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If the galvanometer current is 10 mA, re...

If the galvanometer current is 10 mA, resistance of the galvanometer is `40 Omega` and shunt of `2 Omega` is connected to the galvanometer, the maximum current which can be measured by this ammeter is

A

0.21A

B

2.1A

C

210A

D

21A

Text Solution

Verified by Experts

The correct Answer is:
A
`I=((S+G)/(S))I_(g) = ((2+40)/(2)) xx 0.01 = 0.21 A`
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