Question

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 1

A - 15 mA 720 -50+ 420 1 :( B, Cos2001 + B, sin2oot De 50 ico) ilo)icot) 75 x10, (+0) 75x103 - Bi 4.25 Given v (0) VE L di di 200x103 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 las x103 tzo i(t)= 1031 75 805 (200+) + las sirda004) 2001) 250+ CS Scanned with CamScanner

v (to 28ms) - L.di(28) dilma dt. op (10%(75605200+ + 125 Sin (200 001) e sot 200+) 2507 - 10 10 (15x200 5x200 (- Sin2001) + 125x200 cos 2007 50 C sot +(1 75 (05 2004 +1255in 200 t) -3 di (1) -500 - 500 2001] xvó 575 Cosaoot 175 Sin 200t -50%a8 xio3 575 COS (200x28x10>) + (-175 (282) (5956 50 e sin ( 200x28x10 vo*]evo 6.84 Als Ev.9) X 684 200 xio 3 1.368 V vi al 28 MS as Power ilasms) = 1ö(15005(2004) + 125 sin(2001) 250) Ht28 ila8ms) - 21.414 MA CS Scanned with CamScanner

a> Power 1.368 x 21.414 x103. Power 29.3 mw Part B : Element Power is positive 50 power. It is absorbing CS Scanned with CamScanner