A galvanometer has a coil of resistance 200 O | vedclass

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(B) Given:Resistance of galvanometer $G = 200 \Omega$Full scale deflection current $I_g = 20 \mu A = 20 	imes 10^{-6} A$Required range of ammeter $I

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$0.20$

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(B) Given:Resistance of galvanometer $G = 200 \Omega$Full scale deflection current $I_g = 20 \mu A = 20 	imes 10^{-6} A$Required range of ammeter $I = 20 mA = 20 	imes 10^{-3} A$To convert a galvanometer into an ammeter,a shunt resistance $S$ is connected in parallel.The formula for shunt resistance is $S = \frac{I_g G}{I - I_g}$.Substituting the values:$S = \frac{20 	imes 10^{-6} 	imes 200}{20 	imes 10^{-3} - 20 	imes 10^{-6}}$$S = \frac{4000 	imes 10^{-6}}{20 	imes 10^{-3} (1 - 0.001)}$$S = \frac{4 	imes 10^{-3}}{20 	imes 10^{-3} 	imes 0.999}$$S = \frac{4}{20 	imes 0.999} = \frac{0.2}{0.999} \approx 0.2002 \Omega$Thus,the required shunt resistance is approximately $0.20 \Omega$.

$A$ galvanometer has a coil of resistance $200 \Omega$ with a full scale deflection at $20 \mu A$. The value of resistance to be added to use it as an ammeter of range $(0-20) mA$ is: (in $\Omega$)

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