Question

The current in a 200 mH inductor is i=75 mA, t< 0; i= (B cos 200t + B, sin 2000) 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 = 24 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 delivering power. The inductor is absorbing power. Submit Request Answer

Answer 1

given that tco i= 75 mA iz TB, Cos (200t) + Ba Sin (200t]] è sot tyo VL Lail dt -sot Yal (B, Cos(2004) + B2 Sinacot) e X-50 [200 B, Sin 2004+200 + by Cozabile me? vez lleset (lemon - Soc.,) calzons) + (-2000, -500.) sin trasy {L= 0.2H} given Veltao) = 4:25 L(200 B2 - 50B,) 4.25 4.25 40 B2 - 10 B, -lo Bit 40 B2 = 4.25" Ilto) Į (tot) Bi 75mA So B2= 125mA So Vaa V= Lesot (21.25 C 2008) – 21.255in (2004) ) e sot ( 4.25 Cos (2004) -4-25 Sin (2001)) - S0x24x102 15 Cos (200x 24x164) +125 Sin (200x248709]] e IL (1=24ms) rad Il (t=24ms) र -35.5283mA (30 x 24x163 V VL (t=24mi) ? 4.29 Cos 200 X 0.034) – 4.25 Sin (200X0024) Vilt-zums) 1.3872 V

So Power a V Izazum > 1.3872 x -35. 5283 ( Uma) Power 는 -49.2838 W here negative Sign represent the delivered power So Power at terminal 49.2838 & (6) Current is exit from inductor, So it is delivering Power.