EECE 210 Problems For Final -with Some Solutions, Fall 2012

I 8. The source current in the circuit shown is3cos(5OUOt) A. What impedance should be connected across the tamin& a,b for maximum average pow= transfer. f k m H a 1 0 - 2 0 j n 4 20-10jSL c. 10+20jQ d. lo-ZOjn 8. None of the above. 2 Problem 2 Find the value of the load impedance ZL such that maximum real power is delivered to it A) 3.0 +j1.0 Ω B) 1.75+ j0.25 Ω C) 1.5 - j0.5 Ω D) 3.0 + j1.1 Ω E) None of the above Problem 16 It is given that the load above has complex power S==20 + 15j KVA. It is required to connect an element Z in parallel with the load so as to correct the power factor to unity (power factor= 1). The source voltage is Vs=200∠50 V rms and the frequency is 60 Hz [that is Vs=200cos(377t+50o) V (rms)]. Determine the value and nature of this element Z. A) Capacitor with value C=994.7 μF B) Inductor with value L=7.07 mH C) Capacitor with value C=663.13 μF D) Inductor with value L=10.61 mH E) None of the above. 1/1 Consider the circuit shown Problem 8 ( 14 pts) a. Determine the value of the load L Z to maximize the average power absorbed by the load. (4 pts) 7 12 7) 4(8 5.78 j j j Zsrc − − + = = 3 + j 5.2 Ω; ZLm = 3 – j 5.2 Ω. b. For the value obtained in (a), determine the average power developed by the voltage source and the average power absorbed by the load. (4 pts) VTh = 1.74 9 7 12 7 12 8 j j j − = − − V; |VTh| = 9.18 V; IL = VTh /6 0.29 1.5 − j = A; V1 = (3 – j 0.58)IL = 4.68 +j 0 V ISRC = = − 4 4.68 12 1.83 +j 0 A PSRC = V1ISRC = 12×1.83 ≅ 22 W; 3 4 (9.18) 2 × PL = ≅ 7 W. c. For a purely resistive load, determine its value Rmax for maximum power transfer and find the power absorbed by the load. (3 pts) Rm = 2 2 (5.2) (3) + ≅ 6 Ω; |I| = |VTh|/|Z| = 2 2 (5.2) 6) 3 ( .18 9 + + = 0.883 A P = |I|2×6 = 4.68 W. d. For a purely resistive load with RL = 2Rmax ( Rmax of part c), determine the power absorbed by the load. (3 pts) |I| = |VTh|/|Z| = 2 2 (5.2) 12) 3 ( .18 9 + + = 0.578 A; P = |I|2×12 ≅ 4 W – + – + 4 Ω -j 7 Ω 8 Ω j5.78 Ω 12∠0° V rms ZL – + – + 4 Ω -j 7 Ω 8 Ω j5.78 Ω 12∠0° V rms -j 5.2 Ω 3 Ω V1 IL ISRC 17. If a capacitor with impedance Z2 is connected in parallel to a load Z1 = 300 +j450 Ω. What should be Z2 in ohms so that the equivalent load is purely resistive? a) -928.6 j b) -1112.5 j c) -650 j d) -750 j e) None of the above 18. What is the power factor of the equivalent load of the previous question? a) 0.8 b) 0.6 c) 0 d) 1 e) None of the above 23. Find the maximum power dissipated in R if: Is=2mA, R1=100kΩ, Vs=50V. a) P= 12.5 mW b) P= 1.25 mW c) P= 50 mW d) P= 56.25 mW e) None of the above 100 ohms 11. assuming L = 1 H. Determine the frequency at which maximum – + L 8 Ω 10 Ω 1 F 5∠0° V power is dissipated in the 10 Ω resistor, Solution: ω ω 1 = 1 Ω. Ma C ximum power is issipated in the 10 Ω resistor when XL = -XC, d which gives ωL = ω 1 , or L ω = 1 rad/s. Add 250 ohms 3 Problem 3 Consider the periodic signal Vs(t) shown below. Assume this signal is applied to the circuit in the figure below, find an expression for the voltage Vo(t) across the terminals a,b as shown. A) ∑ ∞ = + = 1 cos 2 8 ( ) 2 sin n o m m o t n V V t V n n ω π π B) ∑ ∞ = + = 1 0 cos 4 8 ( ) 4 sin n o m m t n V V t V n n ω π π C) ∑ ∞ = + = 1 4 cos 2 4 ( ) 2 sin n o m m t n V V t V n n ω π π D) ∑ ∞ = + = 1 4 cos 2 4 ( ) 2 sin n o m m t n V V t V n n ω π π E) None of the above TF = 1/50, G(t) = 1 rms A0=1.5, A5=1.15, B5=0 10 ohms Derive the trigonometric form of the FSE of the waveform shown Solution: The function is even, and a0 = C0 = 4 4 1 A T A T = × × ; ∫ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + − = 4 / 0 0 1 cos 4 4 T n tdt n T t T A a ω = 4 / 0 0 0 0 0 0 2 2 0 1 sin sin 4 cos 1 4 4 T t n n t n n t T t n T n T A ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − − − ω ω ω ω ω ω ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = ⎥⎦ ⎤ ⎢⎣ ⎡ − = cos 2 1 4 cos 2 1 16 2 2 2 0 2 2 π π π ω n n A n T n A . Hence, ⎜ ⎝ ⎛ + + + + = t t t A A f t 0 0 0 2 9 cos3 1 2 cos 2 1 cos 4 4 ( ) ω ω ω π ⎟ ⎠ + ...⎞ 25 cos5 1 ω0t . f (t) t A T T/2 -T/4 T/4 Problem 9 1. in the circuit shown, each source is 15cos10t V. The power dissipated in R is 50 W. If the frequency of one of the sources is doubled, the power dissipated in R is: – + 1 Ω 1 Ω – + 1 Ω R A. 100 W B. 50W C. 25 W D. 12.5 W E. None of the above. Solution: The current due to each source is 1.5 cos10t 15 2 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = 5cos10t A. The power is 12.5 2 5 2 = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ W. the power dissipated due to both sources is 25 W. 2. f2(t) is the function f1(t) lowered by 1 unit, as shown. If F1rms and F2rms are the rms values of f1(t) and f2(t), respectively, then: A. F1rms = F2rms B. F1rms > F2rms C. F1rms < F2rms Solution: The AC components of f1(t) and f2(t) are the same. The DC component of f1(t) is larger than that of f2(t). Hence, F1rms > F2rms. f1(t) t T f2(t) t T 3 2 5. The Fourier coefficients ak and bk for the periodic function shown are: t v V 10π 2.5π -2.5π -10π T 3T/4 T/4 T/2 A. ak = 0 for all k; bk = 0 for k odd and is non-zero for k even B. bk = 0 for all k; ak = 0 for k even and is non-zero for k odd C. bk = 0 for all k; ak = 0 for k = 0, ak = 0 for k odd and is non-zero for k even D. ak = 0 for all k; bk = 0 for k even and is non-zero for k odd E. None of the above. Solution: The function is odd, half-wave symmetric. Its average is zero; it contains no cosine terms, only odd sine terms. 1/1 11. The current through an inductor of 1 H is given by the periodic triangular wave. The amplitude of the fundamental component of the voltage across the inductor is: t, s -Ip i, A Ip T =1/2π A. 4Ip B. 8Ip C. 16Ip D. 32Ip E. None of the above. Solution: p p I I dt L di v π π 8 / 4 1 2 1 = = × = , which is the amplitude of the square waveform representing v. the amplitude of the fundamental is π π p p I I 32 8 4 = × . 15. The voltage and current at the terminals of a circuit are: t V t v 100 sin1500 400 cos500 15 + + = + – v i 15 ) 3cos(1500 60 ) 5 sin(500 2 o o − + + + = t t i A 3% a) Calculate the average power delivered to the circuit. ∑ = − + = 3 1 ) cos( 2 n in vn m m dc dc V I V I P θ θ = 15×2 + cos( 75 ) 3 100 2 1 5cos(30 ) 400 2 1 o o − × × × + × × = 934.85 W 3% b) Calculate the rms value of v. 291.93 2 (100) 2 (400) 15) ( 2 2 2 = + + Vrms = V 2% c) Calculate the rms value of i. 4.58 2 (3) 2 (5) 2) ( 2 2 2 = + + Irms = A 2/2 12. For n = 1, 2, 3,…, the function shown has: A. an and bn nonzero for all n B. an and bn are zero for even n C. an and bn are zero for odd n D. an = 0 for all n E. bn = 0 for all n Solution: When the dc value is removed, the ac component has half-wave symmetry but is neither even nor odd. Hence, an and bn are zero for even n. f(t) t T -T/2 T/2 A A 13. Determine the total power dissipated if vI is a full-wave rectified waveform given by: vI = 6|sin(500t)| V. Solution: 5 10 2 100 500 1 1 5 << × = × = − ωC ohms; π m dc V V = 2 ; 2 2 2 0.04053 10 4 m m dc V V P = = π ; 2 2 2 2 2 2 2 2 0.09472 4 2 1 ; 4 2 m m ac ac m m V V V V V V ⎟ = ⎠ ⎞ ⎜ ⎝ ⎛ − = + = π π ; 5 0.09472 2 m ac V P = ; . 0.01894 2 m V 0.05947 2 Vm P = 100 F 5 Ω 5 Ω – + vI 14. A period of a periodic function f(t) is given by: K(4 + 2sint), 0 < t < 2π. Determine the rms value of f(t), if K = 0.5. Solution: The square of f(t) is ( ) = + + t t K 2 2 4sin 16sin 16 ( t t K 2cos2 16sin 2 2 16 − + + ( 2 0 2 2cos 16 sin 2 16 t K π − + + ∫ ) ) . The area under the square is 2 ; the mean square is 36 2 K t dt π = 2 2 18 2 36 K K = π π and the rms value is 3 2K . f(t) t π 4 2 2π 1/2 16. Derive the trigonometric Fourier expansion of the given periodic function f(t). 1 -1 2 -2 1 2 -1 -2 t f(t) Solution: Since f(t) is odd, a0 = 0 = an; T = 2, ω0 = 2π/T = π; f(t) = t + 1; ∫ = 2 / 0 0 ) ( )sin( 4 T n t dt n f t T b ω = = ⎥⎦ ⎤ ⎢⎣ ⎡∫ 2 / 0 0 ( ) 2Im T jn t dt f t e ω = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + − 1 0 0 2 0 0 0 0 0 ) ( 2Im ω ω ω ω ω ω jn e jn e jn te t jn t jn t jn = ⎥⎦ ⎤ − + − π π jn n 1 1 0 2 2 ⎢⎣ ⎡ + + π π π π π π jn e n e jn e jn jn jn ) ( 2Im 2 ⎥⎦ ⎤ ⎢⎣ ⎡ + − π π π n n n 1 2cos 2 = ( )⎥⎦ ⎤ ⎢⎣ ⎡ − π π n n 2cos 1 1 2 = ( )⎥⎦ ⎤ ⎢⎣ ⎡ − + )1 +1 2( 1 1 2 n n π f(t) = ⎥⎦ ⎤ ⎢⎣ ⎡ + − + − ... 4 sin 4 sin3 2 sin2 2 3 sin t t t t π π π π π . 2/2 Problem 18 The device D in the following circuit is powered by a voltage v(t) = 2 + 2 cos(1000t) + cos(2000t)(V ). The current across is given by i(t) = 1 + sin(1000t) + 0.5 sin(2000t)(A) find the average power associated with D. − + v(t) D i(t) A) 2W B) -2W C) 4W D) -4W E) None of the above Problem 19 The following circuit is powered by a periodic voltage source that has the following Fourier expansion: v(t) = 21 + 20 cos(1000t) + 10 cos(2000t). Find the RMS value of the current i(t). − + v(t) 3Ω 4mH i(t) A) 6.68A B) 8.14A C) 7.3A D) 7.6A E) None of the above Problem 20 For the function v(t) given below, find the value of a5. You may use the following: � t cos(αt) = cos(αt) α2 + t sin(αt) α . 0 1 2 3 4 5 0 1 2 t(s) v(t) A) 0 B) -0.0162 C) 1.27 D) -16.7 E) None of the above The following given is used in the next 5 problems. 3 electrical elements are powered by a 240VRMS, 60Hz source: I L1 I1 L2 I2 L3 I3 - + 240VRMS∠0o The following is given for the three elements: L1: 240W, PF=0.6 Lag L2: 200VARS, PF=0.5 Lag L3: 100VA, PF=0 Lead Problem 11 Find the total apparent power. A) 725.67VA B) 626.33VA C) 550.2VA D) 888.8VA E) None of the above Problem 12 Find the total power factor. A) 0.567 Lag B) 0.646 Lag C) 0.808 Lag D) 0.747 Lag E) None of the above Problem 13 Find the magnitude of the total current I. A) 5.21A B) 3.02A C) 2.292A D) 7.40A E) None of the above Problem 14 Find the capacitor that needs to be placed in parallel with the loads to adjust the power factor to 0.9 Lag. A) 62.7µF B) 147µF C) 49µF D) 11.4µF E) None of the above Problem 15 Find the magnitude of I again after the power factor is adjusted as in the previous problem. A) 1.64A B) 3.29A C) 0.91A D) 2.13A E) None of the above C=2.74 mF Problem 10 Find X such that the maximum power transfer constraint is satisfied. 6jΩ −2jΩ 5jΩ − + 5j(V ) −4jΩ 6jΩ 2jΩ X 1Ω (5 + 5j)(A) A) 1 − 2.5jΩ B) 1 + 2.5jΩ C) 2 − 2.5jΩ D) 2 + 2.5jΩ E) None of the above Problem 8 Find N2 and X such that maximum power is delivered to the 1000Ω resistor. − + 5V 10Ω 0.16jΩ jX 1000Ω 5 : N2 A) N2 = 50, X = −16 B) N2 = 10, X = −16 C) N2 = 10, X = −0.64 D) N2 = 2, X = −0.64 E) None of the above Problem 3 Find R that satisfies the maximum power transfer constraint. − + 100V − + 100V 10Ω 20Ω 20Ω 10Ω 20Ω 3A R A) 14Ω B) 15Ω C) 16.33Ω D) 17.46Ω E) None of the above Problem 2 Find the average power associated with the 6A current source between A and B. 10jΩ 3A∠0o B 2Ω A 2jΩ 6A∠0o 5Ω A) -23.86W B) 23.86W C) 28.71W D) -28.71W E) None of the above delivered assume rms values 7/7 8. Determine Z so that maximum power is transferred to it and calculate this power given that the source voltage is 10 V peak value. s = ancel peak va h re as shown. F KVL: 10∠0° – 5(I + I/2) – 5(I – I/2) = 0. Again, the terms 1 Solution: We will determine TEC as seen by Z. On open circuit, the currents are a shown. From KVL: 10∠0° – 5I/2 + 5I/2 VTh. In This particular problem, the voltages across the 5 Ω resistors c out. Hence, VTh = 10∠0° V When Z is replaced by a s circuit, the currents a sc sc involving I cancel out. Hence, Isc = 1∠0° A, and ZTh = lue ort rom 10 1∠0 o 10 0 = ∠ o Ω. It follows that for aximum power transfer, Z = 10 . The power dissipated in the load is m Ω 1.25 10 4 1 2 2 = × ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ VTh W. – + 10∠0° V 1:2 5 Ω j2 Ω 5 Ω Z – + 10∠0° V 5 Ω j2 Ω 1:2 5 Ω – + VTh I I/2 I/2 I/2 – + 10∠0° V j2 Ω 1:2 5 Ω 5 Ω I I/2 ISC – I/2 Isc Isc + I/2 – + 5∠0° V R – + 2∠0° V 1:2 1:2 RL 10 20 R Rx = , or 2 Rx = R . 7. Determine the maximum power that can be delivered to RL, assuming R = 0.5 Ω. Solution: The primary voltage of the upper transformer is always 1 V. On 3/5 open circuit, the source current is zero, the primary voltage is 5 – 1 = 4 V, and VTh = 8 V. On short circuit, the primary voltage of the lower transformer is zero, the source current is (5 – 1)/R and the short circuit current is 2/R. This gives, RTh = 4R. The maximum power delivered is (8)2/(4×4R) = 4/R. -j30 Ω L RLine Supply voltage 8. Given that the load L consumes 1200 W at 0.8 p.f. lagging and the magnitude of the voltage across L is 300 V rms. Determine the power dissipated in the resistance Rline, if Rline = 0.5 Ω. Solution: The reactive power absorbed by the load is 900 0.6 0.8 = 1200 × VAR. The reactive power absorbed by the capacitor is 3000 30 2 = − − V VAR. The total complex power is 1200 + j(900 – 3000) = 1200 – j2100 VA. The magnitude of the line current 65 300 (2100) (1200) 2 2 = + A. The power dissipated in Rline is 65Rline. 4. Determine the reactive power absorbed in the circuit, given that I = 1∠0° A rms. Solution: The equivalent series impedance is 5 5 5 5 5 5) 5(5 j j j j j + = − + − . The reactive power is 5|I|2 VAR. As a check, the current in the capacitive branch is 5 5 5 5 j j j − + I = jI; the reactive power absorbed by the capacitor is -5|jI|2 = -5|I|2 VAR. The current in the inductive branch is j5 Ω -j5 Ω 5 Ω I 5 5 5 5 5 j j j − + − I = (1 – j)I = 2 ∠-45°I; the reactive power absorbed by the inductor is 5| 2 assuming VSRC = 200∠0° V rms. I|2 = 10|I|2 VAR. The total reactive power absorbed is 10|I|2 – 5|I|2 = 5|I|2 VAR. 5. In the circuit shown, L1 consumes 160 W at 0.8 p.f. lagging and L2 consumes 320 VAR at 0.6 p.f. lagging. Determine I when XC is chosen for unity power factor, – XC VSRC + L1 L2 I 1/5 So nce seen by the source is zero and the source applies lution: Ay unity p.f. the total reacta only real power. The real power consumed by L2 is 240 0.6 320 = × W. The total real .8 0 power supplied by the load is 160 + 240 = 400 W. The current is o o 0 0 ∠ = ∠ SRC SRC V V A rms. 400 400 At 9 Problem 14 It is given that the complex power of L1 is 5+j10 VA. It is also given that L2 absorbs 20W at lagging power factor of 0.8. What is the phase difference between I and V as shown in figure? A) 45.00o B) 39.81o C) 63.33 o D) 60.00 o E) None of the above. Problem 15 What is the impedance of a load if it absorbs 20KVAR at lagging power factor of 0.6 when a current of magnitude 50 A rms flows through it? A) 6 + 8j ohms B) 3 + 4j ohms C) 4+ j3 ohms D) 8+ j6 ohms E) None of the above. – + 5Ω 100∠20 Vrms L1 L2 + V _ I 4. How mu& complex power is delivered by the 5630' A (rms) currd source to fhe circuit shown in figure. a 7.5~137.48' VA. b. 0 VA. C. 100 VA d. 15.35~137.48~ VA. e- None of the above. and L are unknown Load 1 absorbs a complex power absorbs a complex power of 1 0 0 ~ 4 5 ~ VA. Determine Rz rms.