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) $