A microammeter has a resistance of 100,Omega | vedclass

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(B) Given: Galvanometer resistance $G = 100\,\Omega$,full-scale current $I_g = 50\,\mu A = 50 	imes 10^{-6}\,A$.For a voltmeter (series resistance $R

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$10\,V$ range with $200\,k\Omega$ resistance in series

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(B) Given: Galvanometer resistance $G = 100\,\Omega$,full-scale current $I_g = 50\,\mu A = 50 	imes 10^{-6}\,A$.For a voltmeter (series resistance $R$):$R = \frac{V}{I_g} - G = \frac{10}{50 	imes 10^{-6}} - 100 = 200,000 - 100 = 199,900\,\Omega \approx 200\,k\Omega$.Thus,option $(b)$ is correct.For an ammeter (shunt resistance $S$):$S = \frac{I_g G}{I - I_g} = \frac{50 	imes 10^{-6} 	imes 100}{10 	imes 10^{-3} - 50 	imes 10^{-6}} = \frac{5 	imes 10^{-3}}{10^{-2} - 0.05 	imes 10^{-3}} = \frac{5 	imes 10^{-3}}{9.95 	imes 10^{-3}} \approx 0.5\,\Omega$.Options $(c)$ and $(d)$ are incorrect.

$A$ microammeter has a resistance of $100\,\Omega$ and a full-scale range of $50\,\mu A$. It can be used as a voltmeter or as a higher-range ammeter provided a resistance is added to it. Pick the correct range and resistance combination.

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