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Question 2 For the circuit shown in Fig. 3, E is a dc source given by E = 15 V, and Is is an ac source given by is = 0.5V2 si
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D is = 0.5.12 Sin (2000€ + 20) A Taking cosine as reference for phaser representation és = 0.5 ja cos (2000ť +20°-90) A = 0.b) E acting alone (Is=o or open circit) Since this was a De source, capacitor = open Inductor = short у 4 р ь 5 V in § 11503 o tomamo KCL at node - 55/ 18 .-15 2 at node - V = 0 V utj8 15 Kel at node 0, V-55LIP+ := + V - VW -0 -0 @, Varvast van ve

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