question_answer

A galvanometer of resistance 50 Q is converted into ammeter by connecting a low resistance (shunt) of value 1 Q in parallel to the galvanometer S. If full scale deflection current of galvanometer is 10 mA, then the maximum current to be measured by the ammeter is -

A) 500 mA                        

B) 510 mA

C) 610 mA                        

D) 490 mA

Correct Answer: B

Solution :

\[S=G\frac{{{I}_{g}}}{I-{{I}_{g}}}\]   \[1=50\times \frac{10}{I-10}\] \[I-10=500\] \[I=510\,mA\]