EECE 210 Problems, Fall 2010

Determine all the currents and voltages in the circuit using superposition and mark them on the circuit diagram. – + 20 A 2 Ω 2 Ω 4 Ω 4 Ω 1 Ω 40 V – + – + + – 20 A 2 Ω 2 Ω 4 Ω 4 Ω 1 Ω 40 V + – 10 V a b c d 10 A 10 A + – 30 V 15 A 25 A 5 A + – 10 V 40 V 0 A + – 0 V Solution: Refer to figures below 10 ohms 10 ohms Problem 1 (10 pts) Consider the circuit shown below. 1. We, first, set the current sources 5 mA and 10 mA to zero. Determine the equivalent resistance seen by the 20 mA current source. (5 pts) R1 = (4 + 1)||4 = 20/9 kΩ; R2 = 20/9 + 2 = 38/9 kΩ, Req = 2||(38/9) = 19/14= 1.36 kΩ 2. Write the node voltage equations by inspections (do not solve) (5 pts) 1.75V1 – 0.25V2 – V3 = 20 -0.25V1 + V2 – 0.25V3 = 5 -V1 – 0.25V2 + 1.25V3 = 5 Problem 2 (10 pts) Consider the circuit shown below 1. We, first, set the 4V source and the 10V source to zero. Determine the equivalent resistor seen by the 8V voltage source. (5 pts) Req = 5 + (2||4||4||1) = 5 + 0.5 = 5.5 Ω 2. Write the mesh-current equations. (do not solve). (5 pts) 7I1 – 2I2 – 0I3 – 0I4 = 8 -2I1 + 6I2 – 4I3 – 0I4 = 4 0I1 – 4I2 + 5I3 – I4 = -10 0I1 – 0I2 – I3 + 5I4 = -4 1. Determine the equivalent resistance between terminals a and b, given that all resistances are 1 Ω. A. 5 Ω B. 4.5 Ω C. 4 Ω D. 3 Ω E. None of the above Solution: The resistances not connected directly to terminals a and b form a balanced bridge. Hence the resistance across the bridge does not carry any current and can be replaced by an open circuit or a short circuit. If replaced by an open circuit, Req = 1 + 2||2 + 1 = 3 Ω. a b Problem 2 Find the power consumed by the 50 ohm resistor in the circuit shown below A) P= 0.246 W B) P= 0.692 W C) P= 2.358 W D) P= 5.100 W E) None of the above Problem 6 Given the circuit above, determine the current Ix if voltage-controlled current source has B=0.2: A) 0 B) 0.285 C) 1.285 D) 0.5 E) None of the above 7 Problem 11 What is I0? A) 1 A B) -1 A C) 2 A D) -2 A E) None of the above Problem 5 a b c d IB IA IC R1 R2 R3 The following values are given: IA= 1 mA, IB= 2 mA, IC = 4 mA, R1 = R2 = R3 = 1 kΩ, What is the value of Vbc? A) 4 V B) -2 V C) 1 V D) 5 V E) None of the above Problem 9 Find Ix. A) 4.2 A B) 3.5 A C) 2.25 A D) 4.75 A E) None of the above Problem 10 Find the Thevenin equivalent across terminals a and b. A) RTH = 40Ω, VTH = 6.67 V B) RTH = 40Ω, VTH = -6.67 V C) RTH = 80Ω, VTH = 12 V D) RTH = 80Ω, VTH = -12 V E) None of the above 20 V 20 Ω 30 Ω 20 Ω Problem 11 Find the current, I0, flowing through the dependant source. A) 9.4A B) 14A C) -3A D) -12A E) None of the above. Problem 12 Find the voltage, V, across the 2A source. A) 400 V B) 140 V C) 6 V D) -300 V E) None of the above Approx 6.6 A P3.1.25 Determine VO in Fig. P3.1.25 using node-voltage analysis. Figure P3.1.25 10 A 2 Ω – + + – + – 2Vx 20 V + – Vx 2 Ω 2 Ω 4 Ω 4 Ω + – VO Solution: The node-voltage equation for node a is: Va – 0.5Vb – 0.5Vc = 10; substituting Vb = 20 V: Va – 0.5Vc = 20. For node c: -0.5Va + 0.75Vc = -Iy; for node d: -0.5Vb + 0.75Vd = Iy; adding and substituting for Vb: -0.5Va + 0.75Vc + 0.75Vd = 10. For the dependent source: Vd – Vc = 2Vx = 2Vb – 2Va, or 2Va – Vc + Vd = 40. Solving, Va = 40 V, Vc = 40 V, Vd = VO = 0. 10 A 2 Ω – + + – 2Vx 20 V + – Vx 2 Ω 2 Ω 4 Ω 4 Ω + – VO a b c d Iy P3.1.26 Determine VO in Fig. P3.1.25 using mesh-current analysis. Solution: For mesh 2: 10 A 2 Ω – + + – 2Vx 20 V + – Vx 2 Ω 2 Ω 4 Ω 4 Ω + – VO I2 I1 I3 I4 -2I1 + 6I2 – 2I4 = -2Vx; substituting I1 = 10 and Vx = -2I2: I2 – I4 = 10. For mesh 3: -4I1 + 8I3 = 2Vx; I2 + 2I3 = 10. For mesh 4: -2I2 – 4I3 + 6I4 = -20, or -I2 – 2I3 + 3I4 = -10. Solving, I2 = 10 A, I3 = 0, and I4 = 0, which gives VO = 0. P3.2.12 Determine VO in Fig. P3.1.19 using superposition and calculate the power dissipated in the 5 Ω resistor. Figure P3.1.19 2 A 4 A 2 Ω 4 Ω 4 Ω 2 Ω 5 Ω + VO – + 3VO V – 2 A 2 Ω 4 Ω 4 Ω 5 Ω + VO – + 3VO V – Solution: With the 2 A source acting alone, the circuit becomes as shown. The source current flows through the 5 Ω resistor, so that VO1 = 10 V. Similarly, when the 4 A source is applied alone, VO2 = 20 V. From superposition, VO = VO1 + VO2 = 30 V. The dependent source does not contribute to VO. Power dissipated in the 5 Ω resistor is 180 5 (30) 2 = W. P3.2.13 Determine IO in Fig. P3.1.21 using superposition and calculate the power dissipated in the 5 S resistor. Figure P3.1.21 4 S 2 V 5 S IO 4 S + – + – + – + – 3IO 4 V 2 S 2 S Solution: With the 2 V source acting alone, and the 4 V source replaced by a short circuit, the circuit becomes as shown. The source voltage is applied across the 5 S resistor, so that IO1 = 10 A. Similarly, when the 4 V source is applied alone, IO2 = 20 A. From superposition, IO = IO1 + IO2 = 30 A. The dependent source does not contribute to IO. Power dissipated in the 5 S resistor is 180 5 (30) 2 = W. 4 S 2 V 5 S IO 4 S + – 3IO 2 S 2 S P4.1.8 Derive TEC between terminals ab in Fig. P4.1.8. Figure P4.1.8 1 A 1 A 10 Ω 10 Ω 20 Ω a b Solution: Let us first remove the 20 Ω resistor and reapply it later. On open circuit, each 1 A source produces a 10 V drop across the resistor in parallel with it. Hence Voc1 = 20 V. On short circuit, Isc = 1 A, so that RTh1 = 20 Ω. When the 20 Ω resistor is added at terminals ab, VTh = 20×(20/40) = 10 V and RTh = (20||20) = 10 Ω P4.1.9 Derive TEC between terminals ab in Fig. P4.1.9. Solution: On open circuit, KVL around the upper mesh gives 20Ix = 20, or Ix = 1 A. It follows that VTh = Voc = 20 + 20Ix = 40 V. On short circuit, the current in the 10 Ω resistor is 2 A. KVL around the upper mesh gives: 20 = 5(Ix – Isc) + 5(3Ix – Isc), or 2Ix – Isc = 2; from KCL at node c: 3Ix – Isc + 2 = Ix, or 2Ix – Isc = -2. This means that Ix and Isc are indeterminate. This suggests that RTh = 0, which would make Isc indeterminate. To verify this, we apply a test source of 1 A, with the voltage source set to zero. Then, 4Ix + 2 = 0, so that Ix = -0.5 A and VT = 2(-0.5) + 1 = 0. Hence RTh = 0. 5 Ω 5 Ω 10 Ω a b – + 20 V 2Ix Ix Ix 3Ix 2Ix 5 Ω 5 Ω 10 Ω a b – + 20 V 2Ix Ix Ix – Isc 2 A Isc 3Ix – Isc c 5 Ω 5 Ω 10 Ω a b 2Ix Ix + – VT 1 A Ix + 1 3Ix + 1 2Ix + 1 P4.1.10 Derive NEC between the short-circuited terminals ab in Fig. P2.1.4. Solution: If a test voltage is applied, KCL at node a gives Iφ = IT + 3. From KCL at node c, IT + 3 + 10 = 5, or IT = −8 A, irrespective of VT. It follows that IN = 8 A. Since IT is independent of VT, it means that there is no resistance in parallel with the current source IN. Figure P2.1.4 – + – + Ix 3 A 10 A 5 A 10 V 10 V a b – + – + IT 3 A 10 A 5 A 10 V 10 V a b – + VT Iφ c P4.1.11 Determine VO in Fig. P3.1.7 using TEC. Figure P3.1.7 – + – + 10 V 10 Ω 20 Ω 40 Ω 80 Ω 20 V + – VO Solution: Open-circuit voltage: When only the 10 V source is applied, VTh1 = 3 10 10 30 10 = × V. When only the 20 V source is applied VTh2 = 20 V. Hence, VTh = 70/3 V. With both sources set to zero, RTh = 10||20 = 20/3 Ω. It follows that VO = 20 3 70 20 / 3 40 40 = × + V. – + – + 10 V 10 Ω 20 Ω 80 Ω 20 V a b P4.1.12 Determine IO in Fig. P3.1.9 using NEC. Figure P3.1.9 10 A IO 20 A a b c 80 S 10 S 20 S 40 S Solution: With the 10 A source acting alone, IN1 = 3 10 10 30 10 = × A. With the 20 A source acting alone, IN2 = 20 A. Hence, IN = 70/3 A. 10 A IN 20 A a b c 80 S 10 S 20 S The conductance between terminals bc is 3 20 10 20 10 20 = + × S. It follows from NEC that 20 3 70 20 / 3 40 40 = × + = OI A. P4.1.15 Determine VO in Fig. P3.1.15 using TEC. Figure P3.1.15 – + 10 V 2 Ω 2 Ω 4 Ω 5Ix A 4 Ω IX – + VO – + 10 V 2 Ω 2 Ω 5ISC 4 Ω ISC Solution: On open circuit, Ix = 0, and the dependent source becomes an open circuit. It follows that VTh = 10 V. On short circuit, the circuit becomes as shown, where Ix = ISC and the dependent source becomes 5ISC. It follows from KCL that: 4 10 5 + = SC SC I I , which gives 8 ISC = − 5 A, and SC = −16 Ω. Hence R 3 10 = − 10 16 4 4 × − VO = V. P4.1.16 Determine IO in Fig. P3.1.17 using NEC. Figure P3.1.17 4 S 10 A 4 S 5Vx + – 2 S 2 S + – Vx IO 4 S 10 A 5Voc + – 2 S 2 S + – Voc 10 A Solution: On open circuit, 10 A flows through the 4 S resistor, so that Voc = 5Voc 4 + 10 , which gives 8 Voc = − 5 A. On short circuit, Vx = 0 and the dependent source is zero, so that IN = 10 A. This makes N = −16 S. It follows that G = × − = 16 4 4 OI 10 3 − 10 A. P4.1.17 Determine VO in Fig. P3.1.19 using NEC. Solution: If the 5 Ω resistor is replaced by an open circuit, the circuit is invalid, as two unequal current sources will be connected in series through the 2 Ω resistors, and . If a test source is applied in place of the 5 Ω resistor and the current sources replaced by open circuits, the resistance seen by the source is infinite. If the 5 Ω resistor is replaced by a short circuit, ISC = 6 A. It follows that the circuit does not possess a TEC between the specified terminals, only an NEC consisting of an ideal current source of 6 A. This gives VO = 30 V. O → ∞ V Figure P3.1.19 2 A 4 A 2 Ω 4 Ω 4 Ω 2 Ω 5 Ω + VO – + 3VO V – P4.1.18 Determine IO in Fig. P3.1.21 using TEC. Solution: If the 5 S resistor is replaced by a short circuit, the circuit is invalid, as two unequal voltage sources will be connected in series, and . If a test source is applied in place of the 5 S resistor and the voltage sources replaced by short circuits, the resistance seen by the source is zero. If the 5 S resistor is replaced by an ope circuit, VTh = 6 V. It follows that the circuit does not possess an NEC between the specified terminals, only a → ∞ OI Figure P3.1.21 4 S 2 V 5 S IO 4 S + – + – + – + – 3IO 4 V 2 S 2 S n TEC consisting of an ideal voltage source of 6 V. This gives IO = 30 A. P4.1.19 Determine IO in Fig. P3.1.23 using NEC. Figure P3.1.23 – + 10 V 4 S 2 S 2 S 4 S 2 S 2 A IO 2Ix Ix Solution: With the 4 S resistor replaced by a short circuit, IO can be obtained from mesh-current analysis. The mesh current equations are the same as those for P3.1.24 but with a coefficient of 0.5 for I4 in the equation for mesh 4. The equations are: – + 10 V 4 S 2 S 2 S 2 S 2 A IN 2Ix Ix I2 I1 I3 I4 I1 – 0.5I2 = 9; -0.5I1 + 0.75I2 + 0.5I4 = -1; and 2I1 – I2 + I4 = -4. Solving, I4 = IO = - 22 A. If a test source is substituted for the 4 S resistor, with the independent source set to zero, it is seen from KCL at the middle node that IT = 0, which means that the source resistance is infinite. The circuit does not possess a TEC between the specified terminals, only an NEC. – + – + 4 S 2 S 2 S 2 S Ix 2Ix Ix IT VT P4.1.20 Determine VO in Fig. P3.1.25 using TEC. Figure P3.1.25 10 A 2 Ω – + + – + – 2Vx 20 V + – Vx 2 Ω 2 Ω 4 Ω 4 Ω + – VO 2 Ω + – 2Vx + – Vx 2 Ω 2 Ω 4 Ω + – VO IT a b c d + – Vx Solution: If a current source is applied at node d, with the independent sources set to zero, it is seen that Vac = Vx, so that Vbd = 0 and VO = 0. In other words the source sees a short circuit and Rsrc = 0. If the resistor between node d and the reference node is replaced by an open circuit, the node-voltage equation at node a is: , and the node voltage equation at node c is: -0.5Va + 0.75Vc = -Iy, where Iy = . Substituting for Iy: 0.25Va + Vc = 30. Solving, gives Va = Vc = 40 V. Hence, Vx = 20 V and Vd = 0. In other words, TEC and NEC are just short circuits, 20 0.5 = − c a V V ( ) = − + 20 2 0.5 x c V V 30 0.5 − + a c V V 10 A 2 Ω – + + – 2Vx 20 V 2 Ω 2 Ω 4 Ω + – VO a b c d Iy 1. Determine VO assuming ISRC = 0.25 A. A. 4 V B. 1 V C. 5 V D. 2 V E. 3 V Solution: The two 5 Ω resistances can be combined in parallel to give a 2.5 Ω resistance, and the two 10 Ω resistances can be combined in parallel to give a 5 Ω resistance carrying a current of 2Ix, as shown. It follows that ISRC – 2Ix = 0.5Ix, or .5 2 SRC x I I = and , so that . x O I V 10 = SRC O I V 4 = _ + VO ISRC 0.5Ix 10 Ω 5 Ω 5 Ω Ix 10 Ω _ + VO ISRC 0.5Ix 2.5 Ω 2Ix 5 Ω 2. Determine ISRC assuming VSRC = 2 V and all resistances are 2 Ω. VSRC ISRC A. 1.5 A B. 3 A C. 2.5 A D. 2 A E. 1 A Solution: From symmetry the two currents Ix are equal and sum to zero. Hence, Ix = 0 and the two resistors can be removed. The equivalent resistance seen by the source is (2 + 2)||(2 + 2) = 2 Ω. It follows that ISRC = VSRC/2. VSRC ISRC Ix Ix 1/5 3. Determine VO assuming ISRC = 1 A. A. 7.5 V _ + VO ISRC 5 Ω 10 Ω 5 Ω ISRC B. 12.5 V C. 5 V D. 15 V E. 10 V _ + VO 5 Ω 10 Ω 5 Ω ISRC Solution: The two current sources are equivalent to a current source ISRC connected as shown, since KCL is the same at the two nodes. The resistance seen by the source is 10||(5 + 5) = 5 Ω. Hence, VO = 5ISRC. 4. Determine Thevenin’s resistance looking into terminals ab, assuming α = 10. A. 50 Ω B. 25 Ω – + 5 V 2 kΩ – + + – a b αIx 2VO RL Ix VO C. 100 Ω D. 200 Ω E. 20 Ω Solution: When a test source VT is applied at terminals ab, with the independent voltage source set to zero, it follows from the circuit that: T O O x V V V I = − = − = − 2 2 mA. IT = -αIx = αVT mA. Hence, α α 1000 1 kΩ ≡ = T T I V Ω. 2 kΩ – + + – a b αIx 2 VO Ix VO – + VT IT 2/5 – + – + V1 2 A 4 Ω 6 Ω 3 Ω 2 Ω 4 Ω – + V2 Vx 5. Determine V2 so that Vx = 0, assuming V1 = 4 V. A. 8 V B. 6 V C. 6.5 V D. 7.5 V E. 7 V Solution: The 6 Ω and 3 Ω resistors do not carry any current. They can removed from the circuit, with nodes a and b being at the same voltage. V1 can be transformed to a current source V1/4 A in parallel with a 4 Ω resistor. The total current is (0.25V1 + 2) A in parallel with 2 Ω. V2 is the voltage of node a, which gives: V2 = 2(0.25V1 + 2) = (0.5V1 + 4) V. – + – + V1 2 A 4 Ω 2 Ω 4 Ω V2 a b 6. Derive the mesh current equations in terms of I1, I2, and I3. DO NOT SOLVE THE EQUATIONS 20 Ω 5 Ω 5 Ω Ix 15 Ω 12 V – + 10 Ω I1 I2 I3 6 A 2Ix a b Solution: Considering the voltage drop Vab as a unit, the equation for mesh 1 is: (10 + 5)I1 – 5I3 = 12 – Vab The mesh-current equation for mesh 2 is: (20 + 5)I2 – 5I3 = Vab Adding these two equations: 15I1 + 25I2 – 10I3 = 12 The remaining equations are: I3 = 6, and I2 – I1 = 2Ix = 2(I2 – I3), or I1 + I2 – 2I3 = 0 Note that if the 15 Ω resistor is denoted by R and the conventional mesh-current procedure is applied, the term in R cancels out. Thus, for mesh 1: 3/5 (10 + 5 + R)I1 – RI2 – 5I3 = 12 – Vx, where Vx is the voltage drop across dependent current source in the direction of I1. For mesh 2, -RI1 + (20 + 5 + R)I2 – 5I3 = Vx. Adding these two equations gives the same equation as before. If these equations are solved, I1 = 22.8 A, I2 = -10.8 A, Ix = -16.8 A, Vx = -804 V. – + – + 10 V 6 A 6 Ω 3 Ω 4 Ω 1V a b 2 Ω 2 Ω 4 Ω Ix – + 2Ix 7. Determine Thevenin’s equivalent circuit seen between terminals a and b Solution: Method 1: Leave the circuit as it is. Considering the mesh on the RHS, 1 = 3Ix + 4Ix + Vcb, where Vcb = 10 + 2Ix. Substituting for Vcb gives Ix = -1 A, so that VTh = 4 V. Applying a test source with the independent sources set to zero, the branch containing the 6 A source is open circuited. The 6 Ω and 4 Ω resistors are in parallel with one terminal at node b and the other terminal connected to an open circuit. They do not carry any current and can be removed. The circuit reduces to that shown. VT = 4Ix + 2Ix = 6Ix, and IT = Ix + VT/3. Substituting for Ix gives VT/IT = RTh = 2 Ω. – + – + 10 V 6 A 6 Ω 3 Ω 4 Ω 1V a b 2 Ω 2 Ω 4 Ω Ix – + 2Ix Vcb c + – – + a b 4 Ω Ix VT 2Ix IT 3 Ω 2 Ω 4/5 If terminals ab are short circuited, KVL around the outermost loop gives: 10 + 2Ix + 4Ix = 0, so that Ix = -5/3 A; Isc = -Ix + 1/3 = 2 A. It follows that RTh = 4/2 = 2 Ω. – + – + 10 V 6 A 6 Ω 3 Ω 4 Ω 1V a b 2 Ω 2 Ω 4 Ω Ix – + 2Ix Vcb c + – Isc 1/3 A Method 2: If the branch consisting of the 1 V source in series with 3 Ω is removed, Ix = 0, the dependent source becomes a short circuit, and the open-circuit voltage between terminals a and b is the same as that of the 10 V source. Hence VTh1 = 10 V. – + a b 4 Ω IT IT 2IT If a test current source IT is applied between terminals a and b, with the independent sources set to zero, as before, and the 2 Ω resistor removed because it is in parallel with the 2Ix ideal voltage source and is redundant as far as Vab is concerned, the circuit reduces to that shown. The 2IT CCVS is equivalent to a 2 Ω resistor, which in series with the 4 Ω resistor gives RTh1 = 6 Ω. When the branch between terminals a and b is reintroduced, the circuit becomes as shown. With terminals a and b open circuited, the current in the circuit is 1 A in the direction shown and Vab = 4 V. If the voltage sources are set to zero, the resistance seen between terminals ab is (6||3) = 2 Ω. Hence, VTh = 4 V and RTh = 2 Ω. – + 3 Ω 1V a 1 A – + 6 Ω 10 V 5/5 Problem 3 Find Ix. − + 10V 2KΩ +− 5V 10Ω 2Ω 5Ω 10Ω Ix − + 5Ix A) 230.8A B) 76.92A C) -76.92A D) -230.8A E) None of the above Problem 4 Find Vx. 6A 0.25Vx 80Ω − Vx + 10Ω 20Ω 16Ω Ix − + 120V − + 3Ix A) 130.9V B) -43.64V C) 43.64V D) -130.9V E) None of the above Problem 5 Find the power associated with the current source. − + 160V 4Ω 5A 16Ω 32Ω 16Ω 32Ω A) 256W B) -200W C) 200W D) -256W E) None of the above Problem 6 In the circuit below R1 is chosen such that I3 = 1A. Find R1. − + 100V R1 I1 75Ω I3 25Ω I2 A) 12.5Ω B) 16Ω C) 25Ω D) 6.25Ω E) None of the above Problem 7 Find Ix. − + 100V 10Ω 10Ω 10Ω −+ 20V − + 30V Ix A) -6A B) 6A C) 16A D) -16A E) None of the above Problem 8 Find I0. 2I0 25Ω 50Ω I0 20Ω − + 38.5V 100Ω 200Ω 5Ω − + 5I0 A) 1.15A B) -0.65A C) -1.15A D) 0.65A E) None of the above Problem 9 Find the Thevenin equivalent resistance between a and b. 0.4A 4KΩ 12KΩ Ix − + 500Ix 500Ω a 1KΩ b A) 333.33Ω B) 250Ω C) 83.33Ω D) 740.46Ω E) None of the above Problem 10 Find the Thevenin equivalent voltage between a and b (Vab). − + 2V 2KΩ a 1KΩ 2mA 1KΩ 3KΩ − + 4V 6mA 2KΩ 2mA + Vab − b Problem 11 Find the Thevenin equivalent resistance between a and b of the previous figure. A) 6KΩ B) 8KΩ C) 4.5KΩ D) 1.5KΩ E) None of the above Problem 12 Find the Thevenin equivalent resistance between a and b. 100A 250KΩ − + 100V 50Ω a 20Ω b 60Ω I0 40Ω − + 60I0 A) 6Ω B) 8.52Ω C) 14.28Ω D) 10.52Ω E) None of the above Problem 4 Find V1. − + 5V 2Ω 6V1 2Ω 2Ω + V1 − − + 6V A B A) -1.22V B) 1.22V C) -1.57V D) 1.57V E) None of the above Problem 5 Find the Thevenin Equivalent Voltage between A and B (VAB). I R i1 − + R · i1 R I R B A A) RI B) -3RI C) -RI D) 3RI E) None of the above Problem 6 Find the Norton equivalent resistance between A and B. R R R I1 −+ V1 I2 +− V2 B A A) 3R B) 3R/2 C) R/3 D) R E) None of the above Problem 7 Find the Norton equivalent current source between A and B. R R R I R R R R R R R I A B A) -2.28I(A) B) -1.24I(A) C) -3.21I(A) D) -6.42I(A) E) None of the above