Two moving coil galvanometers A and B having | vedclass

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Question

(A) The current sensitivity $(I_s)$ of a moving coil galvanometer is given by the formula: $I_s = \frac{NBA}{k}$,where $N$ is the number of turns,$B$

Vedclass · Accepted Answer

$3: 16$

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(A) The current sensitivity $(I_s)$ of a moving coil galvanometer is given by the formula: $I_s = \frac{NBA}{k}$,where $N$ is the number of turns,$B$ is the magnetic field,$A$ is the area of the coil,and $k$ is the torsional constant of the spring.Since the springs are identical,$k_A = k_B = k$.The ratio of current sensitivities is $\frac{I_{sA}}{I_{sB}} = \frac{N_A B_A A_A}{N_B B_B A_B}$.Given values:$N_A = 36, B_A = 0.25 \ T, A_A = 2.4 	imes 10^{-3} \ m^2$$N_B = 48, B_B = 0.5 \ T, A_B = 4.8 	imes 10^{-3} \ m^2$Substituting these values:$\frac{I_{sA}}{I_{sB}} = \frac{36 	imes 0.25 	imes 2.4 	imes 10^{-3}}{48 	imes 0.5 	imes 4.8 	imes 10^{-3}}$$\frac{I_{sA}}{I_{sB}} = \left(\frac{36}{48}ight) 	imes \left(\frac{0.25}{0.5}ight) 	imes \left(\frac{2.4}{4.8}ight)$$\frac{I_{sA}}{I_{sB}} = \left(\frac{3}{4}ight) 	imes \left(\frac{1}{2}ight) 	imes \left(\frac{1}{2}ight) = \frac{3}{16}$.Thus,the ratio is $3: 16$.

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