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.....$Ω$
AIPMT 2010, Medium
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Resistance of galvanometer, $G=100 \,\Omega$
Current for full scale deflection, $I_{g}=30 \,\mathrm{mA}$
$=30 \times 10^{-3}\, \mathrm{A}$
Range of voltmeter, $V=30\, \mathrm{V}$
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, $V=I_{g}(G+R)$
$ \text { or } R=\frac{V}{I_{8}}-G =\frac{30}{30 \times 10^{-3}}-100\, \Omega $
$=1000-100=900\, \Omega$
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