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

Question

If the galvanometer current is 10 mA, resistance of the galvanometer is 40 Ω and shunt of 2 Ω is connected to the galvanometer, the maximum current which can be measured by this ammeter is (A) 0.21 A (B) 2.1 A (C) 210 A (D) 21 A

Tardigrade Admin · Accepted Answer

I=((S+G/S)) (Ig=(2+40/2))× 0.01 =0.21 A

Q. If the galvanometer current is $10 m A$ , resistance of the galvanometer is $40 Ω$ and shunt of $2 Ω$ is connected to the galvanometer, the maximum current which can be measured by this ammeter is

$0.21 A$

$2.1 A$

$210 A$

$21 A$

$I = (\frac{S + G}{S})$
$(I_{g} = \frac{2 + 40}{2}) \times 0.01$
$= 0.21 A$

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

0.21A

2.1A

210A

21A