A microammeter has a resistance of 100,Omega | vedclass

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(D) Given: Galvanometer resistance $G = 100\,\Omega$,full-scale current $I_g = 50\,\mu A = 50 	imes 10^{-6}\,A$.For Voltmeter conversion: $A$ high re

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Both $(B)$ and $(C)$

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(D) Given: Galvanometer resistance $G = 100\,\Omega$,full-scale current $I_g = 50\,\mu A = 50 	imes 10^{-6}\,A$.For Voltmeter conversion: $A$ high resistance $R$ is connected in series. The formula is $R = \frac{V}{I_g} - G$.For option $(B)$: $V = 10\,V$. $R = \frac{10}{50 	imes 10^{-6}} - 100 = 200,000 - 100 \approx 200\,k\Omega$. Thus,$(B)$ is correct.For Ammeter conversion: $A$ shunt resistance $S$ is connected in parallel. The formula is $S = \frac{I_g \cdot G}{I - I_g}$.For option $(C)$: $I = 5\,mA = 5 	imes 10^{-3}\,A$. $S = \frac{50 	imes 10^{-6} 	imes 100}{5 	imes 10^{-3} - 50 	imes 10^{-6}} = \frac{5 	imes 10^{-3}}{4.95 	imes 10^{-3}} \approx 1.01\,\Omega \approx 1\,\Omega$. Thus,$(C)$ is correct.Since both $(B)$ and $(C)$ are correct,the answer is $(D)$.

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

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