EECE 210 Exam 2 with Solution, Fall 2010

EECE 210 Quiz 2 – Nov 20, 2010 4% 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. Version 1: R = 2 Ω, P = 16R = 32 W Version 2: R = 3 Ω, P = 16R = 48 W Version 3: R = 4 Ω, P = 16R = 64 W Version 4: R = 5 Ω, P = 16R = 80 W Version 5: R = 6 Ω, P = 16R = 96 W. 4% 1 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 − = A. – + 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 Version 1: VS = 1 V, 1 4 1 5 = − = OI A Version 2: VS = 2 V, 0.75 4 2 5 = − = OI A Version 3: VS = 3 V, 0.5 4 3 5 = − = OI A Version 4: VS = 4 V, 0.25 4 4 5 = − = OI A Version 5: VS = 5 V, 0 4 5 5 = − = OI A. 4% 10 Ω 5 Ω ISRC – + αISRC a b R 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 Ω. Version 1: α = 1, R = 1 Ω Version 2: α = 2, R = 2 Ω Version 3: α = 3, R = 3 Ω Version 4: α = 4, R = 4 Ω Version 5: α = 5, R = 5 Ω. 4% 4. Determine the power delivered by the source, assuming all resistances are 1 Ω and VSRC = 1 V. – + VSRC 2 er 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 pow delivered by the source is .8 0 2 VSRC W. – + VSRC 1 Ω 1 Ω 0.5 Ω 0.5 Ω 0.5 Ω 0.5 Ω 0.5 Ω 0.5 Ω Version 1: VSRC = 1 V, 1.25 0.8 1 = P = W Version 2: VSRC = 1.2 V, ( ) 1.8 0.8 1.2 2 = P = W Version 3: VSRC = 1.4 V, ( ) 2.45 0.8 1.4 2 = P = W Version 4: VSRC = 1.6 V, ( ) 3.2 0.8 1.6 2 = P = W Version 5: VSRC = 1.8 V, ( ) 4.05 0.8 1.8 2 = P = W 4% – + 2 Ω 2 Ω 4 Ω 4 Ω 1 Ω – + 2 Ω 4 V VS + – VX 5.  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 V. Version 1: VS = 1 V, 1.75 4 1 2 = − V X= V Version 2: VS = 2 V, 1.5 4 2 2 = − V X= V Version 3: VS = 3 V, 1.25 4 3 2 = − V X= V Version 4: VS = 4 V, 1 4 4 2 = − V X= V Version 5: VS = 5 V, 0.75 4 5 2 = − V X = V. 4% ISRC – + 0.5VX – + 5 Ω 10 Ω 2Io Io a b + – VX 6. 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. Version 1: ISRC = 0.5 A, IN = 0.5 A Version 2: ISRC = 1 A, IN =1 A Version 3: ISRC = 1.5 A, IN = 1.5 A Version 4: ISRC = 2 A, IN = 2 A Version 5: ISRC = 2.5 A, IN = 2.5 A. 2 kΩ 3 – + VSRC – + + – a b 4% 2Ix 2VO RL Ix VO 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 V. Version 1: VSRC = 5 V, VTh =VSRC/2 = 2.5 V Version 2: VSRC = 10 V, VTh =VSRC/2 = 5 V Version 3: VSRC = 15 V, VTh =VSRC/2 = 7.5 V Version 4: VSRC = 20 V, VTh =VSRC/2 = 10 V Version 5: VSRC = 25 V, VTh =VSRC/2 = 12.5 V. 4% 4 8. Determine the power delivered by ISRC assuming ISRC = 0.5 A. 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). _ + 6 V 6 Ω 3 Ω 5 Ω ISRC 1 Ω _ + 6 V 6 Ω 3 Ω ISRC a b Version 1: ISRC = 0.5 A, P = 0.5(1.5 + 4) = 2.75 W Version 2: ISRC = 1 A, P = 1(3 + 4) = 7 W Version 3: ISRC = 1.5 A, P = 1.5(4.5 + 4) = 12.75 W Version 4: ISRC = 2 A, P = 2(6 + 4) = 20 W Version 5: ISRC = 2.5 A, P = 2.5(7.5 + 4) = 28.75 W. 4% 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. 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. + – + – a b + – VTh 4 Ω 2 A 8 V VS R 2R 2R 4R + – 4 Ω 2 A 8 V + – 8 V c d + – + – a b + – VTh 4 Ω 2 A 8 V VS R 2R 2R 4R + – VX VX /6R VX /3R d c e f Version 1: VS = 1 V, VTh = 1 V Version 2: VS = 1.5 V, VTh = 1.5 V Version 3: VS = 2 V, VTh = 2 V Version 4: VS = 2.5 V, VTh = 2.5 V Version 5: VS = 3 V, VTh = 3 V. R 2R 2R 4R d c a b 4 Ω – + VT IT 4% 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. Version 1: R = 1 Ω, RTh = 2 Ω 10 V + – 5 A – + 0.5IX 2 S 2 S 4 S 3 S 3 S 3 S IX a b c Version 2: R = 1.5 Ω, RTh = 3 Ω Version 3: R = 2 Ω, RTh = 4 Ω Version 4: R = 2.5 Ω, RTh = 5 Ω Version 5: R = 3 Ω, RTh = 6 Ω. 20% 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. 10 V + – 5 A – + 0.5Ix 2 S 4 S 1 S Ix a b c 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. 5 20% 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 6 – + 12 V a b c d 20% 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 = 4.8 12 10 20 / 3 20 / 3 = × + V, and VTh = 2.4 4.8 20 10 = × V. Alternatively, VTh can be found by scaling. a – + b 12 V 5 A a b c d 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.