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Transcript
TUTORIAL 1 DC CIRCUIT ANALYSIS TUTORIAL 1 CH. 4 Problems 13, 29, 39 CH. 5 Problems 7, 13, 38, 42 CH. 6 Problems 10, 15, 18, 22 TUTORIAL 1 CH. 7 Problems 6, 7, 8,16, 20 CH. 8 Problems 6, 7, 10 CH. 9 Problems 3, 4, 9, 13(II), 14(II), 18, 20, 23(a) CH. 4 Ans. R = 12.63 Ω Ans. W = 4.1 x 106 J Ans. I = 2.14 mA, P = 25.65 mW, W = 92.34 J Ans. W = 59.80 kWh CH. 5 7. For the series configuration in Fig. 5.91: a. Find the total resistance. b. Calculate the current. Ans. R = 40 Ω Ans. I = 3 A c. Find the voltage across each resistive element. Ans. V1 = 30 V, V2 = 36 V, V3 = 54 V CH. 5 13. For the circuit in Fig. 5.97: a. Find the total resistance, current, and voltage across each element. b. Find the power delivered to each resistor. c. Find the power delivered by the source. d. How does the power delivered by the source compare to that delivered to all the resistors. Ans. a. RT = 82 Ω, Is = 250 mA, VR1 = 5.50 V, VR2 = 2.50 V, VR3 = 11.75 V, VR4 = 0.75 V b. PR1 = 1.38 W, PR2 = 625 mW, PR3 = 2.94 W, PR4 = 187.50 mW c. PE = 5.13 W Fig. 5.97 CH. 5 38.For the network in Fig. 5.121 determine the voltages: a. Va, Vb, Vc, Vd, Ve b. Vab, Vdc, Vcb c. Vac, Vdb Ans. Not provided CH. 5 Ans. Not provided CH. 6 10. For the parallel network in Fig. 6.81: a. Find the total resistance. b. What is the voltage across each branch? c. Determine the source current and the current through each branch. d. Verify that the source current equals the sum of the branch currents. Ans. a. RT = 6 Ω b. VR1 = VR2 = 36 V c. Is = 6 A, I1 = 4.5 A, I2 = 1.5 A d. – Fig. 6.81 CH. 6 Ans. I’ = 12 A, I” = 8 A, V = 12 V CH. 6 Ans. a. I = 4 mA b. V = 24 V c. Is = 18.4 mA CH. 6 Ans. I1 = 5 A, I2 = 3.33 A, I3 = 1.67 A, I4 = 0.92 A Ans. Is = 10.92 A Ans. RT = 10.99 Ω Ans. Ps = 1.31 kW CH. 7 Ans. RT = 0.8 kΩ Ans. Is = 60 mA Ans. V = 19.2 V CH. 7 7. For the network in Fig. 7.67: 7 a. Find currents Is, I2 and I6. Ans. Is = 16 mA, I2 = 2.33 mA, I6 =2 mA b. Find voltages V1 and V5. Ans. V1 = 28 V, V5 = 7.2 V c. Find the power delivered to the 3 kΩ resistor. Ans. P3 = 261.33 mW CH. 7 Ans. a. Is = 16 A b. I3 = I9 = 4 A c. I8 = 1 A d. Vx = 14 V CH. 7 Ans. I = 0.6 A Ans. V1 = 28 V CH. 7 Ans. Vab= 14 V Ans. I = 9 A CH. 8 Ans. I1 = 12 A, Is = 11 A Ans. Vs = 24 V, V3 = 6 V CH. 8 Ans. I = 3 A, Rp = 6 Ω Ans. I = 4.09 mA, Rp = 2.2 kΩ CH. 8 Ans. a. E = 13.6 V, R = 6.8 Ω b. I1 = 458.78 mA (CW) c. Vab = 17.89 V CH. 9 Ans. I 24 V = 3.17 A, direction = ?? CH. 9 Ans. I1 = 4.45 mA, direction = ?? CH. 9 9. a. b. Find the Thevenin equivalent circuit for the network external to the resistor R for the network in Fig. 9.127. Find the power delivered to R when R is 2 Ω and 100 Ω. Ans. a. RTh = 7.5 Ω, ETh = 10 V b. P2Ω = 2.22 W, P100Ω = 0.87 W CH. 9 13. Find the Thevenin equivalent circuit for the portions of the networks in Fig. 9.131 external to points a and b. Ans. RTh = 2.06 kΩ, ETh = 16.77 V CH. 9 14. Find the Thevenin equivalent circuit for the network external to resistor R in both networks in Fig. 9.132. Ans. RTh = 4.03 kΩ, ETh = 12 V CH. 9 18. Find the Norton equivalent circuit for the network external to resistor R for the network in Fig. 9.126. Ans. RN = 14 Ω, IN = 2.57 A CH. 9 20. Find the Norton equivalent circuit for the network external to resistor R for the network in Fig. 9.129. Ans. RN = 1.58 kΩ, IN = 0.73 mA CH. 9 23(a). Find the Norton equivalent circuit for the portions of the network in Fig. 9.136(a) external to branch a-b. Ans. RN = 3 Ω, IN = 5 A