Question

Problem 6.5 1 of 8 The current in a 200 mH inductor is Part A i=75 mA, i= (Bics 200t+B2 sin 200t)ese A, > 0, where t is in seconds. The voltage across the inductor (passive sign convention) is 4.25 Vatt-0 ts0 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 p= 20.04 mW SubmitPre Previous Answers Request Answer X Incorrect, Try Again; 3 attempts remaining

This answer was from an expert response, I got wrong for that as you could see. Can any expet on the filed help me here please. Hint: the power is a delivering power which is negative. Thanks  

Answer 1

Given that Inductor (d) = 200 m mu i_L(0) = 75 mA i_L(t) = (B_1 cos 200t + B_2 sin 200t)e^-50t V_L(0) = (200m) di_L(0)/dt = 4.45 V di_L(0)/dt = 21.25 n/s power(P) = The inductor is observing (or) delivering power Differentiating i_L(L) w_0r_0 + 't' di_L(t)/dt = d/dt [(B_1 cos 200t + B_2 sin 200t)e^-50t] = [B_1((-200)sin 200t - 50 cos 200t)e^-50t B_2 ((200) cos 200t - 50 sin 200t) e^-50t] After that di_L(0)/dt id substituted in eq. t = 0 di_L(0)/dt = d/dt [(B_1 cos 200(0) + B_2 sin 200 (0))e^-50(t)] \r\n = [B_1((-200)sin 200(0) - 50 cos 200(0))e^-50(0) B_2 ((200) cos 200(0) - 50 sin 200(0)) e^-50(0)] diL(0)/dt = -50B_1 + 200B_2 = 21.25 i_L (0) = B_1 = 0.075 B_2 = 21.25 + 50 (0.075)/200 B_2 = 21.25 + 3.75/200 B_2 = 25/200 B_2 = 0.125 After that Substitute B_1 = 0.075 &B_2 = 0.125 in the expression of i(t)