question_answer

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\[6.6\times {{10}^{-34}}\]

Correct Answer: A

Solution :

                                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,  \[=\frac{1}{4}m{{R}^{2}}+m{{R}^{2}}=\frac{5}{4}m{{R}^{2}}\] \[g=g\left( 1-\frac{h}{R} \right)\]              \[\Rightarrow \]                 \[w=w\left( 1-\frac{h}{R} \right)\]