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KING FAHD UNIVERSITY OF PETROLEUM & MINERALS ELECTRICAL ENGINEERING DEPARTMENT FIRST SEMESTER 2009-2010 (091) Course Title: Fundamentals of Electric Circuits Course Number: EE 204 Exam Type: MAJOR EXAM II Date: December 23, 2009 Time: 07:00 pm – 8:30 pm (1 & 1/2 hours) Indicate your lab section number from the list below (lab section/instructor name/lab starting time/day) 51/Salim/2:10/S 52/Al-Saihati/8:00/U 53/Salhab/11:00/U 54/Al-Saihati/11:0/U 55/Alfaouri/2:10/U 56/Ali/2:10/U 57/Mahmoud/2:10/M 58/Akber/2:10/M 59/Al-Saihati/8:00/T 60/Salhab/11:00/T 61/Ismail/11:00/T 62/Alfaouri/2:10/T 63/Akber/2:10/T 64/Ghani/2:10/W 65/Al-Safadi/2:10/W Student Name: ________________________________ Student ID: ________________________________ Section: ________________________________ Serial Number: ________________________________ GRADING Question 1: 16 Question 2: 20 Question 3: 24 Total: 60 Be neat, organized, and show all your work and results. Question 1: For the given ‘is’, a) Find the equations for 'VC' for the following intervals. (i) 0 < t ≤ 1 (ii) 1 < t < ∞ b) Plot the result in part a. c) Plot the voltage 'VR'. Solution: (a) Find the equivalent capacitance: Ceq = [4 in series with 4] in parallel with 2 = [(4*4)/(4+4)]+2 = 2+2 = 4 F. (i) 0 < t ≤ 1 1 VC C eq t t 1 1 2 t 1 2 i ( ) d d ( )0 t s 4 0 8 8 (ii) 1 < t < ∞ 1 VC C eq t 1 1 1 i ( ) d i ( ) d V (1) s C s 4 0 8 (b) and (c) Question 2: Using only the superposition principle, determine the following: a) Thevenin's equivalent circuit that is connected to the resistor RL. b) RL that absorbs maximum power. c) The maximum power absorbed by RL. (a) We need to find RTh and VOC Rth: Deactivate the two sources RTh [20 // 60] 5 1200 5 15 5 20 80 VOC: use superposition The voltage source circuit 48 3 0.6 A 80 5 V OC 1 60I 60(0.6) 36 V I Or using voltage division rule V OC 1 60 3 48 48 36 V 80 4 The current source circuit V OC 2 6(20) 120 V V OC V OC 1 V OC 2 36 120 84 V Draw the equivalent circuit as shown (b) RL RTh 20 (c) (2 marks for the power calculation) Pmax V OC2 4RTh (84) 2 4(20) 7056 441 88.2 W 80 5 Question 3: For the circuit shown below: a) Write the three node equations for node A, B, and C and put them in the standard system of equations. b) Find the node voltages VA, VB, and VC. c) What is the power dissipated in (absorbed by) R1. d) Which method you prefer to use node voltage method or mesh current method? Why? Solution: V A V B I1 I 2 0 a) R1 V A V B 18 1 V A 1 V B 0 V C 18 (1) V B V A V B V C V B 0 0 R1 R2 R3 V A 6V B 3V C 0 (2) KVL for V1 0 VA 0 V B 1 V C 4 (3) 1 1 0 VA 18 1 6 3 V 0 B (1 mark) 0 0 1 VC 4 b) Substitute from 3 into 2 V A 6V B 12 (4) Solve (1) and (4) V B 6V ,V A 24V VA 24V VB 6V VC 4V c) (2 marks) V R 1 V A V B 18V I R 1 I 1 I 2 3A P I R1 9*6 54W 2 R1 OR P V R21 54W R1 d) Mesh is better. As the current in two meshes are known by using KCL directly (I1, I2). We will have only one equation in one unknown for the third mesh contains R2, R3, and V1. (2 marks)