i=I0sin(ωt−ϕ0)=5sin(100πt−0.435) .
Comparing left and right side, find:
a) The amplitude is I0=5A , frequency is f=ω/2π=50Hz , periodic time is T=1/f=0.02s and phase angle is ϕ0=0.435rad=0.435⋅180/π=24.9°.
b) The value of current at t = 0: i(0)=5sin(100π⋅0−0.435)=5sin(−0.435)=−2.107A
c) The value of current at t =8ms: i(8ms)=5sin(100π⋅8⋅10−3−0.435)=4.37A
d) The time when the current is first a maximum is a time, when dtdi=0 . Thus, dtdi=5⋅100πcos(100πt−0.435)=0⇒100πt−0.435=π/2⇒t=6.38ms.
e) The time when the current first reaches 3A: i(t)=5sin(100π⋅t−0.435)=3⇒100π⋅t−0.435=arcsin53⇒t=3.43ms
Sketch one cycle of the waveform showing relevant points:
Comments
Thanks for awesome explanation.