Two moving coil galvanometers,X and Y,have co | vedclass

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(A) Current sensitivity $I_S$ is given by $I_S = \frac{NBA}{k}$. Assuming the spring constant $k$ is the same for both galvanometers,the ratio of curr

Vedclass · Accepted Answer

$2 : 3, 14 : 15$

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(A) Current sensitivity $I_S$ is given by $I_S = \frac{NBA}{k}$. Assuming the spring constant $k$ is the same for both galvanometers,the ratio of current sensitivities is:$\frac{I_{SX}}{I_{SY}} = \frac{N_X B_X A_X}{N_Y B_Y A_Y} = \frac{30 	imes 0.25 	imes 4.8 	imes 10^{-3}}{45 	imes 0.50 	imes 2.4 	imes 10^{-3}} = \frac{30}{45} 	imes \frac{0.25}{0.50} 	imes \frac{4.8}{2.4} = \frac{2}{3} 	imes \frac{1}{2} 	imes 2 = \frac{2}{3}$.Voltage sensitivity $V_S$ is given by $V_S = \frac{I_S}{R} = \frac{NBA}{kR}$. The ratio of voltage sensitivities is:$\frac{V_{SX}}{V_{SY}} = \frac{I_{SX}}{I_{SY}} 	imes \frac{R_Y}{R_X} = \frac{2}{3} 	imes \frac{14}{10} = \frac{2}{3} 	imes \frac{7}{5} = \frac{14}{15}$.Thus,the ratios are $2 : 3$ and $14 : 15$.

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