Magnetic field due to currentHard
Question
A moving coil galvanometer has 150 equal divisions. Its current sensitivity is 10 divisions per milli ampere and voltage sensitivity is 2 divisions per millivolt. In order that each division reads 1 V, the resistance in Ohm′s needed to be connected in series with the coil will be -
Options
A.103
B.105
C.99995
D.9995
Solution
Ig = 
× 10-3 = 0.015 A
Vg =
× 10-3 = 0.075 V ∴ G = 
= 5Ω
If R = resistance to be added in series, I (G + R) = V
⇒ 0.015 (5+R) = 150 × 1 ⇒ R = 9995Ω
× 10-3 = 0.015 AVg =
× 10-3 = 0.075 V ∴ G = = 5ΩIf R = resistance to be added in series, I (G + R) = V
⇒ 0.015 (5+R) = 150 × 1 ⇒ R = 9995Ω
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