0

4.5kviews

A voltage of 150 V, 50 Hz is applied to a coil of negligible resistance and inductance 0.2H. Write current

written 9.4 years ago by teamques10 ★ 70k

•   modified 5.6 years ago

basic electrical engineering

ADD COMMENT FOLLOW SHARE EDIT

1 Answer

1

674views

written 9.4 years ago by teamques10 ★ 70k

Given that:

Vrms = 150V

F= 50Hz

L= 0.2H

XL= 2πf L= 2π * 50 * 0.2 = 62.83 Ω

Vm= Vrms $\sqrt{2}$ = 150 $\sqrt{2}$ = 212.13 V

Im= Vm/XC = 212.13/62.83 = 3.38 A

V= Vm sin 2πft

= 212.13 sin 2 * π 50 t

=212.13 sin 100 πt

I$= Im \sin (2πfE-90^o) \\ = 33.8 \sin (100πt – 90^o)$

ADD COMMENT SHARE EDIT