Electrical Eng. Problem Set: Power Absorption & Avg. Power Calculation in Circuits, Assignments of Electrical and Electronics Engineering

Download Electrical Eng. Problem Set: Power Absorption & Avg. Power Calculation in Circuits and more Assignments Electrical and Electronics Engineering in PDF only on Docsity! 1 Problem set #10a, EE 221, 11/12/2002 – 11/19/2002 Chapter 11, Problem 5. The circuit shown in Fig. 11.24 has reached steady-state conditions. Find the power being absorbed by each of the four circuit elements at t = 0.1 s. rad rad , 8 8 40 53.135 0 A, C 4 , 8 (3 4) 11 4 3.417 33.15 17.087 33.15 , 17.087cos (25 33.15 ) V P (0.1) 17.087cos (2.5 33.147 ) 5cos 2.5 23.51 W 17.087 cos (25 33.15 ) 8 (0.1) 2.136c s in s s s abs i j j j v t i t i ∠− ° → ∠ ° → − Ω = − = − = ∠ − °∴ = ∠− ° = − ° ∴ = − − ° × = − = − ° ∴ = Z V rad 2 8, 3 rad 3 2 3, os (2.5 33.15 ) 0.7338 A P 0.7338 8 4.307 W; 17.087 33.15 3.417 19.98 A 3 4 (0.1) 3.417cos (2.5 19.98 ) 3.272A P 3.272 3 32.12 W 4(3.417 19.983 ) 13.67 70.02 , (0.1) 13.67 abs abc c c j i j v − ° = − ∴ = × = ∠− ° = = ∠ ° − ∴ = + ° = − ∴ = × = = − ∠ ° = ∠ − ° = I V rad , 0cos (2.5 70.02 ) 3.946V P 3.946( 3.272) 12.911 W ( 0)c abc − ° = ∴ = − = − Σ = Chapter 11, Problem 7. Calculate the average power generated by each source and the average power delivered to each impedance in the circuit of Fig. 11.26. Chapter 11, Solution 7. 10, 10, 2 8 20 40 30(10 10) 52.44 69.18 V 5 50 8 20 1P 10 52.44cos 69.18 93.20 W 2 1P 10 52.44cos (90 69.18 ) 245.08 W 2 1 52.44P 8cos ( 20 ) 161.51 W ( ) 2 8 gen j gen abs gen abs j ∠− ∠ ° = + = ∠ ° ∠ ° + ∠ − ° = × × ° = = × × ° − ° =  = − ° = Σ = Σ    V P5∠50º, abs = ½ (52.44/5)2 5 cos(50º) = 176.76 W 2 Chapter 11, Problem 10. Determine the average power supplied by the dependent source in the circuit of Fig. 11.29. Chapter 11, Solution 10. 20 2 , 2 3 3 60 2 2 12 5 14 60, 0 3 2 2 2 3 0, 2 (2 3) 0 60 14 0 2 3 120 180 9.233 83.88 V 5 14 10 15 28 2 2 3 5 60 2 0 5.122 140.9 V 18 15 1P 9.233 2 5 2 x x c c x x c c c x c x c c x c x c x c gen j j j j j j j j − − + = − + − = − ∴ − = + = − ∴ − + = − + + = − + + = = = ∠ − ° − + − − + − = = ∠ − ° ∴ − + = × × × V V V V V V V V V V V V V V V V V V V V .122cos ( 83.88 140.19 ) 26.23 W− ° + ° = Chapter 11, Problem 15. For the network of Fig. 11.32: (a) what impedance ZL should be connected between a and b so that a maximum average power will be absorbed by it? (b) What is this maximum average power? (a) (b) max 480 80 6080 60 80 60 80 60 28.8 38.4 28.8 38.4 th L j jj j j j j − = = + − = + Ω∴ = − Ω Z Z 2 2 ,max 2 5(28.8 38.4) 144 192V, 144 192 2 28.8 1 144 192and P 28.8 250 W 2 4 28.8 th L L j j j = + = + + ∴ = × + = × = × V I