question_answer

A galvanometer of resistance 22.8 \[\Omega \] measures 1 A. How much shunt should be used, so that it can be used to measure 20 A?

A) 1\[\Omega \]

B) 2\[\Omega \]

C) 1.2\[\Omega \]          

D) 2.2\[\Omega \]

Correct Answer: A

Solution :

Key Idea: The voltage across shunt is same to that across galvanometer. Shunt is a low resistance used in parallel with the galvanometer to make it ammeter. The circuit is shown in figure. Voltage across galvanometer = voltage across shunt i.e.,        \[{{i}_{g}}G=(i-{{i}_{g}})S\] or            \[S=\frac{{{i}_{g}}G}{i-{{i}_{g}}}\] Given,   \[G=22.8\,\,\Omega ,\,i=20\,A,\,\,{{i}_{g}}=1\,A\] \[\therefore \] \[S=\frac{1\times 22.8}{20-1}=\frac{22.8}{19}=1.2\,\,\Omega \]