A galvanometer coil of resistance 50 ,Omega s | vedclass

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(D) Given: Galvanometer resistance $G = 50 \,\Omega$,full-scale deflection current $I_g = 100 \,\mu A = 100 	imes 10^{-6} \, A = 10^{-4} \, A$,and th

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$0.5 \,\Omega$ in parallel

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(D) Given: Galvanometer resistance $G = 50 \,\Omega$,full-scale deflection current $I_g = 100 \,\mu A = 100 	imes 10^{-6} \, A = 10^{-4} \, A$,and the desired range $I = 10 \, mA = 10 	imes 10^{-3} \, A = 10^{-2} \, A$.To convert a galvanometer into an ammeter,a shunt resistance $S$ is connected in parallel with the galvanometer.The formula for shunt resistance is $S = \frac{I_g 	imes G}{I - I_g}$.Substituting the values: $S = \frac{10^{-4} 	imes 50}{10^{-2} - 10^{-4}} = \frac{50 	imes 10^{-4}}{100 	imes 10^{-4} - 1 	imes 10^{-4}} = \frac{50 	imes 10^{-4}}{99 	imes 10^{-4}} = \frac{50}{99} \approx 0.505 \,\Omega$.Rounding to the nearest provided option,the shunt resistance is $0.5 \,\Omega$ in parallel.

$A$ galvanometer coil of resistance $50 \,\Omega$ shows full-scale deflection for a current of $100 \,\mu A$. The shunt resistance to be added to the galvanometer to convert it into an ammeter of range $10 \, mA$ is:

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