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2 ES 1 and XC = XC C
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Therefore, the capacitance required to obtain a power factor of unity is: C= QC 11,525 = = 1327 F 2 120 4802 ES
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The correct answer is c
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Transients The FE Exam is primarily focused on rst-order transients, as covered in 5 The concepts of initial and nal value and of time constant are essential to the solution of problems of this nature
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t=0 12 V _ 4H 2
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B14 The expression for the current in the 2- resistor in Figure B7 for time greater than zero is: a 3e 05t + 3 A b 3e 05t + 3 A c 3e05t + 3 A d 6e05t + 6 A e 6e 05t + 6 A
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Solution: Applying KVL to the circuit when the switch is closed (t 0), we have di + 2i = 12 dt Solving the differential equation: 4 i(t) = Ke 05t + 6 A i(0 ) = i(0+ ) = 0 Therefore, i(t) = 6e 05t + 6 A t 0 t 0
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Therefore, the correct answer is d
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Appendix B
Fundamentals of Engineering (FE) Examination
Three-Phase Circuits Three-phase circuits are covered in 7 The FE exam requires pro ciency in the analysis of simple delta and wye circuits, as illustrated in the following examples
Check Your Understanding
B15 A 3-phase circuit is shown in Figure B8 Load resistors (66 ) are connected in delta and supplied by a 220-volt balanced three-phase source through three lines of 2-ohm resistance The magnitude of the root-mean-square, line-to-line voltage across each 66-ohm resistor is most nearly a 198 V b 110 V c 201 V d 220 V e 120 V
Solution: Since the load in this problem is -connected, it must rst be converted to an equivalent Y-form The phase impedance of the -connected load is 66 , so the equivalent phase impedance of the corresponding Y-form is: 66 Z = = 22 3 3 The phase voltage is: ZY = 208 V = 0 V 3 The per-phase equivalent circuit of this problem is shown in Figure B9 The load voltage is obtained by the voltage divider rule: VL = 22 208 22 20167 V = = 0 V 2 + 22 3 24 3 resistor therefore is:
Figure B9
2 220 Vrms 66 2 2 66 66
Figure B8
22
The rms line-to-line voltage across the 66 V66 = 3VL = 20167 0 V The correct answer is c
B16 A three-phase load is composed of three impedance of 90 + j 90 ohms and connected in wye The balanced three-phase source is 208 volt (line to line) The current in each line is most nearly a 40 A b 163 A c 133 A d 9 A e 6 A
Solution: The phase voltage is 208/ 3 Therefore, the magnitude of phase current Ian is Ian 208 1 = = 943 A = 9 + j9 3 1273
208 3
VAB VBC
The line current is equal to the phase current in a Y-connected system The answer is d
Figure B10
IA + VAB B + VBC C A IAB IB IC
The next four questions refer to a three-phase system with line-to-line voltage of 220 V rms, with ABC phase sequence, and with phase reference VAB shown in the phase diagram of Figure B10 The load is a balanced delta connection, shown in Figure B11 with branch impedances Z = 30 j 40 ohms, j = 1
IBC ICA
Figure B11
Appendix B
Fundamentals of Engineering (FE) Examination
Check Your Understanding
B17 The phase current is most nearly a 44 5313 A b 24 5313 A c 44 0 A d 44 5313 A e 24 5313 A
Solution: The load current IAB is given by IAB = 220 0 220 0 VAB = = = 44 5313 A Z 30 j 40 50 5313
The correct answer is a
B18 The line current IA (in amperes) is most nearly a 44 18687 b 44 23 c 7 d 76 23 e 7 18687
IC ICA IAB IB IBC IBC IAB IA ICA
Solution: The current phasor diagram for this -connected system is shown in Figure B12 From the relationship among three-phase currents, we have ICA = IAB 240 = 44 (5313 240 ) = 44 18687 A From the phasor diagram, we have IA = IAB ICA = 44 5313 44 18687 = 264 + j 352 ( 437 + j 053) = 762 231 A Therefore, the correct answer is d
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