AC Circuit Exercise

User Generated

nygnjvy89

Engineering

Description

I have done this exercise, I just need to make sure that my solution is correct for # 7..

Note: Please check out # 7. Find the sinusoidal expression and time domain for the voltages and current if the frequency is 60 Hz.

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Find 1Z » I 2) dieg vant) y OP 5) Y 8.62 no orom + Ve - VL - IV Er 10020 6. Bonus --#7 Verify KVL KVL=E=1p+ V nga rectangular forma Vp - 6014 – 53-13° -360-3 48v. V = 800236.87 = 6484 ; 480 E = Ve+ v = 36v-J45v)+ (54» +j48V) - 2 + 2 = 60 2048 s (90° - 62 + j&r 2= 105 L 53.130) $2 = 1000+ jo Polar form: Os 53.18 7* tant ( tant (100) = 0 √ 100²0² 100²+0² = rio 10000 100 1000 Loo R=60 y cont 7. Bonus #8 Find the sinusoidal expressions and time domain for the voltages and current if the frequency is 60 Hz 3) I - E loovt0° =10 A 2-53.13° 7) ZT 10325313 4) Ve= IZR= (104.45313) (64220°) = 60v2-53-13° 5) V - IZ = (104 L53439) (832L90) = 80v 236.87 (6P = I cos Of (100)(101) COS 53:13 600w V I The sinusoidal erpression for voltage and curent: VR = 12 I'm sin (wt +O) = SE (601) sin (27607 +1-5343) 84-85 sin (12077-53.130) => 84.85 sin ( 3777-53.13°) V = 12 Vm sin (wt +O) V2 (80 v) sin(27607 + 36187) = = 113.13 sin (12016+36,87") = 113./3 sin (3776+ 36.87°) I = 52 Im sin (wt+o). 12 (104) sin (27760t+(-5333) 14.14 sin (1200t – 13.13) = 14.14 sin (377 - 53.13) V I The time domain for Voltage and current: V₂ = √2 (60)sin (wt - 53.13°) = 84 85 sin (wt - 53.13) V = 52 (80) sin (ut + 36.87°) = 113.13 sin (wt + 36,87) I= √2 (10) sin (wt – 53,13%) = 14.14 sin (wt - 53,13)
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Explanation & Answer

I have attached the complete solution in the document.

For voltage, 𝑉(𝑡) = 𝑉𝑚 sin(𝜔...


Anonymous
Very useful material for studying!

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