question_answer

A galvanometer gives full scale deflection with 0.2 A in a coil shown in the figure. The resistance of its coil is\[10\,\Omega \]. How much shunt resistance is required to convert it into an ammeter to read current upto 1.8 A? [BHU PMT-2004]

A)  \[1.25\,\Omega \]                         

B)  \[2.5\,\Omega \]

C)  \[5.0\,\Omega \]                                            

D)  \[10.0\,\Omega \]

Correct Answer: A

Solution :

                 Key Idea: Potential difference across galvanometer and shunt is same. For G being resistance of galvanometer and \[{{I}_{g}}\] the current across it, the current across S is \[\left( I-{{I}_{g}} \right)\]. Since, \[G\] and \[S\] are in parallel, potential across them is same                                                 \[{{I}_{g}}\times G=\left( I-{{I}_{g}} \right)\times S\] \[\Rightarrow \]                               \[S=\frac{{{I}_{g}}G}{\left( I-{{I}_{g}} \right)}\] Given, \[{{I}_{g}}=0.2\,A,\,G=10\,\Omega ,\,I=1.8\,A\] \[\Rightarrow \]                               \[S=\frac{0.2}{1.6}\times 10\]                                                 \[=1.25\,\,\Omega \]