A coil of inductance 0.12 H and resistance 4(Omega), is connected across a 240V, 50Hz supply. Calculate the current through it. [(pi) = 3.142]
-
A.
6.3A -
B.
33.3A -
C.
37.2A -
D.
40.0A
Correct Answer: Option A
Explanation
XL = (2pi f L)
= = 2 x 3.142 x 50 x 0.12
= 37.68(Omega)
Z = (sqrt{R^2 + X_L^2} = sqrt{4^2 + 37.68^2})
I = (frac{V}{z} = frac{240}{37.89})
= 6.3A