A galvanometer of resistance 10 Omega changes | vedclass

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(A) The galvanometer resistance is $G = 10 \Omega$. The full-scale deflection current is $I_g = 1 	ext{ mA} = 1 	imes 10^{-3} 	ext{ A}$. The new ra

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$10$

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(A) The galvanometer resistance is $G = 10 \Omega$. The full-scale deflection current is $I_g = 1 	ext{ mA} = 1 	imes 10^{-3} 	ext{ A}$. The new range is $I = 101 	ext{ mA} = 101 	imes 10^{-3} 	ext{ A}$.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: $S = \frac{1 	imes 10^{-3} 	imes 10}{101 	imes 10^{-3} - 1 	imes 10^{-3}} = \frac{10 	imes 10^{-3}}{100 	imes 10^{-3}} = 0.1 \Omega$.The resistance of a wire is given by $R = ho \frac{L}{A}$,where $ho = 1 	imes 10^{-6} \Omega 	ext{ m}$ and $A = 1 	ext{ mm}^2 = 1 	imes 10^{-6} 	ext{ m}^2$.Setting $R = S = 0.1 \Omega$,we get $0.1 = (1 	imes 10^{-6}) 	imes \frac{L}{1 	imes 10^{-6}}$.$0.1 = L$.Thus,$L = 0.1 	ext{ m} = 10 	ext{ cm}$.

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