Figure B12

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B19 The power factor is most nearly

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a 10 b 06 leading c 0866 leading d 0 e 08 lagging Solution: The impedance angle is: = tan 1 40 = 5313 30

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Therefore, the power factor is: pf = cos = 06, leading We can also get the answer directly from the expression for the current IAB The correct answer is b

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B20 The total real power delivered from the source to the load is most nearly a 1496 W b 580 W c 1742 W d 2904 W e 850 W

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Solution: The total power P delivered to the balanced load is: P = 3VAB IAB cos = 3 220 44 06 = 17424 1742 W The answer is c

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Diode Applications 8 covers all of the diode material that is relevant to the FE exam The following examples illustrate typical questions

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Appendix B

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Fundamentals of Engineering (FE) Examination

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Check Your Understanding

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B21 The circuit of Figure B13 is a a Peak detector b Half-wave recti er d Voltage doubler e Full-wave recti er

Solution: The correct answer is e c Bridge recti er

L vin C vout C

B22 The inductor L and the capacitor C serve the function of

a b c d e Converting the AC input to DC output Increasing the peak value of the output voltage Protecting the diodes A high-pass lter Reducing the ripple component of the output voltage

Figure B13

Solution: The correct answer is e

B23 The ideal diode D in Figure B14 will always conduct if:

a V1 is greater than V2 c V1 is greater than 1 V e R1 is an open circuit b V2 is greater than V1 d R2 is an open circuit

R1 V1

i D V2 R2

Solution: Using the methodology developed in 8, we assume that D conducts, resulting in the following expression for the diode current i= V1 V2 R1

Figure B14

This expression will result in a positive current only if V1 is greater than V2 When this condition applies, the assumption that the diode conducts is correct; thus, a is the correct answer

Operational Ampli ers The coverage of operational ampli ers in this book is well beyond the scope of the FE Exam Sections 121 to 123 cover all of the required material on ideal op-amps

Check Your Understanding

B24 In the circuit of Figure B15, which value is closest to v3 if R1 = 22 k , R2 = 15 k , R3 = 18 k , v1 = 120 mV and v2 = 40 mV a 250 mV b 500 mV c 500 mV d 146 V e 146 V

Solution: Using the summing ampli er equation in 12, we calculate: v3 = 18 R3 R3 18 (120) ( 40) = 5018 mV v1 v2 = R1 R2 22 15

R3 R1 v1 v2 R2 v3 +

Figure B15

Thus, the nearest answer is c

B25 In Figure B15, if R1 = 22 k , R3 = 18k , v1 = 120 mV and v2 = 40 mV, choose the value of R2 such that v3 = 0 a 12 k b 5 k c 733 k d 0733 k e 05 k

Appendix B

Fundamentals of Engineering (FE) Examination

Solution: Using the summing ampli er equation in 12, we calculate: v3 = 18 R3 R3 18 (120) v1 v2 or 0 = ( 40) R1 R2 22 R2

40 120 = 22 R2 22 40 = 0733 k 120 Thus, the nearest answer is d R2 =

Electric and Magnetic Fields Some of the basic ideas on electric elds, voltage, charge, and work are covered in 2 Faraday s law and other magnetic eld concepts are covered in 16 Two examples are given below

Check Your Understanding

B26 Which of the following is a true characteristic of magnetic ux lines

a b c d e They cross each other They begin and end on electric charges They are parabolic They are continuous None of the above

Solution: As discussed in 15, magnetic ux lines are continuous Thus, d is the correct answer area = 5 in2 , and the mean ux path length = 2 in, the total reluctance R of the magnetic circuit in (A t in2 )/Wb is a 1 105 b 2 105 c 15 105 d 35 104 e 2 105 Solution: From 16, the relationship between magnetomotive force and ux is