Download Problem Set #3 EE 221: Electrical Engineering Solutions and more Assignments Electrical and Electronics Engineering in PDF only on Docsity! Problem set #3 EE 221, 09/05/2002 โ 09/12/2002 1 Chapter 3, Problem 4. In Fig. 3.42, (a) How many nodes are there? (b) How many branches are there? (c) If we move from B to F to E to C, have we formed a path? A loop? Chapter 3, Solution 4. (a) Five nodes; (b) seven branches; (c) path, yes โ loop, no. Chapter 3, Problem 6. Referring to Fig. 3.44, (a ) Find ix if iy =2 A and iz =0 A. (b) Find iy if ix = 2 A and iz =2 iy (c) Find iz if ix=iy= iz . Chapter 3, Solution 6. KCL, we may write: 5 + iy + iz = 3 + ix (a) ix = 2 + iy + iz = 2 + 2 + 0 = 4 A (b) iy = 3 + ix โ 5 โ iz iy = โ2 + 2 โ 2 iy Thus, we find that iy = 0. (c) 5 + iy + iz = 3 + ix (let ix = iy = iz = i) 5 + i + i = 3 + i i = -2 A Chapter 3, Problem 22. Find the power absorbed by each of the six circuit elements in Fig. 3.56. Problem set #3 EE 221, 09/05/2002 โ 09/12/2002 2 Chapter 3, Solution 22. Applying KVL about this simple loop circuit (the dependent sources are still linear elements, by the way, as they depend only upon a sum of voltages) -40 + (5 + 25 + 20)i โ (2v3 + v2) + (4v1 โ v2) = 0 [1] where we have defined i to be flowing in the clockwise direction, and v1 = 5i, v2 = 25i, and v3 = 20i. Performing the necessary substition, Eq. [1] becomes 50i - (40i + 25i) + (20i โ 25i) = 40 so that i = 40 V/-20 โฆ = -2 A Computing the absorbed power is now a straightforward matter: p40V = (40)(-i) = 80 W p5โฆ = 5i 2 = 20 W p25โฆ = 25i 2 = 100 W p20โฆ = 20i 2 = 80 W pdepsrc1 = (2v3 + v2)(-i) = (40i + 25i) = -260 W pdepsrc2 = (4v1 โ v2)(-i) = (20i - 25i) = -20 W and we can easily verify that these quantities indeed sum to zero as expected. Chapter 3, Problem 32. (a) Apply the techniques of single-node-pair analysis to the upper right node in Fig. 3.66 and find ix .(b) Now work with the upper left node and find v8.(c) How much power is the 5-A source generating? Chapter 3, Solution 32. Define a voltage v9 across the 9-W resistor, with the โ+โ reference at the top node. (a) Summing the currents into the right-hand node and applying KCL, 5 + 7 = v9 / 3 + v9 / 9 Solving, we find that v9 = 27 V. Since ix = v9 / 9, ix = 3 A. (b) Again, we apply KCL, this time to the top left node: 2 โ v8 / 8 + 2ix โ 5 = 0 Since we know from part (a) that ix = 3 A, we may calculate v8 = 24 V. (c) p5A = (v9 โ v8) ยท 5 = 15 W. (generating +15 W!)
