EECE 210 Problems For Quiz 3 with Solution, Fall 2011

1/1 Problem (12 pts) Consider the circuit shown a. Determine the open-circuit voltage at terminals a and b. (3 pts) 15 = 1.5I; I = 10 A; VTh = 10(2 – j 4) – 5(4 – j 3) = -j 55 V. b. Determine the current flowing in the short circuit when there is a short between terminals a and b. (3 pts) 15 = 1.5I + ISC; I(2 – j 4) = (0.5I + ISC)(4 + j 3); solving these two equations gives ISC = 6) 74 (1 165  j A c. Determine the equivalent impedance ZTh as seen by the terminals a and b. (2 pts) ZTh = |VTh|/|ISC| = 3 2 4 6) 165(1 74 55 j j j      d. Evaluate ZTh again using a different method then that employed in part (c). (4 pts) IT = 1.5I; VT = I(6 – j ) dividing gives: ZTh = 3 2 4 1.5 6 j j    . 150 A rms 2  -j 4  4  j 3  a b I 0.5I 150 A rms 2  -j 4  4  j 3  a b I 0.5I ISC 6  -j  I 0.5I IT + – VT + - Hint: redraw lattice circuit as a bridge Hint: determine current in (5 - j5) ohms, assuming autotransformer is ideal Assume dot markings are both up 1. Find the correct node-equation for the voltage V1. a) 0 8 6 1 = − + b a V V V b) 0 4 2 1 = − + b a V V V c) 0 1 = + − b a V V V d) 0 2 3 1 = + − b a V V V e) 0 4 7 1 = + − b a V V V 4. Find the Thevenin equivalent circuit with respect to the terminals a-b. What are the values of VTh in V and ZTh in Ω? a) VTh =-10 V, ZTh = j10/3 Ω b) VTh =-8 V, ZTh = j3 Ω c) VTh =-6 V, ZTh = j14/5 Ω d) VTh =-4 V, ZTh = j8/3 Ω e) VTh =-2 V, ZTh = j5/2 Ω 5. What is the expression for Vx? a) (Z1+Z2) b) 5 Z1 c) Z1/4 d) 2 Z1 e) Z1/2 6. What is the correct set of equations for the mesh currents Ia and Ib? a) 0 5 ,0 ) 4 (4 ) 4 ( 3 2 2 1 = − = + − + − a b a I Z Z I Z Z I b) 0 2 ,0 ) 2 (2 ) 2 ( 3 2 2 1 = − = + − + − a b a I Z Z I Z Z I c) 0 1 ,0 ) ( ) ( 3 2 2 1 − = = + − + − a b a I Z Z I Z Z I d) 0 1/ 2 ,0 ) ( ) 2 ( 3 2 2 1 = − = + − + − a b a I Z Z I Z Z I e) 0 1/ 4 ,0 ) ( ) 4 ( 3 2 2 1 = − = + − + − a b a I Z Z I Z Z I 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 22. Assuming that the voltage V across inductance L1 is as shown in figure below and that the mutual inductance between ‐ L1 and L2 is M12 ‐ L1 and L3 is M13 ‐ L2 and L3 is M23 Use the mesh technique to find the expression of the voltage V. a) V=- L1dI1/dt - M12dI1/dt + M13dI2/dt b) V=- L1dI1/dt + M12dI1/dt + M13dI2/dt c) V=- L1dI1/dt + M12dI1/dt - M13dI2/dt d) V=- L1dI1/dt + M12dI1/dt + M13dI2/dt e) None of the above 1 Problem 1 Find Zth across a and b A) Zth = 3.85- j 0.77  B) Zth = 1.65- j 5.50  C) Zth = 5.29- j 8.82  D) Zth = 6.50- j 1.65  E) None of the above Problem 2 What are the impedances in this circuit if v(t)=20 cos(10t+50o) Volts and i(t)=50cos(10t+20o) Amperes. A) 10 j j  B) 10 j1.0j C) 10 j j  D) 10 j j  E) None of the above 2 Problem 3 Find the node equations for the following circuit A)       0 2.0 .0 133 2.0 1.0 0 2.0 8.3 .0 05 05 .0 0 .6 43 .7 66 1.0 .0 05 5.0 15 .0 3 2 1 3 2 1 3 2 1                 V j V j V V j V j V j V V V j B)       0 2.0 .0 0833 2.0 05 .0 0 2.0 8.3 1.0 1.0 0 .6 43 .7 66 .0 05 1.0 5.0 15 .0 3 2 1 3 2 1 3 2 1                 V j V j V V j V j V j V V V j C)       0 2.0 .0 133 2.0 1.0 0 2.0 8.3 .0 05 05 .0 0 15.32 12.85 1.0 .0 05 2.0 15 .0 3 2 1 3 2 1 3 2 1                 V j V j V V j V j V j V V V j D)       0 2.0 .0 0833 2.0 05 .0 0 2.0 8.3 1.0 1.0 0 15.32 12.85 .0 05 1.0 2.0 15 .0 3 2 1 3 2 1 3 2 1                 V j V j V V j V j V j V V V j E) None of the above 3 Problem 4 Find V0 (t) given =120 rad/sec. A) V0 = -0.99 cos( 120t +94.29. o ) Volts B) V0 = -1.99 cos( 120t +194.29o ) Volts C) V0 = -1.99 cos( 120t -25.7 o ) Volts D) V0 = -0.99 cos( 120t -115.71 o ) Volts E) None of the above Problem 5 Find io in the circuit above. A) .8 178 104.620  B) 23.14 89.620  C) 16.36 104.620  D) 11.57 89.620  E) None of the above 4 Problem 6 Given the circuit above, what are the two mesh equations? A)     0 50 50 190 ;0 190 280 100 40 120 2 1 2 1 0           i j i j i i j B)      0 50 50 250 ;0 250 400 100 40 120 2 1 2 1 0           i j i j i i j C)      0 50 50 150 ;0 150 200 100 40 120 2 1 2 1 0           i j i j i i j D)      0 50 50 210 ;0 210 320 100 40 120 2 1 2 1 0           i j i j i i j E) None of the above Problem 7 In the circuit shown above, what is the value of the reflected impedance of the 50 ohms resistor from the primary to the secondary side? A)  B)  C)  D)  E) None of the above 5 Problem 8 In the circuit shown above, find the Thevenin voltage across a,b A)  V 39.800 76.82   B)  V 55 0. 0 57 3.   C)  V 63 0. 0 28 6.   D)  V 48 3. 0 65.99   E) None of the above Problem 9 For the same circuit of previous problem, find Zth across a,b A)  Zth= 4.9- j4.1  B)  Zth= 1.08- j2.12  C)  Zth= 4.7- j6.71  D)  Zth= 3.7- j5.1  E) None of the above 6 Problem 10 Find the Thevenin equivalent resistance and capacitance/inductance with respect to the terminals a,b in the circuit shown above A)  R = L B)  R = L C)  R = L D)  R = L E) None of the above Problem 11 In the circuit shown above, find the value of the impedance Z if . A j V j V j g g 30 20 , and I 50 100 , V 30 40 V1       A) j  B) 58+j14  C) 68+j24  D) 5+j20  E) None of the above 7 Problem 12 Find the input impedance Zi at the terminals a,b in the circuit shown above A)       C j jL Zi 25 1 40 B)       C j jL Zi 40 1 25 C)       C j jL Zi 40 1 15 D)       C j jL Z i 15 1 40  E) None of the above Problem 13 In the circuit of the previous problem, find the frequency such that the input impedance Zi is purely resistive. A) rad/s 1 1000 40 1   L C C  B) rad/s 1 1000 25 1   L C C  C) rad/s 1 600 40 1   L C C  D) rad/s 1 600 15 1   L C C   E) None of the above 8 Problem 14 In the circuit shown above, it is given that R=1 , and Vs= 0 10 volts. Find the current i1 as indicated. A)  53.130 8   A B) 53.130 .7 07   A C) 450 .7 07   A D) 450 8   A E) None of the above 4 Problem 5 Find the magnitude of the Thevenin Voltage Vth across terminals a,b in the circuit above. Given A=1/4. A) 15.0 V B) 37.5 V C) 14.29 V D) 7.15 V E) None of the above 5 Problem 7 In the circuit shown above, given V 90 48 0 ∠ Vs = , Ω + = 4 3 1 j Z , Ω − = 6 8 2 j Z , Ω − = 4 3 3 j Z and Ω + = 6 8 4 j Z . The Thevenin equivalent circuit values for the voltage souce and the internal impedance across terminals e and f are: A) V 0 14 ∠ 0 , Ω − 5.3 5.3 j B) V 0 50 ∠ 0 , Ω + 5.2 5.2 j , C) V 0 14 ∠ 0 , .7 29 Ω D) V 0 50 ∠ 0 , 10.42 Ω E) None of the above Problem 8 In the circuit shown above, the phasor form of the current I in amperes is: A) V 45 0. 0 .1 833 ∠ − B) V 45 0. 0 .0 917 ∠ − C) V 53.130 5.1 ∠ − D) V 53.130 0.3 ∠ − E) None of the above I 8 Problem 12 Find V0 if the source voltage is Vs = 20∠ 60 o Volts. A) 14.14 ∠ 15o V B) 7.07∠ 15 o V C) 20∠ 60 o V D) 10 ∠ 60 o V E) None of the above Problem 13 Find V0. A) 400 ∠ 0 V B) - 400 ∠ 0 V C) 100 ∠ 0 V D) - 100 ∠ 0 V E) None of the above + V0 + – – Vs -5j 5j 50 -2j 2j 5 10j 2 2j – + 5:10 2:1 + V0 – 10Ω 10Ω 50Ω 20∠50 V 10∠0 A 3. ~ ~ T h e v ~ ~ c e t o t h . e l & o f t h e ~ T l - T 2 o f ~ ~ in figure- a. 15~50' Q t b. 5 + 5 j 6 2 c. I.O8G+ 2.5351 fi I d. 9.69~51.07~ SZ e. None of the above. a 1.60cos(2f) V. b. l.GOsin(t) V. c. 3.2ms(f) V. d. 2.26cos(t) V. ' Nom of the above. 5. Find the vohage v, (t) across the - - - - ,r' . - - - - e. None of the above. 10. In problem 9, find the average power delivered to the 375 a resistor. a. 99.2 W. b. 50.3 W. c. 49.2 W. d. 62.7 W. e. None of the above. 11. Calculate the reading of the two wattmeters fW,+ WZ) in the circuit shown in figure. The value of Z, = 60L30° a. c. -33.2 kW. d. -15.4 kW. e. None ofthe above. 6 - - - - t Find tbe steady-state expression for v,(r) in the circuit shown if a 4cos(SOOQ) V . b. Scos(5000t) V . c. 16cos(5000t) V . d. 12cos(S000t) V . e. None of the above. 22.3cos(5,000t + 29.7) 1/2 Consider the circuit shown Problem 8 ( 14 pts) a. For ZL = 3 – j 5.2 Ω ,determine the average power developed by the voltage source and the average power absorbed by the load. (4 pts) – + – + 4 Ω -j 7 Ω 8 Ω j5.78 Ω 12∠0° V rms ZL 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. – + – + 4 Ω -j 7 Ω 8 Ω j5.78 Ω 12∠0° V rms -j 5.2 Ω 3 Ω V1 IL ISRC 1. If M = 5 mH, determine the ratio v1/v. M 40 mH 60 mH i _ + v1 _ + v Solution: Leq = L1 + L2 – 2M; dt di L v eq = ; dt M di dt L di v − = 1 1 ; hence, = − + − = M L L M L v v 2 2 1 1 1 M M 2 100 60 − − 2. Determine VO, given that I = 1∠0° A and ω = 10 rad/s. 1 H k = 0.5 I VO _ + 1 H Solution: 0.5 1 2 = = k L L M H; secondary voltage is jωMI, with the dotted terminal positive with respect to the undotted terminal. Hence, VO = -jωMI = -j10×0.5I = -j5I. 3. The lamp glows brighter when the dots are at coil terminals 200 turns 100 turns 1 1′ 2 2′ 12 V ac Solution: The lamp glows brighter when the voltage across it is largest. This occurs when the voltages across the windings are additive, that is, when the dots are at terminals 1 and 2 or 1′ and 2′. . Derive the time-domain expression for vC, 6 _ + vC 5ix 3 Ω _ + 2 mH 100 μF 2 mH vSRC ix given that vSRC = 10sin(2,000t) V. Solution: ωL = 2×103×2×10-3 = 4 Ω; 5 10 100 10 2 6 3 × × × − ωC SRC = 10∠0°. The node-voltage method can be a x – V 5(10 – VC)/j4 at . Derive VTh and ZTh as s . Ω; ωL2 1 1 = = Ω; V pplied, the circuit being as shown. At the middle node: VC/-j5 + (VC – Vx)/j4 + (VC – 10)/j4 = 0 5Ix 3 Ω _ + 10∠0° Ix -j5 Ω j4 Ω j4 Ω VC Vx At the right-hand node: (V C)/j4 + (Vx – 10)/3 = 5Ix = Solving, VC = 11.98 + j1.44 = 12.1∠6.86°, so th vC = 12.1sin(2,000t + 6.86°) V. – + vSRC 5 Ω 50 μF 20 mH 30 mH 20 mH a b 1:2 7 seen between terminal ab, given that vSRC = 10cos(1,000t + 45°) V Solution: ωL1 = 103×30×10-3 = 30 = ωM =103×20×10-3 = 20 Ω; 6 3 10 50 10 − × × C ω = 20 Ω; VSRC = 10∠45°. will be as shown, where ω(L1 – M 0 Ω and is omitted. 1 1 = The circuit in the frequency domain The j20 Ω in parallel with -j20 Ω is effectively an open circuit. The current in the (5 + j10) Ω impedance is zero, Vcd = 10∠45°, and Vab = VTh = 20∠45°. ) = 10 Ω; ω(L2 – M) = – + 5 Ω a b 1:2 j20 Ω -j20 Ω c d 10∠45° j10 Ω 2/5 If the independent voltage source is replaced by a short circuit, the impedance on the primary side is (5 + j10) Ω and ZTh = 4(5 + j10) = 20 + j40 Ω. 3. Determine the impedance seen by the source, assuming a = 2. Solution: Reflection of the (5 – j5) Ω through the RH transformer gives (20 – j20) Ω. The impedance on the secondary side of the LH transformer is (25 – j10) Ω. Reflected to the primary side, this becomes (25 – j10)/a2 Ω. – + 4∠0° V 5 Ω j10 Ω 2:1 5 Ω -j5 Ω 1:a 4. If vsrc = 10cos(1,000t) V, determine the energy stored in the circuit in the sinusoidal steady state at t = 0, assuming C = 1 μF. 1H 1H k = 0.5 – + vsrc C Solution: At t = 0, the voltage across C is 10 V and the current through the inductors is zero, being proportional to the integral of vsrc. The energy stored is W = 2 = 2 1Cv 50C. – + 10∠0° V 8 Ω 6 Ω -j5 Ω – + 30∠0° V -j10 Ω R Rx I 5. Determine Rx given that I = 0 and R = 2 Ω. Solution: Since I = 0, the voltage across Rx is 10 V, and the same current R o o 10 0 30 0 ∠ − ∠ flows through R and Rx. It follows that – + 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 9. Two identical coils, each having an inductance of 10 mH, are connected in series. When the connections to one of the coils are reversed, the total inductance is multiplied by a factor a. Determine the coupling coefficient of the coils. Solution: (10 + 10 + 2M) = a(10 + 10 – 2M); 2M(a + 1) = 20(a – 1); – + 10∠0° V R 2 Ω – + 400 turns 100 turns 200 turns 4∠0° V Ix 1 )1 10( + − = a a M ; 1 )1 ( 10 + − = = a a M k 10. Determine Ix, assuming R = 4 Ω. Solution: The voltage across all windings is zero. Hence, I1 = R 10 A, and I2 = 2 2 4 = A. Setting the net mmf to zero, 400I1 – 100I2 + 200I3 = 0, or – + 10∠0° V R 2 Ω – + 400 turns 100 turns 200 turns 4∠0° V Ix I1 I2 I3 2 2 10 4 + − × R I3 = 0, which gives I3 = R 1− 20 ; IX = I2 – I3 = R 1+ 20 . 4/5 6. Determine iO, given that VSRC1 is 15 V dc and VSRC2 = 10cos(3,000t) V. 1 – + 2Vx 2 Ω 5 Ω 5 Ω – + Vx VSRC2 – + 2 Ω 10 mH 50 μF V + SRC1 – iO Solution: With VS d RC1 applied an VSRC2 set to zero, the circuit ecomes as shown. 15 = 3Vx, so Vx = 5 V and – + 2Vx 2 Ω 5 Ω – + Vx 5 Ω 2 Ω 15 V – + IO1 b 1 5 1 = − − = x O V I that A. With VSRC2 applied and VSRC1 comes as shown. It follows that: -2V = Vx + set to zero, the circuit be x – + C jX 2Vx , or 0 2 3 = ⎟⎠ ⎜⎝ C x jX ⎟ ⎜⎛ + V , which ives Vx = 0. Hence, IO2 ⎞ g o o 2 0 5 10 0 ∠ = ∠ = A. Thus, iO = -1 + 2cos(3,000t) A. 17. Determine k so that the input resistance is purely resistive. transformer by actance is Solution: Disregarding the 10 Ω resistance and replacing the linear its T- equivalent circuit, the circuit becomes as shown. The input re ( ) 0 20 20 10 = + − + j jX j jX jX j M M M , or 0 20 10 2 = − − + M M M X X X , which gives 200 = 10 2 XM = . Hence, = = = 2 1 400 10 2 k 0.71. 2Vx IO2 2 Ω 5 Ω 5 Ω – + Vx 10∠0° V – + jXL 2 Ω jXC 10 Ω k j40 Ω j10 Ω -j20 Ω Zin j(40 + XM) Ω j(10 + XM) Ω -j20 Ω -jXM Ω 6/7 Problem 1 Given 3 elements R = 10KΩ, L = 10mH and C = 625nF powered by a source v = 90sin(10, 000t + π 4) (V). Find the impedance of each element ZR, ZL and ZC. A) ZR = 10KΩ, ZL = 100jΩ, ZC = −160jΩ B) ZR = 10KΩ, ZL = 10jΩ, ZC = −16jΩ C) ZR = 10jKΩ, ZL = 10jΩ, ZC = −1600jΩ D) ZR = 10KΩ, ZL = 10jΩ, ZC = −160jΩ E) None of the above Problem 2 Find the Thevenin equivalent voltage between A and C. (Impedances are in Ω) C −5j −2j −3j 10j 20j B A 50j 5j(A) 10 + 10j(A) 10 + 10j(A) 35Ω A) 285-190j V B) -741+494j V C) -741-494j V D) 285+190j V E) None of the above Problem 3 Find the Thevenin equivalent voltage between A and B. − + 10j(V ) 10Ω i1 −5jΩ 20Ω 2i1 B A A) 39.7V ∠21.6o B) 18.6V ∠7.1o C) 18.6V ∠ − 7.1o D) 39.7V ∠ − 21.6o E) None of the above Problem 4 Find the nature of Z such that the Thevenin equivalent impedance between A and B is 1Ω. 10A jΩ Z 2jΩ A B A) 0.8 − 1.4jΩ B) 0.8 + 1.4jΩ C) 0.5 − 2.5jΩ D) 0.5 + 2.5jΩ E) None of the above Problem 5 Find the Thevenin voltage between A and B. 10jΩ 10Ω 5jΩ 5j(A) A B A) 50+25j V B) -100+50j V C) 100+50j V D) -50+25j V E) None of the above Problem 6 Find the Thevenin impedance between A and B. − + 50V ∠0o 2Ω −8jΩ 4jΩ −6jΩ 10Ω A B A) 1.49-0.55j Ω B) 0.96+3.21j Ω C) 0.96-3.21j Ω D) 1.49+0.55j Ω E) None of the above Problem 7 All the inductors are equal to 5jΩ, all the capacitors are equal to −6jΩ, all the resistances are equal to 10Ω. Find ZAB. A B A) 30.56+16.98j Ω B) 30.56-16.98j Ω C) 27-9j Ω D) 27+9j Ω E) None of the above Problem 8 Find i1. 6mA∠0o 1kΩ + v1 − −+ 10V ∠0 0.1v1 4kΩ 6000jΩ i1 A) 0.76-1.14j mA B) -0.26+1.6j mA C) 0.26-1.6j mA D) -0.76+1.14j mA E) None of the above A Problem 9 Find i1. jΩ 6A∠45o −2jΩ 2Ω i1 2A∠0o A) 3.38∠ − 29.2o (A) B) 7.55∠ − 82.4o (A) C) 7.55∠82.4o (A) D) 3.38∠29.2o (A) E) None of the above Problem 10 Write the two mesh current equation for I1 and I2 Don’t solve. (Impedances are in Ω). − + 50V ∠0o 40j 80j 60j 100Ω 20j 10j 30j I1 I2 A) 100jI1 + 60jI2 = 50 60jI1 + (100 + 80j)I2 = 0 B) 120jI1 − 80jI2 = 50 −80jI1 + (100 + 80j)I2 = 0 C) 100jI1 − 80jI2 = 50 −80jI1 + (100 + 80j)I2 = 0 D) 100jI1 − 60jI2 = 50 −60jI1 + (100 + 80j)I2 = 0 E) None of the above Problem 11 If Ig = 20cos(10, 000t + π 3 )(A) find the energy associated with the 2 coils at the time t = 100πµs . Ig 5Ω 1mH 1mH 15Ω Il 0.5mH A) 65.3mJ B) 261.3mJ C) 40.7mJ D) 163mJ E) None of the above Problem 12 Find the Thevenin equivalent impedance between A and B. (Impedances are in Ω.) A) 10j Ω B) 8j Ω C) 7.5j Ω D) 12j Ω E) None of the above 10j 10Ω 20j 20j −10j D 10j 10j C 10j B 20j A −20j −10j −10j −+ 10V ∠45 Problem 1 Find the Thevenin equivalent voltage between A and B (VAB). − + 2V0 20jΩ −3jΩ 5∠0 3jΩ + V0 − A B A) 15j V B) -20j V C) 20j V D) -15j V E) None of the above Problem 9 Given ZL = 100 + 100j, N2=90, N1=10, find V. 5A + V − N1 N2 ZL A) 7.07V ∠ − 135o B) 63.64V ∠45o C) 63.64V ∠ − 135o D) 7.07V ∠45o E) None of the above Problem 16 Consider the linear transformer of the figure below, given ω = 1rad/s, find the coupling coefficient K, such that the Thevenin impedance between A and B is purely resistive. 10Ω 10jΩ B 10Ω −20jΩ 40jΩ K A A) 0.79 B) 0.82 C) 0.85 D) 0.88 E) None of the above Problem 17 Find I. − + 10j(V ) 2jΩ 2jΩ 4Ω 2jΩ −2jΩ I 2Ω j A) 0.0389 - 0.6226j A B) -0.0778 + 1.2452j A C) -0.0389 + 0.6226j A D) 0.0778 - 1.2452j A E) None of the above . Derive the time-domain expression for vC, 6 _ + vC 5ix 3 Ω _ + 2 mH 100 μF 2 mH vSRC ix given that vSRC = 10sin(2,000t) V. Solution: ωL = 2×103×2×10-3 = 4 Ω; 5 10 100 10 2 6 3 × × × − ωC SRC = 10∠0°. The node-voltage method can be a x – V 5(10 – VC)/j4 at . Derive VTh and ZTh as s . Ω; ωL2 1 1 = = Ω; V pplied, the circuit being as shown. At the middle node: VC/-j5 + (VC – Vx)/j4 + (VC – 10)/j4 = 0 5Ix 3 Ω _ + 10∠0° Ix -j5 Ω j4 Ω j4 Ω VC Vx At the right-hand node: (V C)/j4 + (Vx – 10)/3 = 5Ix = Solving, VC = 11.98 + j1.44 = 12.1∠6.86°, so th vC = 12.1sin(2,000t + 6.86°) V. – + vSRC 5 Ω 50 μF 20 mH 30 mH 20 mH a b 1:2 7 seen between terminal ab, given that vSRC = 10cos(1,000t + 45°) V Solution: ωL1 = 103×30×10-3 = 30 = ωM =103×20×10-3 = 20 Ω; 6 3 10 50 10 − × × C ω = 20 Ω; VSRC = 10∠45°. will be as shown, where ω(L1 – M 0 Ω and is omitted. 1 1 = The circuit in the frequency domain The j20 Ω in parallel with -j20 Ω is effectively an open circuit. The current in the (5 + j10) Ω impedance is zero, Vcd = 10∠45°, and Vab = VTh = 20∠45°. ) = 10 Ω; ω(L2 – M) = – + 5 Ω a b 1:2 j20 Ω -j20 Ω c d 10∠45° j10 Ω 2/5 If the independent voltage source is replaced by a short circuit, the impedance on the primary side is (5 + j10) Ω and ZTh = 4(5 + j10) = 20 + j40 Ω. 3. Determine the impedance seen by the source, assuming a = 2. Solution: Reflection of the (5 – j5) Ω through the RH transformer gives (20 – j20) Ω. The impedance on the secondary side of the LH transformer is (25 – j10) Ω. Reflected to the primary side, this becomes (25 – j10)/a2 Ω. – + 4∠0° V 5 Ω j10 Ω 2:1 5 Ω -j5 Ω 1:a 4. If vsrc = 10cos(1,000t) V, determine the energy stored in the circuit in the sinusoidal steady state at t = 0, assuming C = 1 μF. 1H 1H k = 0.5 – + vsrc C Solution: At t = 0, the voltage across C is 10 V and the current through the inductors is zero, being proportional to the integral of vsrc. The energy stored is W = 2 = 2 1Cv 50C. – + 10∠0° V 8 Ω 6 Ω -j5 Ω – + 30∠0° V -j10 Ω R Rx I 5. Determine Rx given that I = 0 and R = 2 Ω. Solution: Since I = 0, the voltage across Rx is 10 V, and the same current R o o 10 0 30 0 ∠ − ∠ flows through R and Rx. It follows that R 1:2 10 20 R Rx = , or 2 Rx = R . 9. Two identical coils, each having an inductance of 10 mH, are connected in series. When the connections to one of the coils are reversed, the total inductance is multiplied by a factor a. Determine the coupling coefficient of the coils. Solution: (10 + 10 + 2M) = a(10 + 10 – 2M); 2M(a + 1) = 20(a – 1); – + 10∠0° V R 2 Ω – + 400 turns 100 turns 200 turns 4∠0° V Ix 1 )1 10( + − = a a M ; 1 )1 ( 10 + − = = a a M k 10. Determine Ix, assuming R = 4 Ω. Solution: The voltage across all windings is zero. Hence, I1 = R 10 A, and I2 = 2 2 4 = A. Setting the net mmf to zero, 400I1 – 100I2 + 200I3 = 0, or – + 10∠0° V R 2 Ω – + 400 turns 100 turns 200 turns 4∠0° V Ix I1 I2 I3 2 2 10 4 + − × R I3 = 0, which gives I3 = R 1− 20 ; IX = I2 – I3 = R 1+ 20 . 4/5 100 ohms 1. A current i = Imcos(ωt + 30°) flows through an impedance (5 – j5) Ω. Determine the rms phasor voltage across the impedance if Im = 2.5 A. Solution: The phasor current is Im∠30° A. The impedance is 45o 5 2 ∠ − Ω. The phasor voltage is 15o 2 5 ∠ − Im V peak value or 15o 5 Im∠ − = 15o 12.5 ∠ − V rms. 2. In the phasor diagram shown, phasor A is a current, the other phasors B, C, and D are voltages. To which of the following combinations of circuit elements does this phasor diagram apply? A. R in series with L B. R in series with C C. R in parallel with C D. L in series with C E. L in parallel with C. Solution: The phasor diagram represents a series RL circuit, Phasor B is the voltage across R. Phasor C is the voltage across L, leading the current by 90°, and phasor D is the voltage across the series combination. 3. Determine the input impedance Zin assuming R = 2 Ω. Solution: If a test voltage source VT is applied, IT = ⎜⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ − − + 2 1 2 1 j R j R VT, or Zin = = − + = − − − − + = ⎜⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ − − + 4 4 2 2 2) 2)( ( 2 1 2 1 1 2 j R j R j R j R j R j R j R ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + 4 1 R2 j Ω. Alternatively, the dependent source is equivalent to an impedance -(R – j2) and Zin = ( ) ( ) [ 2 ] 2 || j R j R − − + = 2 1 4 2 4 2 2 j j R j R ⎟⎟ = ⎠ ⎞ ⎜⎜ ⎝ ⎛ + = + − Ω. A B C D j2 R VX /(R – j2) + – VX Zin – + VSRC 1 μF 1 μF 1 kΩ 1 kΩ – + VO 4. Determine ZY so that VL is in phase with VSRC, assuming X = -5 Ω with RL and VSRC unknown. Solution: TEC as seen from terminals a and b will have ZTh = jX/2. For VL to be in phase with VSRC, ZY = -jX/2 = j2.5 Ω 5. Determine VO if ω = 1 krad/s and VSRC = 3 V. Solution: ωCR = 1; C j R j CR R j CR R ω ω ω / ) /(1 ) /(1 − + + + = SRC O V V = = − + + + = − + + + j j j C j R j R j R 1 ) 1/(1 ) 1/(1 / ) /(1 ) /(1 ω 3 1 1 1 1 ) )(1 (1 1 1 − 2 = + = − + + j j j ; VO = VSRC/3 = 1 V. 6. Two coils are wound on a core of high permeability. Coil 1 has 100 turns and coil 2 has 400 turns. A current of 1 A in coil 1, with coil 2 open circuited, produces a core flux of 0.5 mWb. Determine the magnitude of the core flux produced by a current of 0.8 A in coil 2, with coil 1 open circuited. Solution: From the definition of mutual inductance, 2 12 1 1 2 21 i N i N φ φ = , so that 21 1 2 1 2 12 φ φ N N i = i = 21 21 3.2 100 400 1 0.8 φ φ = = 3.2×0.5 = 1.6 mWb 7. Determine Zin, assuming R = 10 Ω, and ω = 1 Mrad/s. Solution: M = k L × L = 2 μH. It follows that the series branches of the T-equivalent circuit are zero and the shunt branch is jωM = j2 Ω; j C j = − − ω . The parallel impedance of j2 and -j Ω is 2 ) 2 /(2 j j = − − Ω. Zin = R – j2 = 10 – j2 Ω. – + VSRC jX jX ZY – + VL RL – + VSRC /2 jX/2 ZY – + VL RL a b R k = 1 2 μH Zin 2 μH 1 μF 8. Determine the stored magnetic energy under dc conditions, assuming VSRC = 1 V. Solution: The branch containing the capacitor carries no current under dc conditions. The current in the other two branches is VSRC /0.5 A and the stored magnetic energy is W = ( ) = + + 2 2 1 1 2 2 2 1 MI L I I L ( × ) = + + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ 1 2 3 2 0.5 2 1 2 VSRC 14 2 VSRC = 14 J. 9. Given vSRC = 6cosωt, where ω is unknown and k = 0.97. Determine X so that no power is dissipated in the circuit. Solution: Considering mesh currents I1 and I2, then for no power dissipation, I2 = 0. The mesh- current equation for mesh 2 is: -(jωM + jX)I1 + (j40 + jX +10)I2 = 0 For I2 to be zero, ωM + X = 0, or X = -ωM = k k L L k k L L 20 400 ) )( ( 2 1 1 2 = − = − = − − ω ω ω = -0.97×20 = -19.4 Ω. Alternatively, it follows from the T-equivalent circuit that if X = -ωM the shunt branch will have zero impedance so I2 = 0. k j40 Ω j10 Ω 10 Ω jX _ + vSRC k j40 Ω j10 Ω 10 Ω jX _ + VSRC I1 I2 + – L1 = 2 H L2 = 3 H M12 = 1 H M23 = 1 H M13 = 1 H L3 = 4 H 0.5 μF 0.5 Ω VSRC jωM (j10 – jωM) Ω 10 Ω jX _ + VSRC I2 (j40 – jωM) Ω