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B7 A sinusoidal voltage whose amplitude is 20 2 V is applied to a 5- resistor The
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root-mean-square value of the current is a 566 A b 4 A c 707 A d 8 A Solution: From Section 42, we know that V 20 2 Vrms = = = 20 V 2 2 Thus, Irms = 20/5 = 4 A Therefore, b is the correct answer e 10 A
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B8 The magnitude of the steady-state root-mean-square voltage across the capacitor in the circuit of Figure B5 is a 30 V b 15 V c 10 V d 45 V e 60 V
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Solution: This problem requires the use of impedances (Section 44) Using the voltage divider rule for impedances, we write the voltage across the capacitor as V = 30 0 j 10 10 j 10 + j 10
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Figure B5
= 30 0 ( j 1) = 30 0 1 90 = 30 90 Thus, the rms amplitude of the voltage across the capacitor is 30 V, and a is the correct answer Note the importance of the phase angle in this kind of problem
IS + Es Load D Load A Load B Load C
Figure B6
The next set of questions (Exercises B9 to B28) pertain to single-phase AC power calculations and refer to the single-phase electrical network shown in Figure B6 In this gure, ES = 480 0 V; IS = 100 15 A; = 120 rad/s Further, load A is a bank of single-phase induction machines The bank has an ef ciency ( ) of 80 percent, a power factor of 070 lagging, and a load of 20 hp Load B is a bank of overexcited single-phase synchronous machines The machines draw 15 kVA and the load current leads the line voltage by 30 degrees Load C is a lighting (resistive) load and absorbs 10 kW Load D is a proposed single-phase capacitor that will correct the source power factor to unity This material is covered in Sections 71 and 72
Appendix B
Fundamentals of Engineering (FE) Examination
Check Your Understanding
B9 The root-mean-square magnitude of load A current, IA , is most nearly
a 444 A b 3108 A c 60 A d 3885 A e 555 A Solution: The output power PO of the single-phase induction motor is: PO = 20 746 = 14,920 W The input electric power Pin is: Pin = 14,920 PO = = 18,650 W 080
Pin can be expressed as: Pin = ES IA cos A Therefore, the rms magnitude of the current IA is found as IA = Pin 18,650 = 555015 555 A = ES cos A 480 070
Thus, the correct answer is e
B10 The phase angle of IA with respect to the line voltage ES is most nearly
a 3687 b 60 c 456 d 30 e 48 Solution: The phase angle between IA and ES is: = cos 1 070 = 4557 456 The correct answer is c
B11 The power absorbed by synchronous machines is most nearly
a 20,000 W b 7,500 W c 13,000 W d 12,990 W e 15,000 W Solution: The apparent power, S, is known to be 15 kVA, and is 30 From the power triangle, we have P = S cos Therefore, the power drawn by the bank of synchronous motors is: P = 15,000 cos 30 = 12,99038 1299 kW The answer is d
B12 The power factor of the system before load D is installed is most nearly a 070 lagging b 0866 leading c 0866 lagging d 0966 leading e 0966 lagging
Solution: From the expression for the current IS , we have pf = cos = cos(0 ( 15 )) = cos 15 = 0966 lagging The correct answer is e
B13 The capacitance of the capacitor that will give a unity power factor of the system is most nearly a 219 F b 187 F c 1327 F d 240 F e 1327 pF
Solution: The reactive power QA in load A is: QA = PA tan A A = cos 1 070 = 4557
Appendix B
Fundamentals of Engineering (FE) Examination
Therefore, QA = 18,650 tan 4557 = 19,025 VAR The total reactive power QB in load B is: QB = S sin B = 15,000 sin( 30 ) = 7, 500 VAR The total reactive power Q is: Q = QA + QB = 19,025 7,500 = 11,525 VAR To cancel this reactive power, we set QC = Q = 11,525 VAR and QC =
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