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, \[G=22.8\,\,\Omega ,\,i=20\,A,\,\,{{i}_{g}}=1\,A\] \[\therefore \] \[S=\frac{1\times 22.8}{20-1}=\frac{22.8}{19}=1.2\,\,\Omega \]You need to login to perform this action.
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