In a moving coil galvanometer,two moving coil | vedclass

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(A) The voltage sensitivity $(S_V)$ of a moving coil galvanometer is given by the formula:$S_V = \frac{	heta}{V} = \frac{NAB}{kR}$where $N$ is the nu

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$1: 1$

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(A) The voltage sensitivity $(S_V)$ of a moving coil galvanometer is given by the formula:$S_V = \frac{	heta}{V} = \frac{NAB}{kR}$where $N$ is the number of turns,$A$ is the area,$B$ is the magnetic field,$k$ is the torsional constant,and $R$ is the resistance.Given that $k$ is the same for both coils,the ratio of voltage sensitivity is:$\frac{S_{V1}}{S_{V2}} = \frac{N_1 A_1 B_1}{N_2 A_2 B_2} 	imes \frac{R_2}{R_1}$Substituting the given values:$\frac{S_{V1}}{S_{V2}} = \frac{15 	imes 3.6 	imes 10^{-3} 	imes 0.25}{21 	imes 1.8 	imes 10^{-3} 	imes 0.50} 	imes \frac{7}{5}$$\frac{S_{V1}}{S_{V2}} = \frac{15 	imes 3.6 	imes 0.25}{21 	imes 1.8 	imes 0.50} 	imes \frac{7}{5}$$\frac{S_{V1}}{S_{V2}} = \frac{13.5}{18.9} 	imes 1.4 = \frac{13.5}{18.9} 	imes \frac{7}{5} = \frac{13.5 	imes 7}{18.9 	imes 5} = \frac{94.5}{94.5} = 1$Therefore,the ratio is $1: 1$.

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