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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
1000 $\Omega$
B
900 $\Omega$
C
1800 $\Omega$
D
500 $\Omega$
Detailed Solution
A galvanometer is converted into a voltmeter by connecting a high resistance R in series with it.
At full scale: $V = I_g(G + R)$
$R = \frac{V}{I_g} - G$
$R = \frac{30}{30\times10^{-3}} - 100 = 1000 - 100$
$R = 900\ \Omega$
At full scale: $V = I_g(G + R)$
$R = \frac{V}{I_g} - G$
$R = \frac{30}{30\times10^{-3}} - 100 = 1000 - 100$
$R = 900\ \Omega$