Question 328 of 380

A galvanometer has a resistance of $50\Omega$ and it allows maximum current of $5mA$. It can be converted into voltmeter to measure upto $100V$ by connecting in series a resistor of resistance.

A$5975\Omega$B$20050\Omega$C$19950\Omega$D$19500\Omega$

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Answer: (C) $19950\Omega$

If we connect a high resistance($R$) in series with the galvanometer, it can be successfully converted into a voltmeter. Therefore,

Voltage drop across voltmeter:
$V={I}_{g}\left({R}_{g}+R\right) \Rightarrow 100={I}_{g}\left(50+R\right)$
$\Rightarrow R=\dfrac{100}{5\times {10}^{-3}}-50$
$=19950\Omega$