A galvanometer has resistance ' $G$ ' $\Omega$ and ' $I_g$ ' is current flowing through it which produces…

A galvanometer has resistance ' $G$ ' $\Omega$ and ' $I_g$ ' is current flowing through it which produces full scale deflection. ' $S_1$ ' is the value of shunt which converts it into an ammeter of range 0 to ' $3 I$ ' and ' $S_2$ ' is the shunt value which converts it into an ammeter of range 0 to ' $4 I$ ', the ratio $S_2: S_1$ is
  1. $\frac{4}{3}$
  2. $\frac{3 I-I_g}{4 I-I_g}$
  3. $\frac{3}{4}$
  4. $\frac{4 I-I_g}{3 I-I_g}$

Solution

$\begin{aligned} & s_1=\frac{I_g G}{3 I-I_g}, s_2=\frac{I_g G}{4 I-I_g} \\ & \therefore \frac{s_2}{s_1}=\frac{3 I-I_g}{4 I-I_g}\end{aligned}$

Asked in: MHT CET 2021 (20 Sep Shift 1)

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