Chapter 3, Problem 42. Problem set #5 EE 221, 09/26/2002 10/03/2002 1 In the circuit of Fig. 3.75, choose v 1 to obtain a current i x of 2 A. Chapter 3, Solution 42. We first simplify as shown, making use of the fact that we are told i x = 2 A to find the voltage across the middle and right-most 1-W resistors as labeled. By KVL, then, we find that v 1 = 2 + 3 = 5 V. Chapter 3, Problem 63. What is the power dissipated by (absorbed by) the 47-kΩ resistor in Fig. 3.93? Chapter 3, Solution 63. The controlling voltage v 1, needed to obtain the power into the 47-kΩ resistor, can be found separately as that network does not depend on the left-hand network. The right-most 2 kω resistor can be neglected. By current division, then, in combination with Ohm s law, v 1 = 3000[5 10-3 (2000)/ (2000 + 3000 + 7000)] = 2.5 V Voltage division gives the voltage across the 47-kΩ resistor: 47 0.5(2.5)(47) 0.5v 1 = = 47 + 100 20 47 + 16.67 0.9228 V So that p 47kΩ = (0.9928) 2 / 47 10 3 = 18.12 µw
Problem set #5 EE 221, 09/26/2002 10/03/2002 2 Chapter 4, Problem 3. Use nodal analysis to find v P in the circuit shown in Fig. 4.25. Chapter 4, Solution 3. The bottom node has the largest number of branch connections, so we choose that as our reference node. This also makes v P easier to find, as it will be a nodal voltage. Working from left to right, we name our nodes 1, P, 2, and 3. NODE 1: 10 = v 1 / 20 + (v 1 v P )/ 40 [1] NODE P: 0 = (v P v 1 )/ 40 + v P / 100 + (v P v 2 )/ 50 [2] NODE 2: -2.5 + 2 = (v 2 v P )/ 50 + (v 2 v 3 )/ 10 [3] NODE 3: 5 2 = v 3 / 200 + (v 3 v 2 )/ 10 [4] Simplifying, 60v 1-20v P = 8000 [1] -50v 1 + 110 v P - 40v 2 = 0 [2] - v P + 6v 2-5v 3 = -25 [3] -200v 2 + 210v 3 = 6000 [4] Solving, v P = 171.6 V Chapter 4, Problem 5. For the circuit of Fig. 4.27, (a) use nodal analysis to determine v 1 and v 2, (b) Compute the power absorbed by the 6-Ω resistor. Chapter 4, Solution 5. Designate the node between the 3-Ω and 6-Ω resistors as node X, and the right-hand node of the 6-Ω resistor as node Y. The bottom node is chosen as the reference node. (a) Writing the two nodal equations, then NODE X: 10 = (v X 240)/ 3 + (v X v Y )/ 6 [1] NODE Y: 0 = (v Y v X )/ 6 + v Y / 30 + (v Y 60)/ 12 [2] Simplifying, -180 + 1440 = 9 v X 3 v Y [1] 10800 = - 360 v X + 612 v Y [2] Solving, v X = 181.5 V and v Y = 124.4 V Thus, v 1 = 240 v X = 58.50 V and v 2 = v Y 60 = 64.40 V (b) The power absorbed by the 6-W resistor is (v X v Y ) 2 / 6 = 543.4 W
Problem set #5 EE 221, 09/26/2002 10/03/2002 3 Chapter 4, Problem 12. With the help of nodal analysis on the circuit of Fig. 4.34, find (a) v A ; (b) the power dissipated in the 2.5-Ω resistor. Chapter 4, Solution 12. Choosing the bottom node as the reference terminal and naming the left node 1, the center node 2 and the right node 3, we next form a supernode about nodes 1 and 3, encompassing the dependent voltage source. At the supernode, 5 8 = (v 1 v 2 )/ 2 + v 3 / 2.5 [1] At node 2, 8 = v 2 / 5 + (v 2 v 1 )/ 2 [2] Our supernode equation is v 1 - v 3 = 0.8 v A [3] Since v A = v 2, we can rewrite [3] as v 1 v 3 = 0.8v 2 Simplifying and collecting terms, 0.5 v 1-0.5 v 2 + 0.4 v 3 = -3 [1] -0.5 v 1 + 0.7 v 2 = 8 [2] v 1-0.8 v 2 - v 3 = 0 [3] (a) Solving for v 2 = v A, we find that v A = 25.91 V (b) The power absorbed by the 2.5-W resistor is (v 3 ) 2 / 2.5 = (-0.4546) 2 / 2.5 = 82.66 mw.
Problem set #5 EE 221, 09/26/2002 10/03/2002 4 Chapter 4, Problem 22. Calculate the power being dissipated in the 2-Ω resistor for the circuit of Fig. 4.44. Chapter 4, Solution 22. We define four clockwise mesh currents. The top mesh current is labeled i 4. The bottom left mesh current is labeled i 1, the bottom right mesh current is labeled i 3, and the remaining mesh current is labeled i 2. Define a voltage v 4A across the 4-A current source with the + reference terminal on the left. By inspection, i 3 = 5 A and i a = i 4. MESH 1: -60 + 2i 1 2i 4 + 6i 4 = 0 or 2i 1 + 4i 4 = 60 [1] MESH 2: -6i 4 + v 4A + 4i 2 4(5) = 0 or 4i 2-6i 4 + v 4A = 30 [2] MESH 4: 2i 4 2i 1 + 5i 4 + 3i 4 3(5) v 4A = 0 or -2i 1 + 10i 4 - v 4A = 15 [3] At this point, we are short an equation. Returning to the circuit diagram, we note that i 2 i 4 = 4 [4] Collecting these equations and writing in matrix form, we have 2 0-2 0 0 4 0 i1 4-6 1 i2 0 10-1 i4 1-1 0 v 4A = 60 20 15 4 Solving, i 1 = 16.83 A, i 2 = 10.58 A, i 4 = 6.583 A and v 4A = 17.17 V. Thus, the power dissipated by the 2-Ω resistor is (i 1 i 4 ) 2 (2) = 210.0 W
Problem set #5 EE 221, 09/26/2002 10/03/2002 5 Chapter 4, Problem 26. Determine each mesh current in the circuit of Fig. 4.47. Chapter 4, Solution 26. We define a clockwise mesh current i 3 in the upper right mesh, a clockwise mesh current i 1 in the lower left mesh, and a clockwise mesh current i 2 in the lower right mesh. MESH 1: -6 + 6 i 1-2 = 0 [1] MESH 2: 2 + 15 i 2 12 i 3 1.5 = 0 [2] MESH 3: i 3 = 0.1 v x [3] Eq. [1] may be solved directly to obtain i 1 = 1.333 A. It would help in the solution of Eqs. [2] and [3] if we could express the dependent source controlling variable v x in terms of mesh currents. Referring to the circuit diagram, we see that v x = (1)( i 1 ) = i 1, so Eq. [3] reduces to i 3 = 0.1 v x = 0.1 i 1 = 133.3 ma. As a result, Eq. [1] reduces to i 2 = [-0.5 + 12(0.1333)]/ 15 = 73.31 ma.
Problem set #5 EE 221, 09/26/2002 10/03/2002 6 Chapter 4, Problem 33. Use the supermesh concept to determine the power supplied by the 2.2-V source of Fig. 4.54. Chapter 4, Solution 33. Define four mesh currents i 1 i 2 i 4 i 3 By inspection, i 1 = -4.5 A. We form a supermesh with meshes 3 and 4 as defined above. MESH 2: 2.2 + 3 i 2 + 4 i 2 + 5 4 i 3 = 0 [1] SUPERMESH: 3 i 4 + 9 i 4 9 i 1 + 4 i 3 4 i 2 + 6 i 3 + i 3 3 = 0 [2] Supermesh KCL equation: i 4 - i 3 = 2 [3] Simplifying and combining terms, we may rewrite these three equations as: 7 i 2 4 i 3 = -7.2 [1] -4 i 2 + 11 i 3 + 12 i 4 = -37.5 [2] - i 3 + i 4 = 2 [3] Solving, we find that i 2 = -2.839 A, i 3 = -3.168 A, and i 4 = -1.168 A. The power supplied by the 2.2-V source is then 2.2 (i 1 i 2 ) = -3.654 W.