EECE 210 Homework 4 with Solution, Fall 2014

60  20  a b 30  + – 20 V + – 60 V 1 A 2 A + – VTh I c d + – 60I 10  5  – +VTh 5 V 10  5 V 1 A 10  Figure P4.1.14 + – + – a b 10  5  – +VTh 5 V 10  5 V 1 A 10  e b c d 1 A – + 5 V + – + – I = 0 I = 0 – + 0 V Homework 4 P4.1.14 Derive TEC looking into terminals ‘ab’ in Figure P4.1.14. Solution: Initialize. All given values and the required VTh are entered. The nodes are labeled. Simplify. No meaningful simplifications can be made. 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 . P4.1.36 Derive TEC between terminals ‘ab’ in Figure P4.1.36. Solution: Initialize. All given values are entered. VTh is marked, and the nodes are labelled. Simplify. No meaningful simplifications can be made. Deduce. No immediate deductions can be made. Explore. Every node has two unknowns and every mesh has two unknowns, so it is not possible to introduce a 1 60  20  a b 30  + – 20 V + – 60 V Figure P4.1.36 1 A 2 A single unknown that can be determined from KCL or KVL. Plan. It is seen that if the current in the 60  resistor is I, then VTh = Vab = 20 + 60I + 60. Hence, if I can be determined, VTh follows. A general procedure for finding a current through a resistor is to derive TEC seen by the resistor. Implement. If the 60  resistor is removed from the circuit and a current IX assigned through the circuit, it follows from KVL starting at node ‘b’ and moving CW that: +20 – 30(IX – 1) – 60 – 20(IX +2) = 0, or, -3IX – 2IX + 2 + 3 – 6 – 4 = 0, or, -5IX – 5 = 0, or, IX = -1 A. It is seen that Vdb = 20 V, and Vcb = 20(-1 +2) = 20 V. Hence nodes ‘c’ and ‘d’ are at the same voltage, which means that I = 0 in the 60  resistor, so that VTh = 80 V. To determine RTh, the sources are set to zero. Node ‘c’ becomes the same as node ‘a’ and node ‘d’ the same as node ‘b’. The three resistors are in parallel, as when they are redrawn between the two nodes. (30||60) = 20 , and (20||20) = 10 . Hence, RTh = 10 . 2 20  a b 30  + – 20 V + – 60 V 1 A 2 A c d Ix Ix Ix – 1 + 30(Ix – 1) – Ix + 2 + – 20(Ix + 2) 60  20  a b 30  a b b a 30  60  20  P4.1.39 Derive NEC looking into terminals ‘ab’ in Figure P4.1.39. Solution: Initialize. All given values are entered. Nodes ‘a’ and ‘b’ are short circuited, and IN indicated. Deduce. Because of the short-circuit, VX = 0, the 0.5 A current flows through the short circuit, which makes IN = 0.5 A. Note that IY = 0, so that both dependent sources become short circuits. We will determine VTh = VX. On open circuit, 0.5 = -IY, and flows through the 5  resistor. From KVL, the voltage across the 5  resistor is 0.5VX. Hence, 0.5VX = 50.5, so that VTh = 5 V. It follows that RN = 5/0.5 = 10 . Check. As a check, RN will be derived by applying a test voltage source VT. This makes VX = VT, so that the voltage drop across the 5  resistor is 0.5VT, and the current through this resistor is IT. It follows that 0.5VT = 5IT, so that VT/IT = RN = 10 , as before. 3 – + 2IY 0.5VX 0.5 A 10  IY – + VX a b + – Figure P4.1.39 5  – + 2IY 0.5VX 0.5 A 10  IY – + VX a b + – 5  IN – + 2IY 0.5VX 0.5 A 10  IY = -0.5 A – + VX a b + – 5  IY + – 0.5VX – + 2IY 0.5VT 0.5 A 10  IY – + VX a b + – 5  + – IT VT IT – 0.5VT + P4.2.1 Determine, according to the substitution theorem, (a) the independent voltage source, (b) the independent current source, and (c) the resistance that can replace the dependent current source in Figure P4.2.1 without affecting the rest of the circuit. Solution: From KCL, 3 = 3Ix, so Ix = 1 A. From KVL, Vx = 2 + 21 = 4 V. It follows that the 2Ix source can be replaced by: (a) an independent voltage source of 4 V, (b) an independent current source of 2 A, or (c) a resistance of 4/2 = 2 . P4.2.9 Determine Rin in Figure P4.2.5 by applying the 4 Figure P4.2.1 3 A 2  + – 2 V Ix 2Ix + – Vx Figure P4.2.1 3 A 2  + – 2 V IX 2IX – + 1220IX V Vx/50 1 k 20  IX – + VX a b Rin Figure P4.2.9 source absorption theorem, where IX is in amperes. Solution: The current of the VCCS is VX/50 A in the direction of a voltage drop VX V across the source. It follows that this source is equivalent VX/(VX/50) = 50 . Consider the CCVS and the 20  resistor. The voltage across the combination is VX = 1000IX, as shown. The voltage drop across the 20  resistor is 220IX, and the current through this resistor is 220IX/20 = 11IX A. This current is in the direction of a voltage rise through the source. The source is therefore equivalent to a resistance -1220/11. In series with 20 , this becomes: . This resistance in parallel with 1000  is: . This resistance in parallel with 50  is: . 5 – + 1220Ix 20  – + a b 1000IX – + 220Ix 11IX