Question

The output voltage of an AC generator is given by V = (1.20×10^2) sin (30 pie t) (V). The generator is connected across a 0.500 H inductor. Find the (a) frequency of the generator, (b) rms voltage across to the inductor (c) inductive reactance, (d) rms current in the inductor, (e) maximum current Indian doctor, and (f) average power delivered to the inductor. (g) find an expression for instantaneous current. (h) At what time after t = 0, does the instantaneous current first reach 1 A? Please answer each question completly with a detailed explanation. Thanks in advance!

Answer 1

from the given data

V_max = 1.20*10^2 V = 120 V
w = 30*pi rad/s
L = 0.500 H

a) frquency of the generator, f = w/(2*pi)

= 30*pi/(2*pi)

= 15.0 Hz <<<<<<<<<--------------Answer

b) rms voltage across the inductor = rms voltage across the generator

Vrms = Vmax/sqrt(2)

= 120/sqrt(2)

= 84.8 V   <<<<<<<<<--------------Answer

c) inductive reactnace, XL = w*L

= 30*pi*0.5

= 47.1 ohms <<<<<<<<<--------------Answer

d) Irms = Vrms/XL

= 84.8/47.1

= 1.80 A <<<<<<<<<--------------Answer

e) Imax = Vmax/XL

= 120/47.1

= 2.55 A   <<<<<<<<<--------------Answer

f) Avarge power delivered to the inductor = 0   <<<<<<<<<--------------Answer

Because, inductor does not consume energy.

g) we know, in inductor current lags voltage by pi/2 radians

so, current in the inductor,

I = Imax*sin(w*t - pi/2)

I = 2.55*sin(30*pi*t - pi/2) A     <<<<<<<<<--------------Answer

h) let time t, I = 1 A

use, I = 2.55*sin(30*pi*t - pi/2) A

1 = 2.55*sin(30*pi*t - pi/2)

1/2.55 = sin(30*pi*t - pi/2)

sin^-1(1/2.55) = 30*pi*t - pi/2

0.403 = 30*pi*t - pi/2

0.403 + pi/2 = 30*pi*t

==> t = (0.403 + pi/2)/(30*pi)

= 0.0209 s <<<<<<<<<--------------Answer