The current in a 200 mH inductor is i = 75 mA, t < 0; -50+...
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Question “The current in a 200 mH inductor is i = 75 mA, t < 0; -50+..."
The current in a 200 mH inductor is i = 75 mA, t < 0; -50+ i= (B1 cos 200t + B2 sin 200t) e A, t>0, where t is in seconds. The voltage across the inductor (passive sign convention) is 4.25 V at t = 0. Part A Calculate the power at the terminals of the inductor at t = 28 ms Express your answer to three significant figures and include the appropriate units. μΑ p = Value Units Submit Request Answer Part B State whether the inductor is absorbing or delivering power. O The inductor is absorbing power. O The inductor is delivering power. Submit Request Answer
Answer
A – 15 mA 720 -50+ 420 1 🙁 B, Cos2001 + B, sin2oot De 50 ico) ilo)icot) 75 x10, (+0) 75×103 – Bi 4.25 Given v (0) VE L di di 200×103 dl (o esot. (B, Cos 200+ + Sin 2001) VE $o e sot {2003, v=6.2 cosaoot) e sot Sin 200 + 200 B2 Cosaoo BSin 2002) B, Cos 200+ + V[t=0): 0.2 I doo Be solar]] v (tuo) – 2 [ 2006, – 50(15 x 10) = 4.25 200 Ba – 25. Ba lasx103 tzo = 1031 75 805 (+ las sirda004)2001) 250+ CS Scanned using CamScanner
v (to 28ms). – L.di(28). dilma.dt. op (10% (75605200+ +125 Sin (200001) e SOT 200+) 2507 – 1010 [-500 – 500 2001] xvo 575 Cosaoot 175 sin 200t -50%a8 (200x28x10> + (-175(282) (5956) xvo 3 575 COS (200x28x10> + (-175(282) xio3 575) xvo 684 Vvi.)evo*]evo3 CS]evo3 x10 (2001) 250) Ht28) ila8ms (2001)
a> Power 1.368×21.414×103. Power 29.3mw Part A : Element Power equals 50 power. It absorbs CS Scanned using CamScanner
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