EECE 210 Homework 2, Fall 2015

1-2 Homework 2 P2.1.4 The voltage shown in Figure P2.1.4 is applied across a 5  resistor. Determine: (a) the resistor current.; (b) p(t), 1min; 0  t  (c) the energy dissipated in the resistor at t = 3 min. Solution: v (V) =    0 5 12 /t s 180 180 s 60 60 s 0      t t t (a) i = v/R; i = t/60 A, 0  t  60 s; i = 1 A, 60  t < 180 s; i = 0, t > 80 s. (b) 60 0  t  s; 720 5 12 1 ) ( 2 2 2 t t R v p t          W, where t is in s. (c)        60 0 180 60 2 180 0 700 5 25 720 dt dt t pdt w J. 1 3 t , min v , V 5 2 Figure P2.1.4 2-2 P2.1.5 A voltage t v t 10 cos100 ( )  V is applied across a 10  resistor. (a) Sketch p(t). (b) Determine the average power dissipated in the resistor and the energy dissipated during half a cycle of v(t). Solution: t v t   10 cos100 ( ) 10cost , where  = 100 rad/s, and so that the supply frequency is    2 100 50 Hz, and the supply period is 20 50 1  ms. 𝑓 = 𝑤 2𝜋 𝑤ℎ𝑒𝑟𝑒 𝑓 𝑖𝑠 𝑖𝑛 𝐻𝑧 𝑎𝑛𝑑 𝑤 𝑖𝑠 𝑖𝑛 𝑟𝑎𝑑 2𝜋 (a)   10 10cos100 2 2 t R v p    cos2 𝛼 = (1 + 𝑐𝑜𝑠2𝛼 + 1 2 t  10cos 100 2 = 5(1 + cos2t) W, as shown. (b 5 2 2 2 10 ) ) ( (1 cos2 2 5 2 0             t t d W     .01 0 0 2 .01 0 0 10cos 100 tdt pdt w     .01 0 0 ) 5 (1 cos200 t dt          .01 0 0 200 sin200 5 t t 0.05 J. Since the average power dissipated is 5 W, the energy dissipated during one half cycle is 5(W)0.01(s) = 0.05 J. Note that the average power, i.e., average energy per unit time, is independent of the time scale, but the energy is the integral of instantaneous power with respect to time. 0 2 4 6 8 10 0 p, W /2  0 3 /2 2 t 3-2 P2.2.6 Determine in Figure P2.2.6 the voltage across each current source, the current through each voltage source, and the power delivered or absorbed by each source. Solution: Vab = 20 V, IX = 0.8Vab = 16 A. From KCL at node ‘a’, net current flowing away from this node is 16 – 10 = 6 A. Hence, 6A must flow into node ‘a’ from the 20 V source. Voltage drop across the CCVS = 0.5Ix = 8 V; let the voltage drop across the VCCS be VX. From KVL the voltage drop Vab is the same whether going through the 20 V source or the two dependent sources. Hence 20 = VX + 8, which gives VX = 12 V. The 20 V source delivers 206 = 120 W; the 10 A source delivers 2010 = 200 W; VCCS absorbs 1216 = 192 W; and CCVS absorbs 816 = 128 W. Total power delivered = 320 W = total power absorbed. Figure P2.2.6 – + 10 A 0.8Vab IX 20 V a b 0.5IX + – 10 A 6 A 16 A – + – + 12 V 8 V Figure P2.2.6-1 – + 10 A 0.8Vab IX 20 V a b 0.5IX + – 4-2 P2.2.7 Determine the total power delivered or absorbed by each source in Figure P2.2.7, assuming the voltage sources are 1 V each, I1 = 2 A, I2 = 1 A, and I3 = 1 A. Solution: The current leaving node ‘a’ through V1 is I2 + I3 = 1 + 1 = 2 A. The source V1 absorbs 12 = 2 W. The current leaving node ‘b’ through V2 is I1 – I3 = 2 – 1 = 1 A. The source V2 delivers 11 = 1 W. The current entering node ‘c’ through V3 is I1 + I2 = 3 A. V3 absorbs 13 = 3 W. From KVL, Vab = 2 V, Vac = 2 V, and Vbc = 0. The source I1 neither absorbs nor delivers power; the source I2 delivers 2 W; the source I3 delivers 2 W. Total power delivered = 5 W = total power absorbed. I1 I2 I3 V1 V2 V3 Figure P2.2.7 + – 2 A 1 A a b c 1 A – + 2 V – + 2 V + – 0 I1 I2 I3 V1 V2 V3 Figure P2.2.7-1 + – 5-2 P2.2.8 Determine in Figure P2.2.8 the voltage across each current source, the current through each voltage source, and the power delivered or absorbed by each source. Solution: From KCL at node ‘d’: 10 = 5 + I, so I = 5 A. From KCL at node ‘b’: IY = 10 + 3 = 13 A. From KCL at node ‘a’: 3 + 5 = 8 A = IX. As a check, KCL at node ‘c’ is: 13 = 8 + 5 A. From KVL around the outer loop: -Vcd + 10 = 0, so that Vcd = 10 V. From KVL around the mesh on the RHS: 10 – Vba = 0, so that Vba = 10 V. From KVL around the mesh ‘abda’: -Vba – 10 + Vbd = 0, so that Vbd = 20 V. As a check, KVL around the mesh ‘bcdb’ gives: -10 – Vcd + Vbd = -10 – 10 + 20 = 0. Upper 10 V source delivers 50 W; lower 10 V source delivers 130 W; 5 A source delivers 50 W; 10 A source absorbs 200 W; 3 A source absorbs 30 W. Total power delivered = 230 W = total power absorbed. Figure P2.2.8 3 A 10 A 5 A 10 V 10 V a b c d I IX IY Figure P2.2.8-1 3 A 10 A 5 A 10 V 10 V 6-2 P2.3.23 Determine the power delivered or absorbed by the dependent source in Figure P2.3.23. Solution: Initialize. The circuit is marked with given values. To determine the power delivered or absorbed by the dependent source, the voltage VX should be determined. Simplify. The circuit is in a simple enough form, although the two batteries can be combined in one 15 V battery. Deduce. From KCL around a surface enclosing the dependent source and the two batteries, the current through the lower 10  resistor is 100 mA in the direction shown. From Ohm’s law, VX = 100.1 = 1 V. The source current of the dependent source is 2 A. Since the voltage across the source is 15 V, and the 2 A is in the direction of a 15 V drop across the source, the source absorbs 152 = 30 W. Figure P2.3.23 100 mA + – + –+ – 10 V 5 V 10 V 10  10  + – VX 2VX 100 mA + – 10 V 10  10  + – VX 2VX 100 mA a Figure P2.3.23-1 b + –+ – 10 V 5 V 7-2 P2.3.31 Determine R, VX, and VY in Figure P2.3.31, given that the current in the top connection is zero. Solution: Initialize. The circuit is marked with given values. The nodes are labeled. To determine R, the voltage across it and the current through it should be determined. VX and VY are determined from KVL around a mesh. Simplify. The circuit is in a simple enough form. Deduce. From KCL at node ‘b’, Ibe = 5 A; from KCL at node ‘e’, Ide = 5 A; from KCL at node ‘c’, Ica = 15 A; from KCL at node ‘a’, Iad = 15 A; as a check, KCL is satisfied at node ‘d’. From Ohm’s law, Vbe = 30 V, Vad = 15 V, and Vde = 5R. From KVL in the outer loop, 5R + 15 – 30 = 0, which gives R = 3 . From KVL around the mesh ‘abca’, -V1 + 20 = 0, or V1 = 20 V; from KVL around the mesh ‘cbec’, V1 – 30 + Vx = 0, or Vx = 10 V. From KVL around the mesh ‘acda’, -20 – Vy + 15 = 0, or Vy = -5 V. a b c d e 5 A 5 A 15 A 15 A 5 A Figure P2.3.31-1 5 A 10 A 10 A 6  1  VX 20 V I = 0 R 5 A 10 A 10 A 6  1  VX VY 20 V I = 0 Figure P2.3.31 R 15 V + – + – 15 V Figure P2.3.31-2 a b c d e 5 A 10 A 10 A 6  1  VX 20 V I = 0 R 30 V + – 20 V 8-2 P2.3.32 Determine IO in Figure P2.3.32. Solution: Initialize. The circuit is marked with given values. The nodes are labeled. Simplify. The circuit is in a simple enough form. Deduce. Vab = 2IO, so that the current in the 2  resistor is IO. From KCL at node ‘a’, Ida = 3IO. From KCL at node ‘b’, Ibc = 0. From KVL around the mesh ‘abcd’, -2IO + 0 + 5 – 3IO = 0, which gives IO = 1 A. IO 2  1  1  1  1  IO 2IO Figure P2.3.32 + – 5 V IO 2  1  1  1  1  IO 2IO IO IO 3IO I = 0 Figure P2.3.32-1 + – 5 V a b c d