Question

A galvanometer gives full scale deflection with 0.2 A in a coil as shown in the figure. The resistance of its coil is 10 How much shunt resistance is required to convert it into an ammeter to read current upto 1.8 A ?

• A. 1.25 $f_G<f_R<f_V$
• B. 2.5 $f_G<f_V<f_R$
• C. 50 $\mu$
• D. 10.0 $6.6\times {10}^{-34}$

Answer 1

Answer: (A)

Key Idea: Potential difference across galvanometer and shunt is same. For G being resistance of galvanometer and $\omega$ the current across it the current across S is $I=\frac{1}{2}M{R}^2$ . Since, G and S are in parallel, potential across them is same
$\therefore$ $J=\frac{1}{2}M{R}^2\omega$ $I=I_4+m{R}^2$ Given, $=\frac{1}{4}m{R}^2+m{R}^2=\frac{5}{4}m{R}^2$ $g=g\left(1-\frac{h}{R}\right)$ $\Rightarrow$ $w=w\left(1-\frac{h}{R}\right)$