# R

## Transcription

R

Solutions to FE Problems Chapter 2 2FE-1 What is the power generated by the source in the network in Fig. 2PFE-1? The correct answer is b. a RS R2 R1 120V c b R3 R4 R5 Resistors R1, R2, and R3 are connected in delta. A delta-wye transformation can be conducted. R1 R2 6k (18k ) Ra = = = 3kΩ R1 + R2 + R3 6k + 18k + 12k R2 R3 18k (12k ) = = 6kΩ R1 + R2 + R3 6k + 18k + 12k R1 R3 6k (12k ) Rc = = = 2kΩ R1 + R2 + R3 6k + 18k + 12k Rb = a RS Ra 120V Rc Rb c b R4 Req = Rs + Ra + [( Rc + R4 ) ( Rb + R5 )] = 5k + 3k + [(2k + 4k ) (6k + 6k )] 6k (12k ) = 12kΩ 6k + 12k (120) 2 = = 1.2W 12k Req = 8k + PVs = Vs2 Req 2FE-2 Find Vab in the circuit in Fig. 2PFE-2. The correct answer is d. I1 10Ω b 15Ω I2 10Ω a 4A 5Ω R5 10 + 5 I1 = ( 4 ) = 1.5 A 10 + 5 + 10 + 15 10 + 15 I2 = ( 4 ) = 2.5 A 10 + 5 + 10 + 15 V5Ω = 5(2.5) = 12.5V V15Ω = 15(1.5) = 22.5V Vab = 12.5 − 22.5 = −10V 2FE-3 If Req=10.8Ω in the circuit in Fig. 2PFE-3, what is R2? The correct answer is a. Req = ( R2 8) + 4 + 2 ( R2 )8 +6 R2 + 8 8 R2 10.8 − 6 = R2 + 8 3.2 R2 = 38.4 R2 = 12Ω Req = 2FE-4 Find the equivalent resistance of the circuit in Fig. 2PFE-4 at the terminals A-B. The correct answer is c. R1 R6 R2 R3 R5 R AB R7 R4 R AB = { [[( R6 + R7 ) R5 ] + R4 ] R AB = { [[24k 12k ] + 4k ] R3 } + ( R1 R2 ) 6k } + (12k 6k ) 24k (12k ) 12k (6k ) R AB = + 4k 6 k + 24k + 12k 12k + 6k R AB = (12k 6k ) + 4k = 12k (6k ) + 4k 12k + 6k R AB = 8kΩ 2FE-5 The 100V source is absorbing 50W of power in the network in Fig. 2PFE-5. What is R? The correct answer is a. A 10Ω I1 10Ω I2 R 100V 5A 200V P = VI 1 50 = −100 I 1 I 1 = −0.5 A KVL around the right loop: 100 + I 2 R = 10 I 1 + 200 KCL at node A: 5 = I 2 + I1 I 2 = 5 + 0.5 = 5.5 A 100 + (5.5) R = 10(−0.5) + 200 R = 17.27Ω 2FE-6 Find the power supplied by the 40V source in the circuit in Fig. 2PFE-6. The correct answer is b. I3 I I1 I2 40V 9.99Ω 50Ω 3A 100V Req = (20 25) 100 20(25) 11.1111(100) = 10Ω 100 = 20 + 25 11.1111 + 100 KCL: I 3 + I 1 = I 2 I 3 = I 2 − I1 KVL: 100 = 40 + 50 I 50 I = 60 6 I= A 5 I1 = I − 3 6 I 1 = − 3 = −1.8 A 5 40 I3 = + 1.8 = 5.8 A 10 P40V = VI 3 = 40(5.8) = 232W Req = 2FE-7 What is the current Io in the circuit in Fig. 2PFE-7? The correct answer is d. R2 I R4 R5 VS IS R1 R3 R6 R7 Io R A = R1 + R2 = 9kΩ RB = R4 + R5 + ( R6 R7 ) = 4k + 12k + 9k (18k ) = 6kΩ 9k + 18k Vs 12 Is = = = 1mA R3 + RC 6k + 6k RC = R A RB = RA I = − R A + RB (I s ) 3k (6k ) = 18kΩ 3k + 6k 1 9k I = − (1m ) = − mA 3 9k + 18k R7 6k 1 (I ) = I o = − m 3k + 6k 3 R6 + R7 I o = −0.22mA 2FE-8 Find the voltage Vo in the network in Fig. 2PFE-8. The correct answer is c. R3 R2 R1 IS Io R4 + Vo − R5 R6 R A = R1 + R2 = 4kΩ 12k (6k ) RB = R5 R6 = = 4kΩ 12k + 6k RA Io = (I s ) R A + R3 + R4 + RB 4k Io = (24m) = 6mA 4k + 2k 6k + 4k Vo = I o R4 = 6m(6k ) = 36V 2FE-9 What is the voltage Vo in the circuit in Fig. 2PFE-9? The correct answer is a. R2 2A R1 Io + R3 Vo − R1 2 2 (2) = ( 2) = A R1 + R2 + R3 3 2 +1+ 3 2 Vo = (3) = 2V 3 Io = 2FE-10 Find the current Ix in Fig. 2PFE-10. The correct answer is d. R1 12V Ix R2 R3 R4 R6 Req = { [[( R5 + R6 ) R4 ] + R3 ] Req = { [[(8 + 2) 10] + 1] 3} + 1 Req = (6 3) + 1 = 3Ω Is = 12 12 = = 4A Req 3 R' = [( R5 + R6 ) R4 ] + R3 = 6Ω 24 8 6 Ix = = A (4) = 9 3 3+ 6 R2 } + R1 R5 Solutions to FE Problems Chapter 3 3FE-1 Find Vo in the circuit in Fig. 3PFE-1. The correct answer is a. A 2Ω 6Ω 1Ω I1 I 12V + 2Ω Vo − KCL at node A: I 1 = I + I 2 I = I1 − I 2 KVL around the left loop: 12 = 2 I 1 + 1( I ) + 2 I 12 = 2 I 1 + 1( I 1 − I 2 ) + 2( I 1 − I 2 ) 12 = 5 I 1 − 3I 2 KVL around the right loop: 6 = 6 I 2 − 2 I − 1( I ) 6 = 6 I 2 − 3( I 1 − I 2 ) 6 = 9 I 2 − 3I 1 Two equations and two unknowns: 12 = 5 I 1 − 3I 2 6 = −3 I 1 + 9 I 2 Therefore, I 1 = 3.5 A and I 2 = 1.833 A V0 = 2 I = 2( I 1 − I 2 ) = 2(3.5 − 1.833) V0 = 3.33V I2 6V 3FE-2 Determine the power dissipated in the 6-ohm resistor in the network in Fig. 3PFE-2. The correct answer is d. V1 V2 4Ω I1 12V V1 = 12V KCL at node 2: V2 − V1 V2 V2 + + = 2I1 4 6 12 V − V2 I1 = 1 4 V2 − V1 V2 V2 V − V2 + + = 2 1 4 6 12 4 3V2 − 3V1 + 2V2 + V2 = 6V1 − 6V2 12V2 = 9V1 12V2 = 9(12) V2 = 9 V P6 Ω = V22 9 2 = = 13.5W 6 6 6Ω text 2I1 12Ω 3FE-3 Find the current Ix in the 4-ohm resistor in the circuit in Fig. 3PFE-3. The correct answer is b. 12V 3Ω I1 6Ω 2A + Vx I2 4Ω − I2 = I x + I3 I x = I2 − I3 I1 + 2 = I 2 I 2 − I1 = 2 12 = 6 I 1 + 4 I x 12 = 6 I 1 + 4( I 2 − I 3 ) 12 = 6 I 1 + 4 I 2 − 4 I 3 KVL around the right loop: 2V x + 4(− I x ) + 3I 3 = 0 − 2V x = −4( I 2 − I 3 ) + 3I 3 − 2(4 I x ) = 4( I 3 − I 2 ) + 3I 3 − 8( I 2 − I 3 ) = 4 I 3 − 4 I 2 + 3I 3 8 I 2 − 8 I 3 + 4 I 3 − 4 I 2 + 3I 3 = 0 4I 2 − I 3 = 0 I 3 = 4I 2 12 = 6 I 1 + 4 I 2 − 4(4 I 2 ) 12 = 6 I 1 − 12 I 2 Two equations and two unknowns: − I1 + I 2 = 2 6 I 1 − 12 I 2 = 12 Therefore, I 1 = −6 A and I 2 = −4 A . I 3 = 4(−4) = −16 A I x = −4 + 16 + 2V x - Ix Vx = 4I x I3 I x = 12 A 3FE-4 Determine the voltage Vo in the circuit in Fig. 3PFE-4. The correct answer is a. 12V 4Ω I1 4Ω 2Ω Ix KCL: I 1 = I x + I 2 I x = I1 − I 2 KVL around the left loop: 12 = 4 I 1 + 2 I x 12 = 4 I 1 + 2( I 1 − I 2 ) 12 = 6 I 1 − 2 I 2 KVL around the outer loop: 12 = 4 I 1 + 4 I 2 + 4 I 3 I 2 = 2I x + I 3 2I x = I 2 − I 3 2( I 1 − I 2 ) = I 2 − I 3 I 3 = −2 I 1 + 3 I 2 12 = 4 I 1 + 4 I 2 + 4(−2 I 1 + 3I 2 ) 12 = −4 I 1 + 16 I 2 Two equations and two unknowns: 12 = 6 I 1 − 2 I 2 12 = −4 I 1 + 16 I 2 Therefore, I 1 = 2.45 A and I 2 = 1.36 A . I 3 = −2(2.45) + 3(1.36) I 3 = −0.82 A V o= 4 I 3 = 4(−0.82) V o= −3.28V I3 I2 text 2I x + 4Ω Vo − 3FE-5 What is the voltage V1 in the circuit in Fig. 3PFE-5? The correct answer is c. I1 V1 1Ω V2 I2 V3 2Ω 10V 3Ω 15V 8A 4A KCL at the supernode: 4 = I 1 + 8 + I 2 I 1 + I 2 = −4 V1 V2 − V3 + = −4 4 2 V1 + 2V2 − 2V3 = −16 V3 = 15V V1 + 2V2 − 2(15) = −16 V1 + 2V2 = 14 V2 − V1 = 10 Two equations and two unknowns: − V1 + V2 = 10 V1 + 2V2 = 14 Therefore, V1 = −2V and V2 = 8V . Solutions to FE Problems Chapter 4 4FE-1 Given the summing amplifier shown in Fig. 4PFE-1, what value of R2 will produce an output voltage of –3V? The correct answer is c. − R2 R Vo = (4) − 2 (−2) 4k 12k Vo = −3V 1 − 3 = −1m( R2 ) + m( R2 ) 6 5 − m ( R2 ) = −3 6 R2 = 3.6kΩ 4FE-2 Determine the output voltage Vo of the summing op-amp circuit shown in Fig. 4PFE-2. The correct answer is b. 36kΩ 18kΩ - V1 + 6kΩ + 6kΩ 12 kΩ 12kΩ 2V 3V 1V 18k 18k V1 = −2 (4) + 1 = −4.5V 6k 12k Vo 36k 36k V0 = −V1 − 3 6k 12k 36k 36k V0 = 4.5 − 3 6k 12k V0 = 18V 4FE-3 What is the output voltage Vo in Fig. 4PFE-3? The correct answer is a. if 2Ω i=0 - Vo + 3Ω 4Ω 6V 2V V1 V2 6 2 + = + = 2.5 A R1 R2 3 4 V0 = −i f R f = −(2.5)(2) if = V0 = −5V 4FE-4 What value of Rf in the op-amp circuit of Fig. 4PFE-4 is required to produce a voltage gain of 50? The correct answer is d. The op-amp is a noninverting op-amp. Rf A = 1+ R1 R f = ( A − 1) R1 R f = (50 − 1)5k = 245kΩ 4FE-5 What is the voltage Vo in the circuit in Fig. 4PFE-5? The correct answer is c. R1 = 6kΩ + + - R2 = 2kΩ 5V R3 = 1kΩ 8kΩ Vo 2 kΩ − The 8kΩ and 2kΩ resistors make up a noninverting op-amp. 8k V1 = 1 + 5 = 25V 2k Use nodal analysis at node A: Vo Vo − Vi Vo − 25 + + =0 R3 R1 2k Vo Vo − 5 Vo − 25 + + =0 1k 6k 2k 6Vo + Vo − 5 + 3Vo − 75 = 0 10Vo = 80 Vo = 8V Solutions to FE Problems Chapter 5 5FE-1 Determine the maximum power that can be delivered to the load RL in the network in Fig. 5PFE-1. The correct answer is d. I' R2 R1 R4 + Voc' R3 4mA − Use superposition. R1 (4m) I ' = R1 + R2 + R3 1k I'= (4m) = 1mA 1 k + 1 k + 2 k ' Voc = I ' R3 = (1m)(2k ) Voc' = 2V R1 12V R2 R4 + Voc" R3 − R3 (12) Voc" = R1 + R2 + R3 2k Voc" = (12) 1k + 1k + 2k Voc" = 6V Voc = Voc' + Voc" = 2 + 6 = 8V R2 R1 R4 R3 RTH = [(R1 + R2 ) R3 ] + R4 RTH = 2 k ( 2k ) + 1k = 2kΩ 2k + 2k RTH = 2kΩ Voc = 8V RL PLmax = I L2 R L RL = RTH for maximum power. 2 PLmax PLmax V = oc RTH 2 RTH V2 82 = oc = = 8mW 4 RTH 4(2k ) RTH 5FE-2 Find the value of the load RL in the network in Fig. 5PFE-2 that will achieve maximum power transfer, and determine the value of the maximum power. The correct answer is c. + Vx − 2kΩ 12V 1kΩ I + Voc + − - 2V x 12 = 2kI + 1kI + 2V x V x = I ( 2k ) 12 = 2kI + 1kI + 2(2kI ) 12 I = mA 7 12 = 2kI + Voc 12 12 = 2k m + Voc 7 Voc = 8.57V + Vx − 2kΩ 12V 1kΩ I1 I sc I2 + - 12 I 1 = = 6mA 2k 2V x 1kI 2 + 2V x = 0 1kI 2 + 2(2kI 1 ) = 0 I 2 = −24mA I 1 = I 2 + I sc I sc = 6m − (−24m) = 30mA Voc 8.57 = = 285.7Ω I sc 30m RL = RTH for maximum power. RTH = RTH = 285 .7 Ω Voc = 8.57V PLmax = RL Voc2 (8.57) 2 = = 64.3mW 4 RTH 4(285.7) 5FE-3 Find the value of RL in the network in Fig. 5PFE-3 for maximum power transfer to this load. The correct answer is a. Ix + R1 I 12V R2 2I x text Voc − I = I x + 2I x I = 3I x 12 = 3I x + 12 I 12 = 3I x + 12(3I x ) 4 Ix = A 13 4 Voc = 12 I = 12(3I x ) = 12(3) = 11.08V 13 Ix R1 I2 12V 2I x R2 I sc text I sc = I x + I 2 + 2 I x = 3I x + I 2 I2 = 0A I sc = 3I x 12 = 3I x I x = 4A I sc = 3(4) = 12 A V 11.08 RTH = oc = = 0.92Ω I sc 12 RTH = 0.92Ω Voc = 11.08V RL = 0.92 + 12 = 12.92Ω RL = 12.92Ω for maximum power transfer. 12Ω RL 5FE-4 What is the current I in Fig. 5PFE-4? The correct answer is c. 2Ω I' 3Ω 2Ω 4Ω 10 A Use superposition. 2 I' = (−10) = −4 A 2 + 3 I" 3Ω 20 = 4A 5 I = I' + I" I = −4 + 4 I = 0A I" = 2Ω 4Ω 20V 2Ω 5FE-5 What is the open circuit voltage Voc at terminals a and b of the circuit in Fig. 5PFE-5? The correct answer is b. Use superposition. a 4Ω I 2Ω 3Ω ' 12 A + Voc' − b 4 I' = (12) = 8 A 2+ 4 Voc' = I ' (2) = 8(2) = 16V a 3Ω 4Ω 12V 2Ω + Voc" − b 2 Voc" = (−12) = −4V 2+ 4 Voc = 16 − 4 = 12V Solutions to FE Problems Chapter 6 6FE-1 Given three capacitors with values 2µF, 4µF, and 6µF, can the capacitors be interconnected so that the combination is an equivalent 3µF? The correct answer is a. Yes. The capacitors should be connected as shown. 6µ F 2µ F 4µ F C eq = 6 µ (6 µ ) = 3µ F 6µ + 6µ Ceq 6FE-2 The current pulse shown in Fig. 6PFE-2 is applied to a 1µF capacitor. What is the energy stored in the electric field of the capacitor? The correct answer is c. q(t ) = ∫ i (t ) dt 0 C , t ≤ 0 q(t ) = 6t C , 0 < t ≤ 1µ s 6 µ C , t > 1µ s q(t ) C 0 V , t ≤ 0 v(t ) = 6 x10 6 t V , 0 < t ≤ 1µ s 6 V , t > 1µ s v(t ) = 1 C v 2 (t ) 2 0 J , t ≤ 0 w(t ) = 18 x10 6 t 2 J , 0 < t ≤ 1µ s 18µ J , t > 1µ s w(t ) = 6FE-3 The two capacitors shown in Fig. 6FE-3 have been connected for some time and have reached their present values. Determine the unknown capacitor Cx. The correct answer is b. The voltage across the unknown capacitor Cx is (using KVL): 24 = 8 + V x V x = 16V q = Cv The capacitors are connected in series and the charge is the same. q = 60µ (8) = 480µ C q 480µ Cx = = = 30µ F v 16 6FE-4 What is the equivalent inductance of the network in Fig. 6PFE-4? The correct answer is d. 2 mH 3mH 12mH Leq 3mH Leq = [ {[(3m + 9m) 12m] 6m} 3m] + 2m Leq = [ {[(12m) 12m] 6m} 3m] + 2m Leq = [ {6m 6m} 3m] + 2m Leq = [ 3m 3m] + 2m Leq = 1.5m + 2m Leq = 3.5mH 6mH 9mH 6FE-5 The current source in the circuit in Fig. 6PFE-5 has the following operating characteristics: 0 A, t < 0 i(t ) = . −2t 20te A, t > 0 What is the voltage across the 10 mH inductor expressed as a function of time? The correct answer is a. di(t ) v(t ) = L dt di(t ) = 20te −2t (−2) + 20e − 2t = 20e −2t − 40te − 2t dt [ ] v(t ) = 10m 20e −2t − 40te −2t 0 V , t < 0 v(t ) = −2t −2t 0.2e − 0.4te V , t > 0 Solutions to FE Problems Chapter 7 7FE-1 In the circuit in Fig. 7PFE-1, the switch, which has been closed for a long time, opens at t=0. Find the value of the capacitor voltage vc(t) at t=2s. The correct answer is b. Find the initial condition: 8kΩ + vc (0−) 6kΩ 12V R' = 6 kΩ − 12k (6k ) = 4kΩ 12k + 6k 8kΩ + 12V 4k v c (0 − ) = (12) = 4V 4k + 8k R ' = 4 kΩ v c (0 − ) − 6 kΩ The t > 0 circuit: 6kΩ + v c (t ) 6kΩ − v c (t ) + 12kic (t ) = 0 ic (t ) = C dvc (t ) dt vc (t ) + 12k (100µ ) dvc (t ) =0 dt dvc (t ) 1 + vc (t ) = 0 dt 1.2 1 r+ =0 1.2 1 r=− 1.2 vc (t ) = Ae −t 1.2 vc (0−) = 4V A=4 −t vc (t ) = 4e 1.2 V , t > 0 vc (2) = 0.756V 7FE-2 In the network in Fig. 7PFE-2, the switch closes at t=0. Find v0(t) at t=1s. The correct answer is d. Find the initial condition: 12kΩ 4kΩ + 12 kΩ v0 (0−) 12V − vo (0−) = 0V The t > 0 circuit: R1 = 12kΩ R 2 = 12 kΩ 12V R3 = 4kΩ + v0 (t ) − R1 = 12kΩ R3 = 4kΩ + voc R 2 = 12 kΩ − 12V voc = R2 12k (12) = (12) = 6V R1 + R2 12k + 12k R1 = 12kΩ R 2 = 12 kΩ RTH = ( R1 R2 ) + R3 RTH = 12k (12k ) + 4k = 10kΩ 12k + 12k R3 = 4kΩ RTH RTH = 10kΩ 100µ F voc = 6V − 10ki(t ) + vo (t ) = 6 dv (t ) ic (t ) = C o dt dvc (t ) RTH C + vo (t ) = voc dt dvo (t ) v 1 + vo (t ) = oc dt RTH C RTH C 1 r+ =0 RTH C 1 r=− RTH C The natural solution is: −t von (t ) = Ae RTH C −t 10 k (100 µ ) von (t ) = Ae = Ae −t The forced solution is: vo f (t ) = k dvo f (t ) dt 0+ =0 1 RTH C k = voc + v0 (t ) k= voc RTH C vo f (t ) = 6V vo (t ) = 6 + Ae −t vo (0−) = 0V 0 = 6+ A A = −6 v o (t ) = 6 − 6e − t V , t > 0 vo (1) = 6 − 6e −1 vo (1) = 3.79V 7FE-3 Assume that the switch in the network in Fig. 7PFE-3 has been closed for some time. At t=0 the switch opens. Determine the time required for the capacitor voltage to decay to one-half of its initially charged value. The correct answer is a. Find the initial condition: 12kΩ + v c (0 − ) 12V − 6k v c (0 − ) = (12) = 4V 6k + 12k The t > 0 circuit: + vc (t ) 6kΩ − vc (t ) + 6kic (t ) = 0 ic (t ) = C dvc (t ) dt vc (t ) + 6k (100µ ) dvc (t ) =0 dt dvc (t ) 1 + vc (t ) = 0 dt 0.6 6kΩ 1 =0 0.6 1 r=− 0.6 r+ −t vc (t ) = Ae 0.6 vc (0−) = 4V A=4 −t vc (t ) = 4e 0.6 V , t > 0 −t 0.5(4) = 4e 0.6 −t 1 = e 0.6 2 1 −t ln = 2 0.6 t = −0.6 ln 1 2 t = 0.416s 7FE-4 Find the inductor current iL(t) for t>0 in the circuit in Fig. 7PFE-4. The correct answer is c. Find the initial condition: i L ( 0 −) 2Ω 2Ω 10V i L (0 − ) = 10 = 5A 2 + voc 1Ω 5A 2Ω − 1 2= 2 Ω 3 + voc 2 Ω 3 6A − 2 voc = 6 = 4V 3 2Ω 2Ω RTH = (2 2 ) 2 = 1 2 = RTH 2 Ω 3 2Ω 1A RTH = voc = 4V 2 Ω 3 iL (t ) 4H di L (t ) 2 + i L (t ) = 4 dt 3 di L (t ) 1 + i L (t ) = 1 dt 6 1 r+ =0 6 1 r=− 6 4 −t i Ln (t ) = Ae 6 i L f (t ) = k di L f (t ) =0 dt 1 0+ k =1 6 k =6 i L f (t ) = 6 i L (0 − ) = 5 A −t i L (t ) = 6 + Ae 6 5=6+ A A = −1 −t i L (t ) = 6 − e 6 A, t > 0 7FE-5 Find the inductor current iL(t) for t>0 in the circuit in Fig. 7PFE-5. The correct answer is d. Find the initial condition: i L (0−) 3Ω 1Ω 12V i L (0 − ) = 12 = 3A 4 L iL (t ) R1 R2 Vs di L (t ) + (R1 + R2 )i L (t ) = Vs dt V diL (t ) R1 + R2 + i L (t ) = s dt L L R1 + R2 r+ =0 L R + R2 r=− 1 L L i Ln (t ) = Ae i Ln (t ) = Ae i L f (t ) = k R +R − 1 2 t L 5 − t 3 di L f (t ) =0 dt 5 0+ k = 4 3 12 k= 5 i L (0 − ) = 3 A 5 − t 12 + Ae 3 5 3 = 2.4 + A A = 0.6 i L (t ) = i L (t ) = 2.4 + 0.6e 5 − t 3 A, t > 0 Solutions to FE Problems Chapter 8 8FE-1 Find V0 in the network in Fig. 8PFE-1. The correct answer is d. 2Ω 12∠0°V j1Ω + 4∠0°V − j1Ω 1Ω V0 - j1Ω + 8∠0° A 2Ω − j1Ω 1Ω V0 - Z = 2(− j1) = 0.4 − j 0.8Ω 2 − j1 I0 j1Ω 8∠0° A 1Ω Z 0.4 − j 0.8 (8∠0°) = 1.6 − j 4.8 A I 0 = 0.4 − j 0.8 + 1 + j1 V0 = (1.6 − j 4.8)(1) = 5.06∠ − 71.6°V 8FE-2 Find V0 in the circuit in Fig. 8PFE-2. The correct answer is c. text 2I x 1Ω V1 I1 2Ω V2 I2 12∠0°V − j1Ω Ix 2∠0° A V0 V2 = 12∠0°V V1 − j1 2I x + I x + I1 = 0 Ix = V V V − 12∠0° 2 1 + 1 + 1 =0 1 − j1 − j1 2V1 + V1 + − j1V1 + j1(12∠0°) = 0 − j1(12∠0°) V1 = = 1.2 − j 3.6V 3 − j1 KCL at node 0: 2 I x + I 2 + 2∠0° = 0 V V − V0 2 1 + 2 + 2∠0° = 0 2 − j1 1.2 − j 3.6 12∠0° − V0 + 2 + 2∠0° = 0 2 − j1 V0 = 30.8∠8.97°V 8FE-3 Find V0 in the network in Fig. 8PFE-3. The correct answer is a. I 2Ω − j1Ω j 2Ω I1 I2 + 4Ω 12 ∠0°V + 2V0 - I1 = I + I 2 I = I1 − I 2 KVL around the left loop: 12∠0° = 2 I 1 + j 2 I + 2V0 V0 − V0 = 4I 2 (2 + j 2) I 1 + (8 − j 2) I 2 = 12∠0° KVL around the right loop: 2V0 = − j 2 I − j1I 2 + 4 I 2 j 2 I 1 + (−4 + j1) I 2 = 0 Two equations and two unknowns: (2 + j 2) I 1 + (8 − j 2) I 2 = 12∠0° j 2 I 1 + (−4 + j1) I 2 = 0 It follows that: I 1 = 4.24∠45° A I 2 = 2.06∠ − 30.96° A V0 = 4(2.06∠ − 30.96°) V0 = 8.24∠ − 30.96° V 8FE-4 Determine the midband (where the coupling capacitors can be ignored) gain of the single-stage transistor amplifier shown in Fig. 8PFE-4. The correct answer is b. Z c is small at midband. Vx 5k 5 = = Vs 5k + 1k 6 V0 6k (12k ) = −40 x10 −3 = −160 Vx 6k + 12k V0 V x V0 = Vs Vs V x 5 = (− 160) = −133.33 6 8FE-5 What is the current I0 in the circuit in Fig. 8PFE-5? The correct answer is c. Is 6∠0°V Zeq (1 − j1)(1 + j3) + 3 = 1.58∠ − 18.43° + 3 = 4.53∠ − 6.34° Ω 1 − j1 + 1 + j 3 6∠0° Is = = 1.3245∠6.34° A 4.53∠ − 6.34° Z eq = 3Ω − j 1Ω Io j 3Ω 1Ω 6∠0° V 1Ω KVL around the outer loop: 6∠0° = 3(1.3245∠6.34°) + I 0 (1 − j1) I 0 = 1.48∠32.92° A Solutions to FE Problems Chapter 9 Find V0 in the network in Fig. 8PFE-1. The correct answer is c. From the power triangle: Qold = Pold tan θ old 9FE-1 Qold = 120k tan 45° = 120k var S old = (120k + j120k ) VA = 170∠45° kVA S new = Pold + jQnew θ new = cos −1 (0.95) = 18.19° Qnew = Pold tan θ new Qnew = 120k tan 18.19° = 39.43k var S new = (120k + j 39.43k )VA S new = 126.31∠18.19° kVA S cap = S new − S old = − j80.57 kVA Qcap = −ω CV 2 − 80.57k = −2π 60(C )(480 2 ) C = 928µ F 9FE-2 Determine the rms value of the following waveform. The correct answer is d. T Vrms 1 2 = v (t ) dt T ∫0 Vrms = Vrms = Vrms = 1 2 3 4 1 2 2 2 1 dt 2 dt 1 dt 0 2 dt + + ∫ ∫ ∫ ∫ 4 0 1 2 3 [ ] 1 1 2 3 t 0 + 4t 1 + t 2 + 0 4 1 [1 + (8 − 4) + (3 − 2)] = 1.22V 4 9FE-3 Find the impedance ZL in the network in Fig. 9PFE-3 for maximum power transfer. The correct answer is b. 2Ω − j1Ω ZTH Z TH = (2 − j1) + j 2 = 0.4 + j1.2Ω * Z L = Z TH = 0.4 − j1.2Ω j 2Ω Solutions to FE Problems Chapter 10 10FE-1 In the network in Fig. 10PFE-1, find the impedance seen by the source. The correct answer is a. M 4Ω 24 ∠0°V I1 − j 5Ω j 8Ω j 2Ω k=0.5 M = k L1 L2 = 1H 24∠0° = I 1 (4 + j 2) − j 2 I 2 − j 2 I 1 + (5 + j3) I 2 = 0 I 1 = 4.92∠ − 19.75° A I 2 = 1.69∠39.29° A 24∠0° Z1 = = 4.88∠19.75°Ω 4.92∠ − 19.75° I2 5Ω N2 N1 to achieve impedance matching for maximum power transfer. Using this value of n, calculate the power absorbed by the 3Ω resistor. The correct answer is b. 10FE-2 In the circuit in Fig. 10PFE-2, select the value of the transformer’s turns ratio n = 48 n 2 Ω j 32n 2 Ω − j 2Ω R L = 3Ω V2 Z TH = 3Ω for maximum power transfer. Reflecting the primary into the secondary: Z TH = 48n 2 + j 32n 2 − j 2 = 3 48n 2 + j 32n 2 − j 2 = 3 1 if n = : 4 Z TH = 48 32 + j − 2 = 3Ω 16 16 3Ω I RL = 3Ω 30∠0°V 30∠0° = 5∠0° A 6 1 1 PL = I M2 R L = (5) 2 (3) = 37.5W 2 2 I = 10FE-3 In the circuit in Fig. 10PFE-3, select the turns ratio of the ideal transformer that will match the output of the transistor amplifier to the speaker represented by the 16Ω load. The correct answer is d. + 1kΩ + 5kΩ Vs 4V x 100 Vx 10kΩ Voc - 5 4 Voc = Vs − x10 4 6 100 1kΩ + 5kΩ Vs Vx - 5 4 I sc = Vs − 6 100 Find Rout of the amplifier: 5 V x = Vs 6 Rout = Voc I sc 5 4 x10 4 Vs − 6 100 = 10kΩ = 5 4 Vs − 6 100 16a 2 = 10k a = 25 4V x 100 I sc 10 kΩ 10FE-4 What is the current I2 in the circuit shown in Fig. 10PFE-4? The correct answer is c. Z1 = ZL = 2 2 (1 − j1) = 4 − j 4Ω 2 n 120∠0° = 11.77∠11.31° A 6 + j2 + 4 − j4 I I2 = 1 n I 2 = 23.54∠11.31° A I1 = 10FE-5 What is the current I2 in the circuit in Fig. 10PFE-5? The correct answer is a. I1 = V1 Z1 ZL n2 N n= 2 =2 N1 10 + j10 Z1 = = 2.5 + j 2.5Ω 22 120∠0° I1 = = 24 − j 24 A 2.5 + j 2.5 I 24 − j 24 I2 = 1 = n 2 I 2 = 12 − j12 A = 16.97∠ − 45° A Z1 = Solutions to FE Problems Chapter 11 11FE-1 The correct answer is d. I = 6 A rms V R = 84.85V rms V L = 84.85V rms R= VR I ωL = 84.85 = 14.14Ω 6 = VL = I 84.85 = 14.14Ω 6 Z load = 14.14 + j14.14 Ω 2 S 3−φ = 3 I Z load 2 S 3−φ = 3(6) (14.14 + j14.14) S 3−φ = 2.16∠45° kVA 11FE-2 The correct answer is a. Z ∆ = 12 + j12 Ω Finding the equivalent per-phase wye: Z Y = 4 + j 4Ω = 5.66∠45°Ω V AB = 230V rms V AN = I aA = 230 3 V AN ZY = 132.79V rms = 132.79 = 23.5 A rms 5.66 P3−φ = 3 V AN I aA cosθ Z = 3(132.79)(23.5) cos 45° = 6.62kW 11FE-3 The correct answer is b. 24 + j18 ZY1 = = 8 + j 6Ω 3 Z Y 2 = 6 + j 4Ω V AB = 208V rms V AN = 208 3 = 120V rms 120∠0° = 12∠ − 36.87° A 8 + j6 120∠0° I aN 2 = = 16.64∠ − 33.69° A 6 + j4 I aA = I aN 1 + I aN 2 = 28.63∠ − 35.02° A rms I aN 1 = 11FE-4 The correct answer is c. S 3−φ = 24∠30° kVA = 20784.61 + j12000 VA P3−φ = 20.78kW P1−φ = 6.93kW 11FE-5 The correct answer is d. By use of the power triangle: S 2 = P2 + Q2 Q = S 2 − P2 Q = (100k ) 2 − (90k ) 2 Q = 43.59k var Solutions to FE Problems Chapter 12 12FE-1 The correct answer is a. 1 1 rad = = 10,000 ω0 = s LC 1m(10 µ ) 1 1 Z ( jω 0 ) = 6∠0° = 60∠ − 90°V V0 ( jω 0 ) = 12∠0° c = 6 ∠ 0 ° 2 ω C ∠ 90 ° 10 , 000 ( 10 µ ) ∠ 90 ° 0 12FE-2 The correct answer is c. ω0 = 1 = 1 LC 20m(50µ ) R rad BW = = 200 L s R = 200(20m) = 4Ω = 1000 rad s 12FE-3 The correct answer is d. 1 1 V0 ZC jωC RC = = = = Gv ( jω ) 1 1 VI Z C + R + R jω + jωC RC at DC, Gv=1=0dB 1 at 3dB down, ω = RC 1 rad ω= = 200 (5k )(1µ ) s ω = 2πf ω f = = 31.83 = 32 Hz 2π 12FE-4 The correct answer is b. 1 rad ω0 = = 1000 s LC 1 1 L= 2 = = 100mH ω 0 C (1000) 2 (10 µ ) R rad BW = = 100 L s R = 100m(100) = 10Ω 12FE-5 The correct answer is a. rad f = 8 Hz , ω 1 = 16π s 1 ZC = = − j 2kΩ jωC 1 1 jωC RC G v ( jω 1 ) = = 1 1 + R jω + jωC RC For f=8Hz and ω1=16π V0 ZC − j 2000 = G v ( jω 1 ) = = = 0.707∠ − 45° VS Z C + R 2000 − j 2000 Gv ( j16π ) = 0.707∠ − 45° fC = 1 1 = = 7.96 = 8Hz 2πRC 2π (2k )(10µ ) Solutions to FE Problems Chapter 13 13FE-1 The correct answer is c. 12 12 V0 ( s ) = = 2 s ( s + 3s + 2) s ( s + 2)( s + 1) A B C V0 ( s ) = + + s s + 2 s +1 12 A B C = + + s ( s + 2)( s + 1) s s + 2 s + 1 let s=-1 12 = C (−1)(1) C = −12 let s=-2 12 = B (−2)(−1) B=6 let s=0 12 = 2 A A=6 6 6 − 12 + + s s + 2 s +1 v0 (t ) = 6 + 6e −2t − 12e −t u (t ) V V0 ( s ) = [ ] 13FE-2 The correct answer is d. 120 V0 ( s ) = s ( s + 10)( s + 20) A B C V0 ( s ) = + + s s + 10 s + 20 120 A B C = + + s ( s + 10)( s + 20) s s + 10 s + 20 let s=0 120 = A(10)(20) 3 A= 5 let s=-10 120 = B (−10)(10) 6 B=− 5 let s=-20 120 = C (−20)(−10) C= 3 5 3 6 3 − 5 + 5 V0 ( s ) = 5 + s s + 10 s + 20 3 3 6 v0 (t ) = − e −10t + e − 20t u (t ) V 5 5 5 3 3 6 v 0 (100m) = − e −10(100 m ) + e − 20(100m ) 5 5 5 v 0 (100m) = 0.24V 13FE-3 The correct answer is b. 12( s + 2) V0 ( s ) = s ( s + 1)( s + 3)( s + 4) lim v 0 (t ) = lim sV0 ( s ) t →∞ s →0 12( s + 2) ( s + 1)( s + 3)( s + 4) 12(2) lim v 0 (t ) = lim = 2V t →∞ s →0 (1)(3)( 4) lim v 0 (t ) = lim t →∞ s →0 13FE-4 The correct answer is a. 2s V0 ( s ) = ( s + 1) 2 ( s + 4) A B C V0 ( s ) = + + s + 4 s + 1 ( s + 1) 2 2s A B C = + + 2 s + 4 s + 1 ( s + 1) 2 ( s + 4)( s + 1) let s=-1 2(−1) = C (−1 + 4) 2 C=− 3 let s=-4 2(−4) = A(−4 + 1) 2 8 A=− 9 let s=0 0 = A + B(1)(4) + 4C 8 B= 9 8 8 2 − − 3 V0 ( s ) = 9 + 9 + s + 4 s + 1 ( s + 1) 2 8 2 8 v0 (t ) = − e − 4t + e −t − e −t u (t ) V 9 3 9 13FE-5 The correct answer is c. 2 s 2 X ( s ) + 6[sX ( s )] + 8 X ( s ) = s+3 2 X ( s) = ( s + 2)( s + 4)( s + 3) A B C X ( s) = + + s+2 s+4 s+3 2 A B C = + + ( s + 2)( s + 3)( s + 4) s + 2 s + 4 s + 3 let s=-3 2 = C (−3 + 2)(−3 + 4) C = −2 let s=-4 2 = B(−4 + 2)(−4 + 3) B =1 let s=-2 2 = A(−2 + 3)(−2 + 4) A =1 1 1 −2 X ( s) = + + s+2 s+4 s+3 [ ] x(t ) = e −2t + e −4t − 2e −3t u (t ) Solutions to FE Problems Chapter 14 14FE-1 A single loop, second-order circuit is described by the following differential equation. 2 dv 2 (t ) dv(t ) +4 + 4v(t ) = 12u (t ) 2 dt dt t >0 Which is the correct form of the total (natural plus forced) response? The correct answer is d. v f (t ) = K 1 2r 2 + 4r + 4 = 0 r 2 + 2r + 2 = 0 − 2± 4−8 r= = −1 ± j1 2 v n (t ) = K 2 e −t cos t + K 3 e − t sin t v (t ) = K 1 + K 2 e − t cos t + K 3 e − t sin t 14FE-2 If all initial conditions are zero in the network in Fig. 14PFE-2, find the transfer function Vo(s)/Vs(s). The correct answer is c. 4 Vs ( s ) = I ( s ) + 2 sI ( s ) + 2 I ( s ) s 4 Vs ( s ) = + 2 s + 2 I ( s ) s V0 ( s ) = 2 I ( s ) V0 ( s ) 2 4 V ( s ) Vs ( s ) = + 2 s + 2 0 s 2 V0 ( s ) s = 2 Vs ( s ) s + s + 2 I (s) = 14FE-3 The initial conditions in the circuit in Fig. 14PFE-3 are zero. Find the transfer function Io(s)/Is(s). The correct answer is a. s+4 I (s) I 0 (s) = 3 s s+4+ s I 0 (s) s ( s + 4) = 2 I s ( s ) s + 4s + 3 14FE-4 In the circuit in Fig. 14PFE-4, use Laplace transforms to find the current I(s). Assume zero initial conditions and that vs(t)=4cos t u(t). The correct answer is b. 4s 4 KVL: 2 = s + + 2 I (s) s s +1 2 4s I (s) = 2 ( s + 1)( s 2 + 2s + 4) 14FE-5 Assuming that the initial inductor current is zero in the circuit in Fig. 14PFE-5, find the transfer function Vo(s)/Vs(s). The correct answer is d. KVL: Vs ( s ) = 4 I ( s ) + 2 sI ( s ) V s ( s ) = ( 2 s + 4) I ( s ) V0 ( s ) = 2sI ( s ) I (s) = V0 ( s ) 2s V s ( s ) = ( 2 s + 4) V0 ( s ) s = Vs ( s ) s + 2 V0 ( s ) 2s Solutions to FE Problems Chapter 15 15FE-1 Given the waveform in Fig. 15PFE-1, determine if the trigonometric Fourier coefficient an has zero value or nonzero value and why. The correct answer is b. By observation, the waveform has odd symmetry. Therefore, an = 0 for all n due to odd symmetry. 15FE-2 Given the waveform in Fig. 15PFE-2, describe the type of symmetry and its impact on the trigonometric Fourier coefficient bn. The correct answer is d. By observation, the waveform has half-wave symmetry. Therefore, bn =0 for n even due to halfwave symmetry and bn is nonzero for n odd. 15FE-3 Determine the first three nonzero terms of the voltage vo(t) in the circuit in Fig. 15PFE-3 if the input voltage vs(t) is given by the expression 1 ∞ 30 v s (t ) = + ∑ cos 2nt V 2 n =1 nπ The correct answer is a. jω Vs (nω ) V0 (nω ) = 1 + jω V0 (0) = 0 j 2 30 ∠0° V0 (ω ) = 1 + j 2 π V0 (ω ) = 8.54∠26.57°V j 4 30 V0 (2ω ) = ∠0° 1 + j 4 2π V0 (2ω ) = 4.63∠14.04°V j 6 30 ∠0° V0 (3ω ) = 1 + j 6 3π V0 (3ω ) = 3.14∠9.46°V j8 30 V0 (4ω ) = ∠0° 1 + j8 4π V0 (4ω ) = 2.37∠7.13°V v0 (t ) = 8.54 cos (2t + 26.57°) + 4.63 cos(4t + 14.04°) + 3.14 cos(6t + 9.46°) + ... V 15FE-4 Find the average power absorbed by the network in Fig. 15PFE-4. The source voltage is as follows: v s (t ) = 20 + 10 cos(377t + 60°) + 4 cos(1131t + 45°) V . The correct answer is b. Vs I (nω ) = 2 + jn3.77 20 I DC = = 10 A 2 10∠60° I1 = = 2.34∠ − 2.05° A 2 + j 3.77 4∠45° I3 = = 0.35∠ − 34.97° A 2 + j11.31 i(t ) = 10 + 2.34 cos(377t − 2.05°) + 0.35 cos(1131t − 34.97°) A ∞ V I P = V DC I DC + ∑ n n cos(θ vn − θ in ) 2 n =1 10(2.34) 4(0.35) cos(45° + 34.97°) P = 20(10) + cos(60° + 2.05°) + 2 2 P=205.61 W 15FE-5 Find the average value of the waveform shown in Fig. 15PFE-5. The correct answer is b. 2 a0 = 4 1 1 10 dt + ∫ − 2 dt ∫ 40 42 1 [10t ]02 + 1 [ − 2t ]42 4 4 1 1 a 0 = (20) + [− 2(4) − (−2)(2)] 4 4 a 0 = 4V a0 = Solutions to FE Problems Chapter 16 16FE-1 A two-port network is known to have the following parameters: 1 1 1 y11 = S y12 = y 21 = − S y 22 = S 14 21 7 If a 2-A current source is connected to the input terminals as shown in Fig. 16PFE-1, find the voltage across this current source. The correct answer is a. I1 = 2 A , I 2 = 0 I 1 = y11V1 + y12V2 I 2 = y 21V1 + y 22V2 1 1 V1 − V2 = 2 14 21 1 1 − V1 + V2 = 0 21 7 21V1 − 14V2 = 588 − V1 + 3V2 = 0 V1 = 36V V2 = 12V 16FE-2 Find the Thevenin equivalent resistance at the output terminals of the network in Fig. 16PFE-1. The correct answer is b. I 1 = y11V1 + y12V2 I 2 = y 21V1 + y 22V2 y11V1 + y12V2 = 0 V1 = − y12 V2 y11 y I 2 = y 21 − 12 V2 + y 22V2 y11 − y y I 2 = V2 21 12 + y 22 y11 Z TH = V2 y11 = I 2 − y 21 y12 + y11 y 22 Z TH 1 14 = = 9Ω − 1 − 1 1 1 − + 21 21 14 7 16FE-3 Find the Y parameters for the two-port network shown in Fig. 16PFE-3. The correct answer is d. 3 V1 I1 3 56 y11 = = = S V1 V =0 V1 56 2 2 V1 14 = − 2 V 1 4 56 2 − V1 2 1 = 56 = − S = − S V1 56 28 V −V I2 = 2 = 4 y 21 = I2 V1 V2 = 0 0− 16 Ω 3 16 16 4 V = 3 (V2 ) = V2 = V2 16 28 7 +4 3 4 0 − V2 V −V 1 7 I1 = 1 = = − V2 16 16 28 1 − V2 I1 1 y12 = = 28 = − S V1 V =0 V2 28 16 8 = 1 4 V2 − V2 V −V 3 7 I2 = 2 = = V2 4 4 28 3 V I 2 28 2 3 y 22 = = = S V2 V2 28 16FE-4 Find the Z parameters of the network shown in Fig. 16PFE-4. The correct answer is c. The 3Ω and 9Ω are in series. 9(12) 36 z11 = = Ω 9 + 12 7 12 4 I= ( I1 ) = I1 12 + 9 7 4 12 V2 = 3I = 3 I 1 = I 1 7 7 12 I1 V2 12 7 z 21 = = = Ω I 1 I =0 I1 7 2 3 3 1 I= (I 2 ) = I2 = I2 3 + 18 21 7 1 12 V1 = 12 I = 12 I 2 = I 2 7 7 12 I2 V1 12 = 7 = Ω z12 = I 2 I =0 I2 7 1 z 22 = 3(18) 18 = Ω 3 + 18 7 16FE-5 Calculate the hybrid parameters of the network in Fig. 16PFE-5. The correct answer is a. h11 = (2 4) + 8 = 28 Ω 3 4 (I 1 ) 4+2 2 I 2 = − I1 3 I2 = − I h21 = 2 I1 I2 = − = V2 = 0 V2 V = 2 2+4 6 2 I1 3 =−2 I1 3 I h22 = 2 V2 I1 = 0 V2 1 = 6 = S V2 6 V 2 V1 = 4 I 2 = 4 2 = V2 6 3 2 V2 V1 2 3 = = h12 = V2 I = 0 V2 3 1