A galvanometer of resistance 240 Omega allows | vedclass

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(A) Given,galvanometer resistance $G = 240 \Omega$.Let the main current be $I$ and the current through the galvanometer be $I_G$.According to the prob

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$10$

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(A) Given,galvanometer resistance $G = 240 \Omega$.Let the main current be $I$ and the current through the galvanometer be $I_G$.According to the problem,$I_G = 4 \% 	ext{ of } I = \frac{4}{100} I = 0.04 I$.The shunt resistance $S$ is connected in parallel with the galvanometer.Since they are in parallel,the potential difference across the galvanometer and the shunt resistance is the same:$I_G G = (I - I_G) S$Substituting the values:$0.04 I 	imes 240 = (I - 0.04 I) S$$9.6 I = 0.96 I 	imes S$$S = \frac{9.6 I}{0.96 I} = 10 \Omega$.Therefore,the value of the shunt resistance is $10 \Omega$.

$A$ galvanometer of resistance $240 \Omega$ allows only $4 \%$ of the main current to pass through it after connecting a shunt resistance. The value of the shunt resistance is (in $\Omega$)

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