EECE 210 Problems For Quiz 2, Fall 2012

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 P3.1.25 Determine VO in F P3.1.25 using node voltage analys ig. - is. Va 3.1.26 Determine VO in Fig. P3.1.25 using mesh-current analysis. I1 = 10 and 3 – I4 = 10. For mesh 4: A, I3 = 0, and I4 = 0, 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 2 10 A 2 Ω – + + – 2Vx 20 V + – Vx 2 Ω 2 Ω 4 Ω 4 Ω + – VO a b c d Iy – Vc + Vd = 40. Solving, Va = 40 V, Vc = 40 V, Vd = VO = 0. P Solution For mesh 2: + – V -2I1 + 6I2 – 2I4 = 10 A 2 Ω – + + – 2Vx 20 V x 2 Ω 2 Ω 4 Ω 4 Ω -2Vx; substituting Vx = -2I2: I2 – I4 = 10. For mesh 3: -4I1 + 8I3 – 4I4 = 2Vx; I2 + 2I -2I2 – 4I3 + 6I4 = -20, or -I2 – 2I3 + 3I4 = -10. Solving, I2 = 10 which gives VO = 0. + – VO I2 I1 I4 I3 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 -76.92 mA 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 176.9 W absorbed 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 Vab = -1 V 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 1. The current components in a resistor R, as found from superposition, are -8 A and +4 A. Determine the power dissipated in the resistor if R = 2 Ω. Solution: The magnitude of the current through R is 4 A. The power dissipated in R is 16R = 32 W. 2. Determine IO if VS = 1 V. Solution: The two terminals of the 5 Ω resistor are at the same voltage. Hence, this resistor carries no current and can be removed. The 1 Ω in series with VS can be split into two parallel branches, each of 2 Ω in series with VS. The circuit can then be split into two halves as shown. It follows that 4 5 S O V I − = = 1 A. 3. Determine R so Norton’s current between terminals a and b is zero, given that ISRC = 5 A and α = 1 V/A. Solution: If a and b are short-circuited and no current flows in the short circuit, ISRC flows through R. Applying KVL, RISRC = αISRC, which gives R = α = 1 Ω. 4. Determine Norton’s current at terminals a and b, assuming ISRC = 0.5 A. Solution: When terminals ab are short circuited, VX = 0 and the dependent voltage source 0.5VX is zero, so that IN = ISRC = 0.5 A. 10 Ω 5 Ω ISRC – + αISRC a b R – + 5 V – + – + 5 V VS 4 Ω 4 Ω 1 Ω 2 Ω 2 Ω 5 Ω IO – + – + 5 V VS 4 Ω 2 Ω 2 Ω IO ISRC – + 0.5VX – + 5 Ω 10 Ω 2Io Io a b + – VX – + VSRC 2 kΩ – + + – a b 2Ix 2VO RL Ix VO 5. Determine the power delivered by the source, assuming all resistances are 1 Ω and VSRC = 1 V. Solution: From symmetry, each top node along the same vertical line is at the same voltage as the corresponding bottom node. Joining these nodes together simplifies the circuit to that shown. The resistance on each side of the source is 0.5||2 Ω = 0.4 Ω. Hence, ISRC = VSRC/0.8 A and the power delivered by the source is .8 0 2 VSRC = 1.25 0.8 1 = W. 6.  Determine VX assuming VS = 1 V. Solution: With the 4 V source applied alone, and VS set to zero, 2 4 4 || 4 2 2 1 = × + VX = V. With VS applied and the 4 V source set to zero, 4 4 / 3 4 4 / 3 2 || 4 4 2 || 4 2 S S S X V V V V = − × + = − × + = − ; VX = 4 2 − VS 1 75 4 1 2 = . − = V. 7. Determine Thevenin’s voltage between terminals a and b, with RL removed, assuming VSRC = 5 V. Solution: When terminals ab are open circuited, IX = 0. To make IX = 0, 2VO = VSRC, so that VTh = VSRC/2 = 2.5 V. 8. Determine the power delivered by ISRC assuming ISRC = 0.5 A. – + VSRC 1 Ω 1 Ω 0.5 Ω 0.5 Ω 0.5 Ω 0.5 Ω 0.5 Ω 0.5 Ω – + 2 Ω 2 Ω 4 Ω 4 Ω 1 Ω – + 2 Ω 4 V VS + – VX – + VSRC _ + 6 V 6 Ω 3 Ω 5 Ω ISRC 1 Ω Solution: To determine Vab, the 5 Ω and 1 Ω resistances are redundant. Transforming the voltage source to a current source, it follows that Vab = 2(2 + ISRC) From the original circuit, the voltage rise across the current source is Vab + 1×ISRC = 3ISRC + 4, and the power delivered is P = ISRC(3ISRC + 4) = 2.75 W. 9. Determine VTh assuming VS = 1 V and R = 2 Ω. Solution: Let VX be the voltage drop Vdc. Applying KCL to node d: 2 4 8 6 3 = + + + X X X V R V R V ; this gives VX = 0, so that Vef = 0 and VTh = VS = 1 V. Alternatively, if the subcircuit between nodes c and d is removed, Vdc in this circuit is zero, and Vdc in the bridge is zero, which makes Vef = 0 and VTh = VS. 10. Determine RTh in the preceding problem. Solution: If a test source is applied with all independent sources set to zero, the circuit reduces to that shown, where the current in the 4 Ω resistor is zero. It follows that RTh = 3R||6R = 2R = 4 Ω. + – + – a b + – VTh 4 Ω 2 A 8 V VS R 2R 2R 4R + – + – a b + – VTh 4 Ω 2 A 8 V VS R 2R 2R 4R + – VX VX /6R VX /3R d c e f + – 4 Ω 2 A 8 V + – 8 V c d R 2R 2R 4R d c a b 4 Ω – + VT IT _ + 6 V 6 Ω 3 Ω ISRC a b 11. For the circuit shown, write the circuit equations in terms of the node voltages Va, Vb, and Vc. You are not required to solve these equations. Solution: The circuit is redrawn as shown with the redundant 2 S resistor removed and the three 3S resistors combined into a 1 S resistor. For node a: Va = 0.5IX, where, for node b, IX = 5 – 1×Vb. Substituting and rearranging, 2Va + Vb = 5. For node c: -2Va + 6Vc = IX = 5 – Vb; or -2Va + Vb + 6Vc = 5. The third equation is: Vb – Vc = 10. 12. Use the substitution theorem and superposition to determine Vy, where R is unknown. Solution: In accordance with the substitution theorem, R can be replaced by a 0.5 A current source. With this current source set to zero, Vy1 = 17.5 225 30 17.5 7.5 = − × − V. With the 30 V source set to zero and the 0.5 A current source applied, Vy2 = 17.5 37.5 7.5 0.5 17.5 10 = − × × − V. It follows that Vy = Vy1 + Vy2 = 15 17.5 262.5 17.5 37.5 225 = − = − + − V. 30 V – + 0.5 A 18 Ω 8 Ω 10 Ω 7.5 Ω R Vy 10 V + – 5 A – + 0.5Ix 2 S 4 S 1 S Ix a b c 10 V + – 5 A – + 0.5IX 2 S 2 S 4 S 3 S 3 S 3 S IX a b c 1 – + 12 V a b c d 13. Determine Thevenin’s equivalent circuit between terminals ab, assuming all resistances are 5 Ω. Solution: With the 5 A source applied alone, it follows from symmetry that VTh1 = 0. With the 12 V source applied alone, it follows from symmetry (as in Example 4.5.1a) that the middle nodes are all at the same voltage. The circuit reduces to that shown, where the effective resistance between nodes c and d is 10||20 = 20/3 Ω, so that Vcd = V, and VTh = V. Alternatively, VTh can be found by scaling. To determine RTh, with the two sources set to zero, it follows again from Example 4.5.1a that the middle nodes are at the same voltage so that the horizontal resistor in the middle carries no current and can be removed. With the 12 V source replaced by a short circuit, the circuit reduces to that shown. It follows that the effective resistance between nodes c and d is 5 Ω, and RTh = 15||10 = 6 Ω. 1. Determine Thevenin’s equivalent circuit (TEC) looking into terminals ab, assuming VSRC = 5 V and ISRC = 2 A. A. VTh = 12 V, RTh = 2 Ω B. VTh = 15 V, RTh = 6 Ω C. VTh = 12 V, RTh = 6 Ω D. VTh = 15 V, RTh = 2 Ω E. TEC does not exist Solution: On open circuit, ISCR flows through the 6 Ω resistor, so that VTh = 6ISRC. If a test source is applied with the voltage source replaced by short circuit and the current source replaced by an open circuit, the resistance looking into terminals ab is RTh = 6 Ω. 4.8 12 10 20 / 3 20 / 3 = × + 2.4 4.8 20 10 = × a b c d a – + b 12 V 5 A – + VSRC 3 Ω 6 Ω a b ISRC 2 2. A circuit has 20 essential branches, 11 essential nodes and 10 meshes. How many independent KCL equations can be written for this circuit? A. 8 B. 6 C. 7 D. 10 E. 9 Answer: The number of independent KCL equations is one less than the number of essential nodes. 3. How many independent KVL equations can be written for the circuit of the preceding question? A. 11 B. 10 C. 9 D. 12 E. 13 Answer: The number of independent KVL equations is same as the number of meshes. 11. Determine IO assuming all resistances are 1 Ω and VSRC = 5 V. Solution: Superposition can be applied. If the 10 V source is applied alone, with VSRC replaced by a short circuit, the component of IO is zero. If VSRC is applied alone, IO is determined by VSRC and what happens in the circuit to the left of VSRC is not relevant to IO. The resistor across VSRC is redundant, and the circuit reduces to that shown. If scaling is applied, assuming IO = 1 A, then Vcd = 3 V, Iac = Ibd = 4A, Vab = 2×4×1 + 3 =11 V, ISRC = 15 A, and VSRC = 2×15×1 + 11 = 41 V. Hence, IO = VSRC×(1/41). Alternatively, VSRC can be divided into two sources, and the left-hand part of the circuit reduces to that shown. – + VSRC 1 A – + 3 A 3 V 4 A – + 11 V 11 A 15 A a c b d – + – + 10 V VSRC IO 3 4. Determine VO. A. 2 V B. 4 V C. 0 D. 1.5 V E. 2.5 V Solution: Since the circuit does not have any independent sources, VO = 0. This can also be shown by analyzing the circuit. Thus, 1.25Vx = VO, so I1 = Vx/1 = VO/1.25 = 0.8VO; I2 = 0.5VO/2 = 0.25VO; IO = VO/5 = 0.2VO. Applying KCL, 0.8VO + 0.25VO + 0.2VO = 0, or 1.25VO = 0, so VO = 0. 5. Given that Va = 25 V and Vb = 12 V. Determine Va if node b is grounded instead of c. A. 13 V B. 37 V C. 14 V D. 15 V E. 16 V Solution: When node b is grounded, Vb = 0. Hence, 12 V is subtracted from all node voltages, and Va = 25 – 12 = 13 V. Alternatively, it can be argued that the voltage drop from a to b is 13 V irrespective of which node is grounded. 6. The power dissipated in R is 8 W if ISRC is applied alone, with VSRC set to zero, and is 0.5 W if VSRC is applied alone, with ISRC set to zero. Determine the power dissipated in R if both sources are applied together, with polarities that will give the largest current in R. A. 8.5 W B. 18 W C. 24.5 W D. 32 W E. 12.5 W – + – + – + VO 0.5VO – + Vx 0.25Vx 5 Ω 2 Ω 1 Ω I1 I2 IO – + VSRC 2 Ω R ISRC 2 Ω 4 Solution: When ISRC source is applied alone, with VSRC set to zero, the current in R is 2 8/ 2 = A. When VSRC is applied alone, with ISRC set to zero, the current in R is Pv / 2 . With both sources applied together to to give the largest current in R, the total current in R is / 2 2 V P + and the power dissipated in R is ( ) V V V T P P P P + + = + = / 2 8 8 / 2 2 2 2 . 7. Determine Norton’s equivalent circuit (NEC) looking into terminals ab, assuming VSRC = 5 V and ISRC = 2 A. A. IN = 2 A, RN = 5 Ω B. NEC does not exist C. IN = 3 A, RN = 5 Ω D. IN = 4 A, RN = 5 Ω E. IN = 2 A, RN = 0 Solution: On open circuit, VTh = VSRC. If VSRC and ISRC are set to zero, the resistance looking into terminals ab is that of a short circuit, so RTh = 0. If terminals ab are short-circuited, IN → ∞. Hence, TEC is the ideal voltage source VSRC, and NEC does not exist. 8. Determine TEC looking into terminals ab, assuming VSRC = 4 V. A. VTh = 4 V, RTh = 75 Ω B. VTh = 4 V, RTh = 300 Ω C. VTh = 4 V, RTh = 100 Ω D. VTh = 3 V, RTh = 300 Ω E. VTh = 3 V, RTh = 75 Ω Solution: The 1 kΩ resistor and the dependent current source are redundant because they are in parallel with an ideal voltage source. The circuit reduces to a voltage source connected to a 100 Ω-300 Ω voltage divider with terminals ab across the 300Ω resistor. It follows that VTh = VSRC×300/400 = 3VSRC/4. If a test source is applied with VSRC replaced by a short circuit, the resistance looking into terminals ab is RTh = 100||300 = 75 Ω. Alternatively, if Ix is the current flowing in the 100 Ω resistor toward node a, with the output open circuited, then Vφ = VSRC – 100Ix and Vφ = 300Ix. Eliminating Ix gives VTh = Vφ = 3VSRC/4. VSRC – + ISRC 5 Ω a b – + VSRC 0.5Vφ 1k Ω a b 100 Ω 300 Ω – + Vφ 5 1. Determine, according to the substitution theorem, which of the following ideal circuit elements can replace the dependent source without affecting the currents and voltages in the rest of the circuit (Mark all correct answers if you believe that more than one answer is correct). A. Independent 4 V source B. Dependent 2 V source C. Dependent 1 A source D. Independent 2 A source E. 2 Ω resistor Solution: From KCL at node a, 3 = 3Ix, so Ix = 1 A, and Vab = 4 V. The voltage across the dependent source is 4 V and the current through this source is 2 A. According to the substitution theorem, the dependent source can be replaced by: i) an independent 4 V source, ii) an independent 2 A source, or iii) a 2 Ω resistor. Replacing the dependent source by any of these elements does not affect the voltages and currents in the rest of the circuit, as indicated by the figures shown. 2. All resistances are 1 Ω, except for the two 4 Ω resistances indicated. Determine the source current, assuming VSRC = 1 V. A. 2 A B. 5 A C. 3 A D. 1 A E. 4 A – + 2 V Ix 2Ix 3 A 2 Ω – + 2 V Ix 2Ix