1.

The current measuring capacity of a galvanometer of resistance `50 Omega` is to be increased from x mA to `101 xxmA`.The shunt resistance isA. `0.5Omega`B. `0.25Omega`C. `1.5Omega`D. `0.35Omega`

Answer» Correct Answer - A
`S=(G)/((n-1))=(50)/((101-1))=(50)/(100)=0.5Omega` .


Discussion

No Comment Found

Related InterviewSolutions