A galvanometer,having a resistance of 50 ,Ome | vedclass

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Question

(C) To convert a galvanometer into an ammeter,a shunt resistance $S$ is connected in parallel with the galvanometer.Given: Galvanometer resistance $G

Vedclass · Accepted Answer

$3$

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(C) To convert a galvanometer into an ammeter,a shunt resistance $S$ is connected in parallel with the galvanometer.Given: Galvanometer resistance $G = 50 \,\Omega$,full-scale current $I_g = 0.05 \,A$,maximum current to be measured $I = 5 \,A$.The formula for shunt resistance is $S = \frac{I_g \cdot G}{I - I_g}$.Substituting the values: $S = \frac{0.05 	imes 50}{5 - 0.05} = \frac{2.5}{4.95} = \frac{250}{495} = \frac{50}{99} \,\Omega$.We know $S = \frac{ho \cdot l}{A}$,where $ho = 5 	imes 10^{-7} \,\Omega \cdot m$ and $A = 2.97 	imes 10^{-2} \,cm^2 = 2.97 	imes 10^{-6} \,m^2$.Rearranging for length $l$: $l = \frac{S \cdot A}{ho} = \frac{50}{99} 	imes \frac{2.97 	imes 10^{-6}}{5 	imes 10^{-7}}$.$l = \frac{50}{99} 	imes \frac{29.7 	imes 10^{-7}}{5 	imes 10^{-7}} = \frac{50}{99} 	imes 5.94 = \frac{297}{99} = 3 \,m$.

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