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 maximu

Question

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 maximu

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

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - A
`I=((S+G)/(S))I_(g) = ((2+40)/(2)) xx 0.01 = 0.21 A`

4 views Answered 2023-02-05 15:29