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# 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 i t C v t dt = . Then KVL gives ( ) ( ) ( ) ( ) ( ) C C C C s 0.0075 d d v t RC v t v t v t v t dt dt = + = + When the source voltage is ( ) ( ) s 25cos 100 15 V v t t = + ° we expect ( ) ( ) C cos 100 v t 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 t A t A t dt A t A t A t A t θ θ θ θ θ θ + ° = + + + = + + = − + + + + Solution using trigonometry: Recall that ( ) ( ) ( ) ( ) ( ) cos cos cos sin sin α β α β α ± = β and ( ) ( ) ( ) ( ) ( ) sin sin cos cos sin α β α β α ± = ± β 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 t t A t t A t A t t θ θ θ θ θ θ θ θ ° ° = − + + + = + − − + 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 θ θ θ θ θ ° = + = = − ° = − + °

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