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Sinusodial RL anaylysis

 
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david90



Joined: 05 May 2004
Posts: 373
Helped: 1


Post10 Oct 2004 4:39   Sinusodial RL anaylysis

I have Source = 20cos(800+25), Resistor = 80, L=75mh and they are all connected in series. Find the current I(t)

My teacher wrote out 20cos(800+25)= 80I+75mh(di/dt). Then I he used euler's ident to convert 20cos(800+25) to 20e^(j25).

so 20e^(j25)= 80I+75mh(di/dt)

after i'm kinda lost on what to do. Can somebody help me finish it by using the method above?
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