Question

The current in a 200 mH inductor is į = 75 mA, t = 0; i = (B1 cos 200t + B2 sin 200t) e -50t 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 = 23 ms. Express your answer to three significant figures and include the appropriate units. MÅ ? p = Value Units Submit Request Answer Part B State whether the inductor is absorbing or delivering power. The inductor is absorbing power. The inductor is delivering power.

Answer 1

Given, h = 200 mit Ĉ= 75mA 9 t ļo 507 dit (B, cos 200t + B2 Sin 200t) é Amp. t>,0 voltage aeross Endue for ot too is ve= 4,25 volt We know, VL at di at ☆ Vilt=0) cl dio) * dico) 4:25 2008 10-3 ► dico) 21025 at At t=0, Current equate'on for tro becomes ¿co) ico) + B4 = By dt at 11 B (Bu toeo 75x1003 > O. 075

ditt) - -200 B1 Differentiate current equation for tyo ė 50 *** [-200 finqoot + 200 B2 cos 2007 at By cos aoot + B2 bin 2004) x 500x-50 At t=0 dilo) [0+ 26082] – 50 [4] of c >> dt 21:25 200 32 200 32 - 50% 0.075 » 200 B2=25 7 B2=00125 induetion The total cument urrent flows through i's, èct) = 100) + ilt) for to Ć (oo) =0 → LCH) ¿(t) 9

-50% top, Elt) = (0.075 cos 2006 +0.125 bin 200t) é At t=23msee * t 0.023 see Ĉz (0.023) = 0.075 cos (200 * 0.023) + 0.125 bin (200 x00022) 50% 0.023 X flow through A -3 Hér (0.023) Co. 000758 -0.112) + (0.125x-019931)]0.3166 → z (0.023) = 0.042 Amps At t= 23 m see, Current Enductor is -0.042 Amp. power at the terminals of the incluctor at t=23msee is, pa Lea à x 200x1oºx (-0.042)2 23 X10-3 p = 7.67 X1003 P = 7,67 m watts 6 Normally inductor absorbs power, when Current through it po is positive. But at t=23msee current Flows through it is negative. Hence Enductor delivering power at that moment (or) Encluctor discharges stored en energy. t ť i

(A) P=7.67 mW

(B) Inductor delivering power because current flows through inductor at t=23 msec is negative or we can say that inductor discharges stored energy at that moment.