EECE 210 Homework 4, Fall 2015

1-2 Figure P4.1.7 VSRC 5 A 6  24  12  8  + – VX + – VY ISRC a b c d + – Figure P4.1.7-1 VSRC 6  24  12  8  a b c d + – + – VTh c a, d 6  24  8  12  b Figure P4.1.7-2 + – 0.4VSRC V 9.6  5 A b c Figure P4.1.7-3 4  120 V + – a bc d + – VX + – VY ISRC 6  Figure P4.1.7-4 Homework 4 P4.1.7 (a) Determine VSRC in Figure P4.1.7 by deriving TEC between terminals ‘bc’. (b) Determine ISRC, VX, and VY. Solution: (a) VTh = Vcd – Vbd, where node ‘c’ is taken as positive with respect to node ‘b’ since current flows from node ‘c’ to node ‘b’. From voltage division, Vcd = VSRC(24/30), and Vbd = VSRC(8/20). Hence, SRC SRC Th V V V 5 2 5 2 5 4          = 0.4VSRC. With VSRC replaced by a short circuit, the resistance seen between terminals ‘bc’ is 8||12 + 6||24 = 4.8 + 4.8 = 9.6 . TEC becomes as shown, where IX is the short circuit current given by: 5 9 6 0 4  . . VSRC . It follows that VSRC = 120 V. (b) The resistance between nodes ‘a’ and ‘b’ and ‘c’ connected together in the given circuit is 12||6 = 4 , and the resistance between nodes ‘b’ and ‘c’ connected together and node ‘d’ is 8||24 = 6 . The circuit can be redrawn as a voltage divider, as shown. It follows that ISRC = 120/(4 + 6) = 12 A; VX = 120(4/10) = 48 V, and VY = 120(6/10) = 72 V. As a check, the current in the 6  resistor is 48/6 = 8 A, and the current in the 24  resistor is 72/24 = 3 A, the difference being 5 A. 2-2 10  5  – +VTh 5 V 10  5 V 1 A 10  Figure P4.1.11 + – + – a b 10  5  – +VTh 5 V 10  5 V 1 A 10  Figure P4.1.11-1 + – + – a b c d e + – 12 V 1  1  P4.1.21 – + 1  IX 2IX a b + – 12 V 1  1  P4.1.21-1 – + 1  IX 2IX a b RL IX V IX 0 c 1  1  P4.1.21-2 – + 1  IX 2IX a b IX V 2IX IX c + – VT P4.1.11 Derive TEC looking into terminals ‘ab’ in Figure P4.1.11. Solution: Initialize. All given values and the required VTh are entered. The nodes are labeled. Simplify. The circuit is in a Simple enough form. Deduce. The 1 A source current flows through the 5  resistor, producing a voltage drop of 5 V. The current through the upper 10  resistor is zero, so that the current through the lower 10  resistor is zero, and the voltage across this resistor is zero. From KVL starting at node ‘b and going CW:-5 + 0 + 5 – VTh = 0, which gives VTh = 0. To determine RTh, the sources are set to zero. The 10  resistor on the left is short-circuited, leaving the remaining resistors in series. It follows that RTh = 25 . P4.1.21 Connect a resistor RL between terminals ‘ab’ in Figure P4.1.21 and show that the voltage Vab is independent of RL. Deduce that TEC looking into terminals ‘ab’ is an ideal voltage source. Verify this deduction by determining VTh and RTh looking into terminals ‘ab’. Solution: Vcb = IX V, Vac = IX V, Iac = IX A, and current through 12 V source is zero, it follows that Vcb = IX V = 12 V, Vab = 2IX V = 24 V, independently of RL, Hence, VTh = 24 V, RTh = 0. On opencircuit Vcb = 12 V = IX so that VTh = Vab = 24 V. If at test source is applied, Icb through the 1  resistor on the LHS Is IX; Iac = 2IX and Vab across the resistors Is 3IX, which equals the source voltage 2IX. It follows that IX = 0 so that the trst source 3-2 Figure P4.1.28 5  5  10  a b 20 V 2IX IX + – Figure P4.1.28-1 5  5  10  a b 20 V 2IX IX + – IX 3IX 2IX c d Figure P4.1.28-2 5  5  10  a b 2IX IX IT + IX IT + 3IX IT + 2IX + – IT VT c d sees a short circuit. P4.1.28 Derive TEC looking into terminals ‘ab’ in Figure P4.1.28. Solution: Initialize. All given values and the required VTh are entered. The nodes are labeled. Simplify. The circuit is in a Simple enough form. Deduce. On open circuit, the currents are as shown. Iac = IX; Icd = 3IX; Idb = 2IX; from KVL around the upper mesh, 20 = 20IX, so that IX = 1 A. It follows that VTh = Vab = 210 + 20 = 40 V. When a test source is applied, with the 20 V source set to zero, Iac = IT + IX; Icd = IT + 3IX; Idb = IT + 2IX;. from KVL in the upper mesh, 5(IT + IX + IT + 3IX) = 0, which gives, IT = -2IX. It follows that VT = 10(IT + 2IX) = 0, so that RTh = 0. 4-2 10 V 2  2  4  5IX A 4  IX – + VO a b c + – Figure P4.1.30 10 V 2  2  4  – + VTh a b c + – Figure P4.1.30-1 d 10 V 2  2  5ISC 4  ISC a b c + – Figure P4.1.30-2 d 2.5 A + – 10 V 4  – + VO Figure P4.1.30-3 -16  P4.1.30 Determine VO in Figure P4.1.30 using TEC. Solution: Initialize. All given values and the required VTh are entered. The nodes are labeled. Simplify. The circuit is in a Simple enough form. Deduce. When the 4  resistor is removed, IX = 0, and the dependent source becomes an open circuit. It follows that VTh = 10 V. When the resistor is replaced by a short circuit, the circuit becomes as shown, where Ix = ISC and the dependent source becomes 5ISC. Iac = 2.5 A. It follows from KCL that: ISC = 5ISC + 2.5, which gives ISC = -2.5/4 = -5/8 A, and RTh = VTh/ISC = -80/5 = -16 . Hence, 3 10 10 16 4 4     VO  V.