A galvanometer having a coil of resistance 30 | vedclass

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(C) The condition for the shunt resistance $r_s$ to convert a galvanometer into an ammeter is given by the parallel circuit relation: $I_g R_g = (I -

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$149$

Vedclass · Answer

(C) The condition for the shunt resistance $r_s$ to convert a galvanometer into an ammeter is given by the parallel circuit relation: $I_g R_g = (I - I_g) r_s$.Given:Galvanometer resistance $R_g = 30 \ \Omega$Full-scale deflection current $I_g = 20 \ 	ext{mA} = 20 	imes 10^{-3} \ 	ext{A} = 0.02 \ 	ext{A}$Maximum current to be measured $I = 3 \ 	ext{A}$Substituting the values into the formula:$0.02 	imes 30 = (3 - 0.02) 	imes r_s$$0.6 = 2.98 	imes r_s$$r_s = \frac{0.6}{2.98} \ \Omega$We are given that $r_s = \frac{30}{X} \ \Omega$. Therefore:$\frac{30}{X} = \frac{0.6}{2.98}$$X = \frac{30 	imes 2.98}{0.6}$$X = 50 	imes 2.98 = 149$.Thus,the value of $X$ is $149$.

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