A galvanometer coil has a resistance of $12\,\Omega $ and meter shows full scale deftection for a current of $3\,mA$ then to convert it into a voltmeter of range $0\,-18\, V$ a resistance should be added
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$\mathrm{R}_{\mathrm{g}}=12\, \Omega, \mathrm{I}_{\mathrm{g}}=3 \mathrm{\,mA}$
$\mathrm{R}_{\mathrm{g}}=\frac{\mathrm{V}_{\mathrm{R}}}{\mathrm{R}_{\mathrm{g}}+\mathrm{R}_{\mathrm{H}}}$
$\Rightarrow \mathrm{R}_{\mathrm{H}}=\frac{\mathrm{V}_{\mathrm{R}}}{\mathrm{I}_{\mathrm{g}}}-\mathrm{R}_{\mathrm{g}}$
$\Rightarrow \mathrm{R}_{\mathrm{H}}=5988\, \Omega$
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