EECE 210 Final, Fall 2010

Final Exam – 2010-2011 3% _ + ISRC 5IX 15 Ω 12 Ω 5 Ω IX 10 Ω _ + 10 Ω 10 V 1. Determine IX, assuming ISRC = 1 A (Hint: write one KVL equation and one KCL equation). A. 4 A B. 2 A C. 3 A D. 1 A E. 5 A Solution: Let the current through the 10 Ω resistor be IY. From KVL around the mesh abcda: 5IX + 5IX – 10 – 10IY = 0, which gives, IY = IX – 1. From KCL at node a, ISRC = IY + IX, or ISRC = 2IX – 1. It follows that IX = (ISRC + 1)/2. Version 1: ISRC = 1 A, IX = (1 + 1)/2 = 1 A Version 2: ISRC = 1.5 A, IX = (1.5 + 1)/2 = 1.25 A Version 3: ISRC = 2 A, IX = (2 + 1)/2 = 1.5 A Version 4: ISRC = 2.5 A, IX = (2.5 + 1)/2 = 1.75 A Version 5: ISRC = 3 A, IX = (3 + 1)/2 = 2 A 3% 8 Ω j20 Ω -j2 Ω R _ + 50∠0° V 2. In the circuit shown, the capacitance absorbs -200 VAR. Determine the average power dissipated in R if R = 5 Ω. A. 57.1 W B. 80 W C. 44.4 W D. 66.7 W E. 50 W Solution: Q = , where Vrms is the rms voltage across R and C, and B = -1/X = 1/2 S. Substituting, -200 = 2 BVrms − 2 2 rms − 1V , and Vrms = 20 V. It follows that PR = R V 2 rms . Version 1: R = 5 Ω, PR = = 5 400 80 W Version 2: R = 6 Ω, PR = = 6 400 66.7 W 1 Version 3: R = 7 Ω, PR = = 7 400 57.1 W Version 4: R = 8 Ω, PR = = 8 400 50 W Version 5: R = 9 Ω, PR = = 9 400 44.4 W 3% 3. Determine IX assuming ISRC = j A. A. j6 A B. -j3 A C. j3 A D. -j6 A E. j4 A Solution: The voltage across the -j3 Ω capacitor is 6 V and the current through this capacitor, directed upwards is j2 A. It follows that IX = ISRC + j2 A. 6 V ISRC j18 Ω -j3 Ω 5 Ω IX Version 1: ISRC = j A, IX = j + j2 = j3 A Version 2: ISRC = j2 A, IX = j2 + j2 = j4 A Version 3: ISRC = j3 A, IX = j3 + j2 = j5 A Version 4: ISRC = j4 A, IX = j4 + j2 = j6 A Version 5: ISRC = j5 A, IX = j5 + j2 = j7 A 3% 4. Determine the resistance between nodes a and b assuming that all resistances are 1 Ω. a b a b 2R 2R R R R R R a b R 0.5R 0.5R 0.5R 0.5R 0.25R 0.25R A. 4 Ω B. 2.4 Ω C. 3.2 Ω D. 1.6 Ω E. 0.8 Ω Solution: The resistances can be split into two halves in parallel, where the resistances in the middle become 2R. Applying Δ-Y transformation, 2 the resistances become as shown. It follows that 2Rab = 0.5R + R||1.5R + 0.5R = R + 0.6R = 1.6R, so that Rab = 0.8R. Version 1: R = 1 Ω, Rab = 0.8×1 = 0.8 Ω Version 2: R = 2 Ω, Rab = 0.8×2 = 1.6 Ω Version 3: R = 3 Ω, Rab = 0.8×3 = 2.4 Ω Version 4: R = 4 Ω, Rab = 0.8×4 = 3.2 Ω Version 5: R = 5 Ω, Rab = 0.8×5 = 4 Ω 3% 5. Two coils are tightly coupled to a high-permeability core, so that the leakage flux is negligibly small. If coil 1 has 100 turns and an inductance of 10 mH, and the mutual inductance is 12.5 mH, determine the number of turns of coil 2. A. 125 B. 250 C. 150 D. 175 E. 200 Solution: From the definitions of self and mutual inductance, with negligible leakage flux, 1 21 1 1 i N L φ = and 1 2 21 i N M φ = . It follows that M L N M N 10 1 1 2 = = . Version 1: M = 12.5 mH, N2 = 10×12.5 = 125 turns Version 2: M = 15 mH, N2 = 10×15 = 150 turns Version 3: M = 17.5 mH, N2 = 10×17.5 = 175 turns Version 4: M = 20 mH, N2 = 10×20 = 200 turns Version 5: M = 22.5 mH, N2 = 10×22.5 = 225 turns 3% 6. Determine the inductance of coil 2 of the preceding problem. A. 22.5 mH B. 30.63 mH C. 15.63 mH D. 40 mH E. 50.63 mH 3 Solution: Since the coils are tightly coupled to the core, k = 1, so that M2 = L1L2, or 2 1 2 2 .1 0 M L M L = = mH. It also follows from the solution of the preceding problem that 2 2 1 L N N = M . Dividing, 2 2 1 2 1 2 .1 0 M N N L L = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = Version 1: M = 12.5 mH, L2 = 0.1×(12.5)2 = 15.625 mH Version 2: M = 15 mH, L2 = 0.1×(15)2 = 22.5 mH Version 3: M = 17.5 mH, L2 = 0.1×(17.5)2 = 30.625 mH Version 4: M = 20 mH, L2 = 0.1×(20)2 = 40 mH Version 5: M = 22.5 mH, L2 = 0.1×(22.5)2 = 50.625 mH 3% 7. A D’Arsonval movement has a resistance of R Ω and a full-scale deflection of 100 μA. Determine the shunt resistance that will result in a full-scale deflection of 150 μA, assuming R = 50 Ω. A. 150 Ω B. 200 Ω C. 300 Ω D. 100 Ω E. 250 Ω Solution: At full-scale deflection, the voltage drop across the movement and shunt is (R Ω)×(100 μA) = 100R μV. The shunt has to pass 50 μA, so its resistance is Rshunt = 100R/50 = 2R Ω. Version 1: R = 50 Ω, Rshunt = 2×50 = 100 Ω Version 2: R = 75 Ω, Rshunt = 2×75 = 150 Ω Version 3: R = 100 Ω, Rshunt = 2×100 = 200 Ω Version 4: R = 125 Ω, Rshunt = 2×125 = 250 Ω Version 5: R = 150 Ω, Rshunt = 2×150 = 300 Ω 3% 8. When a 9950 Ω resistance is connected in series with a D’Arsonval movement of unknown resistance and full-scale deflection current, a voltage of 1 V across the series combination gives a certain full-scale deflection. If an additional 10,000 Ω is connected in series with the combination, 2 V are required for full-scale deflection. Determine the resistance of the D’Arsonval movement. 4 A. 150 Ω B. 100 Ω C. 75 Ω D. 125 Ω E. 50 Ω Solution: Let the resistance of the movement be Rm, its FSD current be IFSD, and the FSD voltage with series resistance be VFSD. Then IFSD(R + Rm) = VFSD, and IFSD(10,000 + R + Rm) = 2VFSD. It follows that R + Rm = 10,000, or Rm = 10,000 – R. Version 1: R = 9950 Ω, Rm = 10,000 – 9950 = 50 Ω Version 2: R = 9925 Ω, Rm = 10,000 – 9925 = 75 Ω Version 3: R = 9900 Ω, Rm = 10,000 – 9900 = 100 Ω Version 4: R = 9875 Ω, Rm = 10,000 – 9875 = 125 Ω Version 5: R = 9850 Ω, Rm = 10,000 – 9850 = 150 Ω 3% 2 H Leq 3 H L 1 H 1 H 1 H 9. Determine Leq if L = 1 H. A. 6 H B. 4 H C. 8 H D. 7 H E. 5 H Solution: Consider that a voltage V is applied, causing a current I to flow. V = jωI[(2 – 1 +1) + (3 – 1 – 1) + (L + 1 – 1)]; Leq = 3 + L. Version 1: L = 1 H, Leq = 4 H Version 2: L = 2 H, Leq = 5 H Version 3: L = 3 H, Leq = 6 H Version 4: L = 4 H, Leq = 7 H Version 5: L = 5 H, Leq = 8 H. 10 Ω _ + VSRC – + VTh 1:2 3% 10. Determine VTh, assuming VSRC = 1∠0° V A. -1∠0° V B. 1∠0° V C. -2∠0° V D. 2∠0° V E. 4∠0° V 5 Solution: On open circuit, no current flows. The primary voltage is VSRC as shown, and VTh = -VSRC. 10 Ω _ + VSRC – + VTh 1:2 _ + VSRC _ + 2VSRC Version 1: VSRC = 1∠0° V, VTh = -1∠0° V Version 2: VSRC = 2∠0° V, VTh = -2∠0° V Version 3: VSRC = 3∠0° V, VTh = -3∠0° V Version 4: VSRC = 4∠0° V, VTh = -4∠0° V Version 5: VSRC = 5∠0° V, VTh = -5∠0° V 3% R _ + 100∠0° V rms IX j10 Ω j10 Ω ZL 11. Determine ZL for maximum average power delivered to it if R = 5 Ω and IX = k∠-45° where k = 2 A rms. A. 10 + j10 Ω B. 5 + j5 Ω C. 5 – j5 Ω D. 10 – j10 Ω E. 15 – j15 Ω Solution: ZTh is (R + j5) Ω. Hence, ZL for maximum power transfer is (R – j5) Ω. Version 1: R = 5 Ω, ZLmax = (5 – j5) Ω Version 2: R = 6 Ω, ZLmax = (6 – j5) Ω Version 3: R = 7 Ω, ZLmax = (7 – j5) Ω Version 4: R = 8 Ω, ZLmax = (8 – j5) Ω Version 5: R = 9 Ω, ZLmax = (9 – j5) Ω 3% 12. Determine the maximum average power delivered to ZL in Problem 11, assuming that R = 5 Ω and IX is as in Problem 11. A. 90 W B. 200 W C. 320 W D. 180 W E. 245 W 6 Solution: VTh as seen by ZL is determined from superposition as ∠ ° × + 100 0 10 10 10 j j j + = = 10 || 10 XI) (5 j + j XI) 5(1 50 0 + j + ∠ o o o o 45 (5 2 45 ) 50 0 × ∠ − ∠ + ∠ k = 50 + 5 2k ; ( ) ( ) 20 5 2 50 2 2 k k + = 4 5 2 50 max R P Th L + = Version 1: k = 2 A, ( ) ( ) 20 60 20 2 5 2 50 2 2 max = × + = PL = 180 W Version 2: k = 2 2 A, ( ) ( ) 20 70 20 2 2 5 2 50 2 2 max = × + = PL = 245 W Version 3: k = 3 2 A, ( ) ( ) 20 80 20 3 2 5 2 50 2 2 max = × + = PL = 320 W Version 4: k = 4 2 A, ( ) ( ) 20 90 20 4 2 5 2 50 2 2 max = × + = PL = 405 W Version 5: k = 5 2 A, ( ) ( ) 20 100 20 5 2 5 2 50 2 2 max = × + = PL = 500 W 3% 13. Load L1 absorbs 15 kVA at 0.6 p.f. lagging, whereas Load L2 absorbs 4.8 kW at 0.8 p.f. leading. If VSRC = 200∠0° V rms at f = 50 Hz, determine the capacitor that must be connected in parallel with L1 and L2 to have maximum magnitude of current through the source. – + L1 L2 VSRC A. 0.67 mF B. 0.55 mF C. 0.34 mF D. 0.46 mF E. 1.24 mF Solution: The reactive power absorbed by L1 is 15×0.8 kVAR = 12 kVAR, whereas the reactive power absorbed by L2 is 3.6 0.6 0.8 4.8 = − × − kVAR. For maximum magnitude of source current, the p.f. should be unity. The capacitor must add a reactive power of -(12 – 3.6) = -8400 VAR. hence, -8400 = -ωC×|VSRC|2, or 2 2 | | 84 | | 100 8400 SRC SRC V V π π = C = . 7 Version 1: |VSRC| = 200 V, 0.67 (200) 84 2 ≡ = π C mF Version 2: |VSRC| = 220 V, 0.55 (220) 84 2 ≡ = π C mF Version 3: |VSRC| = 240 V, 0.46 (240) 84 2 ≡ = π C mF Version 4: |VSRC| = 260 V, 0.40 (200) 84 2 ≡ = π C mF Version 5: |VSRC| = 280 V, 0.34 (200) 84 2 ≡ = π C mF 3% 14. A periodic current is shown, where over a period, t π 6 A 2π -4 A i = 6 + Asin2t 0 ≤ t ≤ π i = -4 + Asin2(t – π) π 0 ≤ t ≤ 2π Determine the rms value of i if A = 1 A. A. 5.83 A B. 5.15 A C. 6.20 A D. 5.29 A E. 5.52 A Solution: The waveform consists of three components: i) a dc component of 1 A, ii) a square wave of 5 V amplitude, and iii) a sinusoidal wave of amplitude A. It follows that the rms value is I = / 2 26 / 2 5 1 2 2 2 2 A A + = + + A Version 1: A = 1; 5.15 26.5 / 2 26 2 = = + = A I A Version 2: A = 2; 5.29 28 / 2 26 2 = = + = A I A Version 3: A = 3; 5.52 30.5 / 2 26 2 = = + = A I A Version 4: A = 4; 5.83 34 / 2 26 2 = = + = A I A Version 5: A = 5; 6.20 38.5 / 2 26 2 = = + = A I A. 3% 15. The current waveform of the preceding problem is applied to a 2 Ω resistor in parallel with a very large capacitor. Determine the voltage across the parallel combination. 8 A. 2.5 V B. 2 V C. 3 V D. 4 V E. 3.5 V Solution: The ac voltage will be negligibly small. The dc voltage is the dc component of current multiplied by R, or V = 1×R. Version 1: R = 2 Ω; V = R = 2 V Version 2: R = 2.5 Ω; V = R = 2.5 V Version 3: R = 3 Ω; V = R = 3 V Version 4: R = 3.5 Ω; V = R = 3.5 V Version 5: R = 4 Ω; V = R = 4 V 11% 16. The period of a periodic function f(t) is defined as: f(t) = cos(t + π) – 2, -π < t < -π/2 f(t) = -cos(t) + k, -π/2 < t < +π/2 f(t) = cos(t – π) – 2, π/2 < t < π Derive the trigonometric Fourier series expansion of f(t), assuming k = 3. Solution: ( ) ⎥⎦ ⎤ ⎢⎣ ⎡ − − + + − = ∫ ∫ π π π π π 2 / 2 / 0 2 ) cos( ) ( cos 1 dt t k dt t ao = ⎥⎦ ⎤ ⎢⎣ ⎡ − − − ∫ ∫ ∫ ∫ 2 / 0 / 2 2 / 2 / 0 cos cos 2 1 π π π π π π π tdt tdt dt k dt = ⎥⎦ ⎤ ⎢⎣ ⎡ − ∫ ∫ π π π π 2 / 2 / 0 2 1 dt k dt = ⎥⎦ ⎤ ⎢⎣ ⎡ π − π π 2 1 k = 2 −1 k . f(t) t π 1 π /2 -π -π/2 k -1 -2 an = ( ) ⎥⎦ ⎤ ⎢⎣ ⎡ − − + + − ∫ ∫ π π π π π 2 / 2 / 0 2 cos ) cos( )cos ( cos 2 ntdt t ntdt k t = ⎥⎦ ⎤ ⎢⎣ ⎡ − + − ∫ ∫ ∫ π π π π π 2 / 2 / 0 0 2cos cos cos cos 2 ntdt ntdt k ntdt t = ⎥⎦ ⎤ ⎢⎣ ⎡ − + + − − − ∫ ∫ ∫ ∫ π π π π π π 2 / 2 / 0 0 0 2 cos cos )1 cos( 2 1 )1 cos( 2 1 2 ntdt ntdt k tdt n tdt n = [ ] [ ] π π π π π π π / 2 / 2 0 0 sin 4 sin 2 1 )1 sin( 1 )1 1 sin( nt n nt n k n t n n t n − + ⎥⎦ ⎤ ⎢⎣ ⎡ + + + − − − = 4 sin 2 0 0 sin 2 2 0 0 π π π π n n n n k + − − + − = 2) sin 2 2( π π n n k + . an is zero for even values, and the 9 odd harmonics alternate in sign. Thus, ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + − + − + + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = ... 7 cos7 1 5 cos5 1 3 cos3 1 2) cos 2( 2 1 1 2 ( ) t t t t k k f t π Version 1: k = 3, ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + − + − + = ... 7 cos7 1 5 cos5 1 3 cos3 1 10 cos 2 1 ( ) t t t t f t π Version 2: k = 4, ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + − + − + = ... 7 cos7 1 5 cos5 1 3 cos3 1 12 cos 1 ( ) t t t t f t π Version 3: k = 5, ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + − + − + = ... 7 cos7 1 5 cos5 1 3 cos3 1 14 cos 2 3 ( ) t t t t f t π Version 4: k = 6, ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + − + − + = ... 7 cos7 1 5 cos5 1 3 cos3 1 16 cos 2 ( ) t t t t f t π Version 5: k = 7, ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + − + − + = ... 7 cos7 1 5 cos5 1 3 cos3 1 18 cos 2 5 ( ) t t t t f t π 11% _ + vSRC iX 1 H iSRC 1 F 1 Ω 17. Given vSRC = cost V and iSRC = sin2t A. 6% (a) Derive the expression for iX in the time domain. 5% (b) Determine the power dissipated in the resistor. Solution: (a) Let the amplitude of vSRC and iSRC be K. With the current source replaced by an open circuit, Ix1 = = + ∠ j K 1 1 1 0 o ) 2 (1 j K − ; iX1 = 45 ) cos( 2 o t − K A. With the current source replaced by a short circuit, Ix2 = = + ∠ 2 1 2 0 j j K o 2) tan (90 5 2 −1 − ∠ o K ; iX2 = 26.57 ) sin(2 5 2 o t + K A. Hence, iX = 45 ) cos( 2 o t − K + 26.57 ) sin(2 5 2 o t + K A. (b) Power dissipated is P = 2 2 2 0.65 5 4 2 2 1 K K K ⎟⎟ = ⎠ ⎞ ⎜⎜ ⎝ ⎛ + W. Version 1: K = 1, + A, P = 0.65 W 45 ) 0.71cos( o t − 26.57 ) 0.89sin(2 o + t Version 2: K = 2, + A, P = 2.60 W 45 ) 1.14cos( o t − 26.57 ) 1.79sin(2 o + t Version 3: K = 3, + A, P = 5.85 W 45 ) 2.12cos( o t − 26.57 ) 2.68sin(2 o + t Version 4: K = 4, + A, P = 10.4 W 45 ) 2.83cos( o t − 26.57 ) 3.58sin(2 o + t Version 5: K = 5, + A, P = 16.25 W 45 ) 3.54cos( o t − 26.57 ) 4.47sin(2 o + t 10 11% 18. Given the circuit shown, with a = 1. – + 5∠0° V -j Ω 1:a ISRC 3% (a) Determine the current in the capacitor 2% (b) Replace the capacitor by a current source, in accordance with the substitution theorem 3% (c) Rearrange the current source as two current sources across the transformer windings 3% (d) Determine ISRC. Solution: (a) The voltage across the capacitor is 5(1 + a) V. The current through the capacitor is j5(1 + a) A directed from primary to secondary. (b) (c) – + 5∠0° V j5(1 + a) A 1:a j5a(1 + a) A 1:a j5(1 + a) A j5(1 + a) A ISRC (d) It follows that ISRC = j5(a + 1)2 A. Version 1: a = 2, ISRC = j5(2 + 1)2 = j45 A Version 2: a = 3, ISRC = j5(3 + 1)2 = j80 A Version 3: a = 4, ISRC = j5(4 + 1)2 = j125 A Version 4: a = 5, ISRC = j5(5 + 1)2 = j180 A Version 5: a = 6, ISRC = j5(6 + 1)2 = j245 A 11% j5 Ω 5 Ω 19. Determine X and R for maximum power transfer to R and calculate this power. Assume VSRC = 4∠0° V rms. 1:2 – + 1:3 jX R VSRC 11 Solution: On open circuit, VTH = V = VSRC. On short circuit, V = 0 and IN = I = ) 5(1 j + VSRC . YN = ) 5(1 1 + j = 0.1(1 – j) S. For maximum power transfer, GL = 0.1 S, or R = 10 Ω, and B = 0.1 S, or X = -10 Ω. 5 Ω – + j5 Ω 1:2 1:3 VSRC – + V – V + – + 2V – + 3V – + VTh Under conditions of maximum power transfer, the current in R is 0.5|IN| = 10 | 0 5 | VSRC . and the power transferred is 10 | 0 25 | VSRC 2 . = 40 | | VSRC 2 W. 5 Ω j5 Ω – + 1:2 VSRC I 2I 1:3 3I – + V – V + + 2V – + 3V – I Version 1: VSRC = 4∠0° rms, P = 0.4 40 16 = W Version 2: VSRC = 5∠0° rms, P = 0.625 40 25 = W Version 3: VSRC = 6∠0° rms, P = 0.9 40 36 = W Version 4: VSRC = 7∠0° rms, P = 1.225 40 49 = W Version 5: VSRC = 8∠0° rms, P = 1.6 40 64 = W. 11% 200 turns 100 turns ZL _ + VSRC ISRC 20. Determine the complex power delivered by each source given that VSRC = 5cosωt, ISRC = -2sinωt, and assuming ZL = k(1 – j) where k = 1. Solution: The currents and voltages are as shown. Equating mmfs: 100×2∠90° + 200IL = 0, or j2 = -2IL, and IL =-j A, I1 = IL – j2 = -j3 A. VL = ZLIL = -jk(1 – j) = -k(1 + j) V. V2 = VL – 5 = -(k + 5) – jk. V1 = V2/2 = -(k + 5)/2 – jk/2 V. VI = 5 – V1 = (15 + k)/2 + jk/2 V. Power delivered by voltage source = Sv = ( ) 2 15 2 5 ( 3) 1 2 2 1 j j Vsrc I = = ∗ VA 12 Power delivered by current source SI = = ⎟ − ⎠ ⎞ ⎜ ⎝ ⎛ + + = ∗ 2) 2 ( 2 15 2 1 2 I 2 V SRC I j jk k ( )) (15 2 1 k j k + − VA I1 – + V1 – + VL IL 200 turns 100 turns ZL _ + – + V2 – + VI j2 A 5 V Total power delivered by sources = ( ) ( )j k jk j k j − = − − + 2 1 ) 15 15 2 1 As a check, SL = ( ) ( j k j j jk I V Lm Lm − = − − = ∗ 2 1 ) ( ) (1 2 1 2 2 ) VA. Version 1: k = 1, Sv = j7.5 VA; SI = 0.5 – j8 VA Version 2: k = 2, Sv = j7.5 VA; SI = 1 – j8.5 VA Version 3: k = 3, Sv = j7.5 VA; SI = 1.5 – j9 VA Version 4: k = 4, Sv = j7.5 VA; SI = 2 – j9.5 VA Version 5: k = 5, Sv = j7.5 VA; SI = 2.5 – j10 VA 13