Question #113247

The current in an a.c. circuit at any time t seconds is given by:

i = 5 sin(100πt − 0.432) A

Determine (a) the amplitude, frequency, periodic time and phase angle (in

degrees), (b) the value of current at t = 0, (c) the value of current at t =8ms,

(d) the time when the current is first a maximum, (e) the time when the current

first reaches 3A. Sketch one cycle of the waveform showing relevant points.

Expert's answer

i=I0sin(ωtϕ0)=5sin(100πt0.435)i = I_0\sin(\omega t-\phi_0) = 5\sin(100\pi t-0.435) .


Comparing left and right side, find:

a) The amplitude is I0=5AI_0 = 5 A , frequency is f=ω/2π=50Hzf = \omega/2\pi = 50 Hz , periodic time is T=1/f=0.02sT = 1/f = 0.02s and phase angle is ϕ0=0.435rad=0.435180/π=24.9°.\phi_0 = 0.435 rad = 0.435\cdot 180/\pi = 24.9 \degree.

b) The value of current at t = 0: i(0)=5sin(100π00.435)=5sin(0.435)=2.107Ai(0) = 5\sin(100\pi \cdot 0-0.435) = 5\sin(-0.435) = -2.107A

c) The value of current at t =8ms: i(8ms)=5sin(100π81030.435)=4.37Ai(8ms) = 5\sin(100\pi \cdot 8\cdot10^{-3}-0.435) =4.37 A

d) The time when the current is first a maximum is a time, when didt=0\frac{di}{dt} = 0 . Thus, didt=5100πcos(100πt0.435)=0100πt0.435=π/2t=6.38ms.\frac{di}{dt} = 5\cdot 100\pi \cos(100\pi t-0.435) = 0 \Rightarrow 100\pi t-0.435 = \pi/2 \Rightarrow t = 6.38ms.

e) The time when the current first reaches 3A: i(t)=5sin(100πt0.435)=3100πt0.435=arcsin35t=3.43msi(t) = 5\sin(100\pi \cdot t-0.435) = 3\Rightarrow100\pi \cdot t-0.435 = \arcsin\frac{3}{5}\Rightarrow t = 3.43ms


Sketch one cycle of the waveform showing relevant points:





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