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A galvanometer of resistance `40 Omega` gives a deflection of 5 divisions per mA. There are 50 divisions on the scale. The maximum current that can pass through it when a shunt resistance of `2Omega` is connected isA. 210mAB. 155 MaC. 420mAD. 75 Ma

Answer» Correct Answer - A
`I_(G) = 50/5=10 mA, R_(G)=40 Omega, R_(S)= 2 Omega`
Maximum current,
`I=(R_(G) + R_(S))/(R_(G)) xx I_(G) = ((40 + 2) xx 10)/(2) = 210 mA`


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