impedances - Analysis of AC Circuits Let's represent this...

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Analysis of AC Circuits Let’s represent this circuit by a differential equation. First, () C d it C v t dt = . Then KVL gives () () CC s 0.0075 dd vt R C vt vt vt vt dt dt =+ = + When the source voltage is ( ) ( ) s 25cos 100 15 V vt t ° we expect ( ) ( ) C cos 100 vt A t θ = + Substituting into the differential equation, we get ( )() 25cos 100 15 0.0075 cos 100 cos 100 0.0075 100 sin 100 cos 100 0.75 sin 100 cos 100 d tA t A t dt At A t A t θθ +°= + + + ⎡⎤ ⎣⎦ =− + + + + + + Solution using trigonometry: Recall that ( ) ( ) ( ) ( ) cos cos cos sin sin α βα β ±= and ( ) ( ) ( ) ( ) sin sin cos cos sin ± so ( ) ( ) () ( ) () ( ) () ( ) () ( ) ( ) () () 25cos 15 cos 100 25sin 15 sin 100 0.75 sin cos 100 cos sin 100 cos cos 100 sin sin 100 0.75sin cos cos 100 0.75cos sin cos 100 tt A A t t °− ° + + + + + t Equating the coefficients of ( ) cos 100 t and ( ) sin 100 t gives ( ) ( ) ( ) ( ) ( ) ( ) 25cos 15 0.75sin cos 20 V and 22 25sin 15 0.75cos sin A A A °= − + ⇒= = −° = + ° That is ( ) ( ) C 20cos 100 22 V t °
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Solution using Euler’s identity: Euler’s identity is: { } cos sin cos Re j j ee j θ θθ =+ = Using , we can write the differential equation as () ( ) sin cos 90 −= + ° () ( ) ( ) 25cos 100 15 0.75 cos 100 90 cos 100 tA t +°= ++°+ + ⎡⎤
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This note was uploaded on 05/10/2008 for the course EEL 3004 taught by Professor Gong during the Spring '08 term at University of Central Florida.

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impedances - Analysis of AC Circuits Let's represent this...

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