Electrical Engineering Problem Set: Average Power Calculation in RLC Circuits, Assignments of Microelectronic Circuits

Download Electrical Engineering Problem Set: Average Power Calculation in RLC Circuits and more Assignments Microelectronic Circuits in PDF only on Docsity! Problem set #8, EE 223, 3/27/2003 – 4/03/2003 Chapter 13, Problem 19. Let ω =100 rad/s in the circuit of Fig. 13.49 and find the average power: (a) delivered to the 10-Ω load; (b) delivered to the 20-Ω load; (c) generated by the source. 100 rad/sω= Use mesh-analysis to determine the three currents (I1, I2, and I3, ) (a) (b) 220P 0.1144 20 0.2617 W= × = (c) P 100 0.5833cos88.92 1.1039 W= × ° =gen or: Use equivalent input impedances .... 1) let X = jωL = j100Ω, R2 = 10Ω, R3 = 20Ω, M2 = 0.5H, M3 = 0.2H 2) Z2 = X + ω2M22/(R2 + X) = 2.476 + j75.248 Ω, Z3 = X + ω2M32/(R3 + X) = 0.769 + j96.153 Ω, and Z1 = Z2 + Z3 = 3.245 + j171.4 Ω 3) I1 = V/Z1 4) V2 = Z2 I1 = 43.913 - j0.613 V and V3 = Z3 I1 = 56.087 + j0.613 V 5) PT = Re{V I1*}, P10 = Re{V2 I1*}, P20 = Re{V3 I1*} 1 2 1 2 3 2 1 3 1 3 1 2 1 1 1 1 2 K 50 , K 20 , 1H j100 100 j200 I 50 I 20 I 0 (10 100) I 50 I 0 (20 100) I 20 I 2 5 5 2I I , I I 10 20 5 2 I 2 10 1 10 1 10 2 10 25 410 20 I I 0.5833 88.92 A, I 1 10 2 10 → Ω → Ω → Ω = − − = + − = + −   ∴ = = ∴ = − − + + + +    ∴ = + + ∴ = ∠− ° = + +  j j j j j j j j j j j jj j j j j j j j j j 2 3 10 0.2902 83.20 A, I 0.11440 77.61 A P 0.2902 10 0.8422 WΩ ∠− ° = ∠− ° ∴ = × = Chapter 13, Problem 33. In the circuit of Fig. 13.60, let ZL be a 100- µF capacitor having impedance -j31.83 Ω. Calculate Zin when k = (a) 0; (b) 0.5; (c) 0.9; (d) 1. Verify with appropriate PSpice simulations. 1/ωC = XC = 31.83 ⇒ ω = 1/CXC = 314 rad/s, i.e. a 50 Hz system M = k √(L1L2) Zin = Z11 + ω2M2 / (R22 + jX22) = 20 + j31.4 + k2 [ω2L1L2 / (2 + j7.85 - j31.83)] = = 20 + j31.4 + k2 [ω2L1L2 / (2 - j23.98)] = = 20 + j31.4 + k2 [ω2L1L2 2 / (22 + 23.982) + jω2L1L2 23.98 / (22 + 23.982) ] = = 20 + j31.4 + k2 [493 / 579.04 + j5910.8 / 579.04 ] = = 20 + j31.4 + k2 [0.8514 + j10.21] (a) Zin ( k = 0 ) = 20.00 + j31.4 Ω (b) Zin ( k = 0.5 ) = 20.21 + j33.97 Ω (c) Zin ( k = 0.9 ) = 20.69 + j39.69 Ω (d) Zin ( k = 1.0 ) = 20.85 + j41.63 Ω