EECE 210 Homework 5, Fall 2015

1-2 Homework 5 P4.1.50 Derive NEC between terminals ‘ab’ in Figure P4.1.50, assuming all resistances are 2 . Solution: Initialize. All given values and the required IN are entered. The nodes are labeled. Simplify. The circuit is in a simple enough form. Deduce. The branches ‘ha’ and ‘hcb’ are in parallel, so that by current division, Icb = 1/3 A. It follows that Ibd = IN + 4/3. Since the voltage across the branch ‘bdf’ is the same as the Vaf, then Iaf = 2(IX +4/3) and Ifg = 3(IN + 4/3) = 3IN + 4; From KCL at node ‘g’, Iga = 3IN + 3. From KVL around the mesh ‘afga’, 22(IN + 4/3) 4(3IN + 4) + 2(3IN + 3) = 0, which gives IN = -41/33 A. When the current sources are set to zero, Raf = 2||6 = 1.5 ; Rfb = 4 ; R of the branch on the RHS is 6 ; it follows that RN = (65.5)/11.5 = 66/23 . 1 A a 1 A b Figure P4.1.50 1 A a 1 A b Figure P4.1.50-1 IN 1/3 A IN + 4/3 3IN + 4 c d f g h 2/3 A 2(IN + 4/3) 3IN + 3 a b Figure P4.1.50-2 a b 6  4  1.5  (a) (b) f f 2-2 P4.2.2 Determine VX in Figure P4.2.2 by using the substitution theorem, where NA is an unspecified circuit that passes a current of 0.5 A. Solution: NA is replaced by a 0.5 A current source. The current in the 7.5  resistor is VX/7.5 A. The current in the 10  resistor is (VX/7.5 – 0.5). From KVL around the mesh on the LHS, 30 – VX – 10(VX/7.5 – 0.5) = 0, which gives IX = 2 A, so that VX = 15 V. P4.2.10 Determine Rin in Figure P4.2.10 by applying the source absorption theorem. Solution: The voltage across the 2  resistor is 4 VO, so that the current in the middle branch is 2VO in the direction of a voltage rise across the VCVS. This source is equivalent to a resistance of -2.5 . In series with 2 , the resistance of this branch is -0.5 , which cancels out the 0.5  resistance of the branch on the RHS. The current in the branch on the LHS is VA A in the direction of a voltage drop through the VCVS, so that this source is equivalent to a resistance of 0.25 . Added to 1 , this gives Rin = 1.25 . Figure P4.2.2-1 + – 30 V 7.5  18  10  8  0.5 A + – VX a b VX / 7.5 VX / 7.5 – 0.5 3-2 P5.1.10 Determine IX in Figure P5.1.10. Solution: The sources can be split in two resulting in two half circuits as shown. These two half circuits are symmetrical about the midline, so they have the same currents at symmetrtical nodes. Thus, if a current I enters node ‘a’. then an equal current I enters node ‘a’. this means that I = -I, so that I is zero. Similarly, no current flows between nodes ‘b’ and ‘b’ and betwee nodes ‘c’ and ‘c’. the two half circuts can therefore be separated. It follows that IX = 0.5 A. P5.1.15 Determine IX in Figure P5.1.15. Note that IX is independent of the 10  resistor and the dependent source because the 5 A source partitions the circuit into two subcircuits. The upper subcircuit is independent of the lower subcircuit, but the lower subcircuit depends on the upper subcircuit through IX. Solution: The current in the diagonal branch is, from KCL, IX – 5. From KVL around the mesh on the RHS, 10 + 5IX + 5(IX – 5) = 0, or 10IX = 15, so that IX = 1.5 A. 20 V + – 1  1  1  1  2  2  4  4  4  4  IX 1 A Figure P5.1.10 20 V + – 1  1  1  1  2  2  4  4  4  4  IX 0.5 A Figure P5.1.10-1 + – 20 V a a b b c c 0.5 A I I 1  1  2  4  4  IX Figure P5.1.10-2 + – 20 V a b c 0.5 A 10 V 5 A 2  10  Figure P5.1.15 3  5  IX 3IX 4-2 P5.1.22 Determine IX in Figure P5.1.22. Solution: VO = -55 =-25 V, so that the dependent source is replaced with an independent voltage source of 5 V, and of reversed polarity, as shown. When either voltage source is applied alone, with the current source replaced by an open circuit, the components of VO are zero. When the current source is applied alone, VO = -25 V. From KCL at the upper node, the current in the left branch is (5 – IX) A. From KVL around the outer loop, starting with the lower node and moving CW: 10(5 – IX) – 5 – 10 – 4IX = 0, or, -14IX + 35 = 0, which gives IX = 2.5 A. P5.1.25 Determine IX in Figure P5.1.25. Solution: The dependent source is replaced by an independent source VY. If the 10 V source is applied alone, with the other sources set to zero. the component of IX due to this source is zero. If the 12 V source is applied alone, the 12 V appears across 10  in series with a parallel combination of 3 and 6 . The source current is 12 V/12  = 1 A. From current division, IX1 = -1/3 A. When VY is applied alone, the source current is VY/12. From current division, IX2 = VY/36. By superposition, and substituting VY = 6IX, IX = -1/3 + IX/6. This gives IX = -0.4 A. 5  5 A 10  + – 4  10 V Figure P5.1.22 + – VO – + 0.2VO IX 5  5 A 10  + – 4  10 V + – 25 V IX + – 5 V IX 5 – IX + – 4IX + – 10(5 – IX) Figure P5.1.22-1 Figure P5.1.25 + – 10  12 V IX 6  3  5  + – 10 V 2  – + 6IX 5-2 P5.1.29 Determine in Figure P5.1.29: (a) VO, and (b) the power delivered or absorbed by the 1 A source. P5.1.29 (a) The 8  resistor is redundant as far as IX and VO are concerned. If the 1 A source is applied alone, IX1 = 1 A. If the 5IX source is replaced by an independent source IY and applied alone, IX2 = -IY. If the 3 V source is applied alone, IX3 = -3/2. If the 10IX source is replaced by an independent source VY and applied alone, IX4 = VY/2. By superposition, and substituting IY = 5IX and VY = 10IX, IX = 1 – 5IX – 3/2 + 10IX/2, or IX = -0.5 A. From the original circuit, VO = 3 – 2(1 – 6IX) = 3 – 2(1 + 3) = -5 V. (b) The current in the 4  resistor is (1 – 5IX) = 3.5 A; hence, the voltage across the 1 A source is 3 + 43.5 + 8 = 25 V. The power delivered by the source is P = 25 W. – + + – 8  4  2  10IX 1 A 3 V IX + – VO Figure P5.1.29 5IX – + + – 8  4  2  10IX 1 A 3 V IX + – VO Figure P5.1.29 5IX – + + – 8  4  2  10IX 1 A 3 V IX + – VO Figure P5.1.29-1 5IX 1 – 5IX 1 – 6IX