The ratios of the voltage sensitivities,resis | vedclass

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(A) The voltage sensitivity $(V_s)$ of a moving coil galvanometer is given by $V_s = \frac{NBA}{kR}$,where $N$ is the number of turns,$B$ is the magne

Vedclass · Accepted Answer

$100$

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(A) The voltage sensitivity $(V_s)$ of a moving coil galvanometer is given by $V_s = \frac{NBA}{kR}$,where $N$ is the number of turns,$B$ is the magnetic field,$A$ is the area,$k$ is the torsional constant,and $R$ is the resistance.Given the ratio of voltage sensitivities: $\frac{V_{sA}}{V_{sB}} = \frac{4}{3}$.Given the ratio of resistances: $\frac{R_A}{R_B} = \frac{3}{4}$.Given the ratio of areas: $\frac{A_A}{A_B} = \frac{1}{2}$.Since $B$ and $k$ are constant,we have $\frac{V_{sA}}{V_{sB}} = \frac{N_A A_A R_B}{N_B A_B R_A}$.Substituting the given values: $\frac{4}{3} = \frac{N_A}{N_B} 	imes (\frac{1}{2}) 	imes (\frac{4}{3})$.$\frac{4}{3} = \frac{N_A}{N_B} 	imes \frac{2}{3}$.$\frac{N_A}{N_B} = \frac{4}{3} 	imes \frac{3}{2} = 2$.Given $N_A = 200$,then $N_B = \frac{N_A}{2} = \frac{200}{2} = 100$.

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