Problem set #5 EE 221, 09/26/ /03/2002 1

Transcription

1 Chapter 3, Problem 42. Problem set #5 EE 221, 09/26/ /03/ 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 = = 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[ (2000)/ ( )] = 2.5 V Voltage division gives the voltage across the 47-kΩ resistor: (2.5)(47) 0.5v 1 = = V So that p 47kΩ = (0.9928) 2 / = µw

2 Problem set #5 EE 221, 09/26/ /03/ Chapter 4, Problem 3. Use nodal analysis to find v P in the circuit shown in Fig 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 / (v P v 2 )/ 50 [2] NODE 2: = (v 2 v P )/ 50 + (v 2 v 3 )/ 10 [3] NODE 3: 5 2 = v 3 / (v 3 v 2 )/ 10 [4] Simplifying, 60v 1-20v P = 8000 [1] -50v v P - 40v 2 = 0 [2] - v P + 6v 2-5v 3 = -25 [3] -200v v 3 = 6000 [4] Solving, v P = 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, = 9 v X 3 v Y [1] = v X v Y [2] Solving, v X = V and v Y = V Thus, v 1 = 240 v X = V and v 2 = v Y 60 = V (b) The power absorbed by the 6-W resistor is (v X v Y ) 2 / 6 = W

3 Problem set #5 EE 221, 09/26/ /03/ 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 v v 3 = -3 [1] -0.5 v v 2 = 8 [2] v v 2 - v 3 = 0 [3] (a) Solving for v 2 = v A, we find that v A = V (b) The power absorbed by the 2.5-W resistor is (v 3 ) 2 / 2.5 = ( ) 2 / 2.5 = mw.

4 Problem set #5 EE 221, 09/26/ /03/ Chapter 4, Problem 22. Calculate the power being dissipated in the 2-Ω resistor for the circuit of Fig 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: i 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 i 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 i i i v 4A = Solving, i 1 = A, i 2 = A, i 4 = A and v 4A = V. Thus, the power dissipated by the 2-Ω resistor is (i 1 i 4 ) 2 (2) = W

5 Problem set #5 EE 221, 09/26/ /03/ Chapter 4, Problem 26. Determine each mesh current in the circuit of Fig 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: i 1-2 = 0 [1] MESH 2: i 2 12 i = 0 [2] MESH 3: i 3 = 0.1 v x [3] Eq. [1] may be solved directly to obtain i 1 = 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 = ma. As a result, Eq. [1] reduces to i 2 = [ (0.1333)]/ 15 = ma.

6 Problem set #5 EE 221, 09/26/ /03/ Chapter 4, Problem 33. Use the supermesh concept to determine the power supplied by the 2.2-V source of Fig 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: i i i 3 = 0 [1] SUPERMESH: 3 i i 4 9 i i 3 4 i 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 i i 4 = [2] - i 3 + i 4 = 2 [3] Solving, we find that i 2 = A, i 3 = A, and i 4 = A. The power supplied by the 2.2-V source is then 2.2 (i 1 i 2 ) = W.