R}$

$100+R=1000$

$R=900\Omega$