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B1 Assuming the connecting wires and the battery have negligible resistance, the voltage across the 25- resistance in Figure B1 is

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a 25 V b 60 V c 50 V d 15 V e 125 V

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25

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60 V + _

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Solution: This problem calls for application of the voltage divider rule, discussed in Section 26 Applying the voltage divider rule to the circuit of Figure B2, we have v25 = 60 25 3 + 2 + 25 = 50 V

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Figure B1

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Thus, the answer is c

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2 v 12 V + _ 12 6

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B2 Assuming the connecting wires and the battery have negligible resistance, the voltage across the 6- resistor in Figure B2 is a 6 V b 35 V c 12 V d 4 V e 3 V

Solution: This problem can be solved most readily by applying nodal analysis (Section 31), since one of the node voltages is already known Applying KCL at the node v, we obtain 12 v v v = + 2 6 12 This equation can be solved to show that v = 4 V Note that it is also possible to solve this problem by mesh analysis (Section 32) You are encouraged to try this method as well

Figure B2

B3 A 125-V battery charger is used to charge a 75-V battery with internal resistance of

15 If the charging current is not to exceed 5 A, the minimum resistance in series with

Appendix B

Fundamentals of Engineering (FE) Examination

the charger must be a 10 b 5

c 385

d 415

e 85

Solution: The circuit of Figure B3 describes the charging arrangement Applying KVL to the circuit of Figure B3, we obtain imax R + 15imax 125 + 75 = 0 and using i = imax = 5 A, we can nd R from the following equation: 5R + 75 125 + 75 = 0 R = 85 Thus, e is the correct answer

+ 125 V

15 + imax 75 V

Figure B3

Capacitance and Inductance The material on capacitance and inductance pertains to two basic areas: energy storage in these elements, and transient response of the circuits containing these elements The examples below deal with the former part (covered in 4); the latter is covered under the heading Transients in this appendix, and in 5 of the book

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B4 A coil with inductance of 1 H and negligible resistance carries the current shown

in Figure B4 The maximum energy stored in the inductor is: a 2 J b 05 J c 025 J d 1 J e 02 J Solution: The energy stored in an inductor is W = 1 Li 2 (see Section 41) Since the maximum 2 2 current is 1 A, the maximum energy will be Wmax = 1 Limax = 1 J Thus, b is the correct 2 2 answer

i (A) 1 0 2 4 6 t (ms)

Figure B4

B5 The maximum voltage that will appear across the coil is:

a 5 V b 100 V c 250 V d 500 V e 5,000 V Solution: Since the voltage across an inductor is given by v = L(di/dt), we need to nd the maximum (positive) value of di/dt This will occur anywhere between t = 0 and t = 2 ms The corresponding slope is: di dt =

1 = 500 2 10 3

Therefore vmax = 1 500 = 500 V, and the correct answer is d

AC Circuits AC circuit analysis emphasis pro ciency in the use of complex algebra (see Appendix A for a review of this subject, including sample exercises), and is primarily concerned with the use of AC circuits in electrical power systems The following exercises illustrate typical problems

Appendix B

Fundamentals of Engineering (FE) Examination

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B6 A voltage sine wave of peak value 100 V is in phase with a current sine wave of peak value 4 A When the phase angle is 60 later than a time at which the voltage and the current are both zero, the instantaneous power is most nearly a 250 W b 200 W c 400 W d 150 W e 100 W

Solution: As discussed in Section 71, the instantaneous AC power p(t) is VI VI cos + cos(2 t + V + I ) 2 2 In this problem, when the phase angle is 60 later than a zero crossing, we have V = I = 0, = V I = 0, 2 t = 120 Thus, we can compute the power at this instant as p(t) = p=

100 4 2 2