Homework 3 solution - Principles and Applications of...

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Unformatted text preview: Principles and Applications of Electrical Engineering, Fourth Edition - Giorgio l Rizzoni Semester -2, Homework 3 Problem 3.1 Solution: Known quantities: Mesh currents: I] = 5 A, 12 = 3 A, 13 = 7 A. Find: The branch currents through: R], R2, R3. Analysis: Assume a direction for the current through RI (e.g., from node A to node B). Then summing currents at node A: KCL: ——I, +1]?l +13 =0 [RI 21' -I3 =—2A This can also be done by inspection noting that the assumed direction of the current through R1 and the direction of I I are the same. Assume a direction for the current through R2 (e.g., from node B to node A). Then summing currents at node B: KCL: 12+IR2-I320 1R2 =I3—IZ=4A This can also be done by inspection noting that the assumed direction 0! the current through R2 and the direction of I 3 are the same. Only one mesh current flows through R3. If the current through R3 is assumed to flow in the same direction, then: 1R,=1,=7A, Problem 3.2 Solution: Known quantities: Source and node voltages: VS] 2 VS2 2110 V, VA = 103 V, VB = —107 V. Find: The voltage across each of the five resistors. Analysis: Assume a polarity for the voltages across R1 and R2 (e.g., from ground to node A, and from node B to ground). R1 is connected between node A and ground; therefore, the voltage across R] is equal to this node voltage. R2 is connected between node B and ground; therefore, the voltage across R2 is equal to the negative of this voltage. V,H = V, = 103 v, VR2 =—VB = +107 v The two node voltages are with respect to the ground which is given. Assume a polarity for the voltage across R3 (e.g., from node B to node A). Then: KVL: VA +VR3 +VB =0 VR3 =VA-VB =210V Assume polarities for the voltages across R4 and R5 (e.g., from node A to ground , and from ground to node B): u“ N-.." ,. .,...,.._... .. .4.“ .“w-—_ W” A...“ .. . . i .. .. ‘. ~_..-.. ,r -..,.A_ KVL: _VSI+VR4+VA =0 1 VR4 2V5] _VA =7V KVL: —VS2 —VB -VRS :0 VRS : _Vsz ‘VI; 2‘3 V Problem 3.4 Solution: Known quantities: Resistance values, current sources values. Find: The currents 1|, [2. Analysis: At mesh (3): i" =1 A At mesh (b): 3(i,, —iu)+i,, +6(i,, —iC)=0 At mesh (c): V " z 2 A m (i; “» l a it» Solving, we find that: ’ s T ib =1.5A .. l Then, i. = (in —i,,)= —0.5 A i, = (i, 4,): —0.5 A Problem 3.8 ‘v "‘ Solution: . Known quantities: / The voltage source value, 15 V, and the four resistance values, indicated in Figure P18. Find: The voltage at nodes a and b, Va and Vb , and their difference, Va — Vb. Analysis: Using nodal analysis at the two nodes a and b, we write the equations V — 15 V ” + —” = 0 1 8 20 Va - 15 Va _ + _ 36 20 Rearranging the equations, 38V, —300 = 0 14V“ —75 = 0 0 Solving for the two unknowns, Vu=5.§6V and V,=7.89V G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 Therefore, Va — V, = —2.54 V Problem 3.14 Known quantities: The values of the resistors and of the voltage sources (see Figure P3.14). Find: The voltage across the 10 Q resistor in the circuit of Solution: Figure P3.14. J / azvfw 1 f Analysis: For mesh (a): i" 50+20+20)—ib(20)—ic(20)=12 For mesh (b): —iu(20)+ib (20+ 10)—ic(10)+ 5 = 0 For mesh (c): —ia(20)—ib(10)+ic(20+10+15)=0 Solving, i” =127.5mA Q=—6T8mA Q=4L6mA and v,“ =10(i,, —ic)= 10(—0.109)= —1.09 mV. Problem 3.15 Solution: Known quantities: 20 Q 103'.) The values of the resistors, of the : , , voltage source and of the current , source in the circuit of Figure P3.16. 3 V g‘ + > , Find: z The voltage across the current source. ‘2 "45.... ..... 0,525» ’ 2 «M25 «3 Vi Analysis: For mesh (a): i” (20 + 30)+ i, (* 3o): 3 For meshes (b) and (c): iu(—30)+ i,(10+ 30)+ir(30+ 20): o For the current source: 1; —i,, = 0.5 Solving, "i, =—133mA, i, =—322mA and ic = 178 mA. Therefore, v=ic(30+20)=8.89V. ...
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This note was uploaded on 10/27/2009 for the course ECTE 290 taught by Professor Zhengli during the Three '09 term at University of Wollongong, Australia.

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Homework 3 solution - Principles and Applications of...

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