JEE PYQ: Magnetism And Matter Question 14
Question 14 - 2019 (12 Apr 2019 Shift 2)
A moving coil galvanometer, having a resistance $G$, produces full scale deflection when a current $I_g$ flows through it. This galvanometer can be converted into (i) an ammeter of range $0$ to $I_0$ ($I_0 > I_g$) by connecting a shunt resistance $R_A$ to it and (ii) into a voltmeter of range $0$ to $V$ ($V = GI_0$) by connecting a series resistance $R_V$ to it. Then,
(1) $R_A R_V = G^2 \left(\frac{I_0 - I_g}{I_g}\right)$ and $\frac{R_A}{R_V} = \left(\frac{I_g}{I_0 - I_g}\right)^2$
(2) $R_A R_V = G^2$ and $\frac{R_A}{R_V} = \left(\frac{I_g}{I_0 - I_g}\right)^2$
(3) $R_A R_V = G^2 \left(\frac{I_g}{I_0 - I_g}\right)$ and $\frac{R_A}{R_V} = \left(\frac{I_0 - I_g}{I_g}\right)$
(4) $R_A R_V = G^2$ and $\frac{R_A}{R_V} = \frac{I_g}{(I_0 - I_g)}$
Show Answer
Answer: (2) $R_A R_V = G^2$ and $\frac{R_A}{R_V} = \left(\frac{I_g}{I_0 - I_g}\right)^2$
Solution
From ammeter and voltmeter equations, solving gives $R_A R_V = G^2$ and $\frac{R_A}{R_V} = \left(\frac{I_g}{I_0 - I_g}\right)^2$.
BETA
