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FE Exam ReviewElectrical Circuits

LT Aaron Still

P.E. in Electrical Engineering

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Exam Strategies• http://www.ncees.org/exams/fundamentals/• Only 4 minutes per problem.

– Don’t dwell on a problem.– If it looks like it may take > 4 minutes, then there is

probably a trick (e.g. short, open, extra info).• Do the ones you know. Make an educated

guess at the ones you don’t know.• Answers are typically in SI unit. Set your

calculator to engineering notation.• Pay attention to units (degrees vs. radians)• On multiple answer problems, look for the

unique set of answers.

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Exam Strategies• Know your calculator. Start using your

exam calculator now.

• The following are allowed for the FE exam:Hewlett Packard – HP 33s Hewlett Packard – HP 9sCasio – 115 MS and 115 MS PlusTexas Instruments – TI 30X IIS and TI 30X IIBTexas Instruments – TI 36X

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Important Concepts• Voltage (V) – electric potential

V = ε●d {Electric field, distance moved between 2 points}

• Current (A) – flow or movement of chargeI = dQ / dt

• Resistance (Ω) – opposition to flow of chargeR = ρL / A {in a conductor}

Rseries = R1 + R2 + …. + Rn

Rparallel = (1/R1 + 1/R2 + … + 1/Rn)-1

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Important Concepts

• Ohm’s Law: V = IR

• Joule’s Law: P = IV = I2R = V2/R

• Capacitance (F) = ability to store electric charge

C = εA / d

Cseries = (1/C1 + 1/C2 + … + 1/Cn)-1

Cparallel = C1 + C2 + …. + Cn

ic(t) = C dvc(t) / dt

W = ½CV2

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Important Concepts• Inductance (H) = ability to store magnetic

energyL = N2 μA / ℓ

Lseries = L1 + L2 + …. + Ln

Lparallel = (1/L1 + 1/L2 + … + 1/Ln)-1

vL(t) = L diL(t) / dtW = ½LI2

• DC TransientsT = RC or T = L / R

v(t) = VF + (VI – VF)e-t/T

i(t) = IF + (II – IF)e-t/T

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Important Concepts• Circuit Analysis: Know 1 or 2 well and use

them– By Inspection (i.e. 2 || 2 = 1)– KVL– KCL– Nodal Analysis (KCL for each node)– Mesh Analysis– Superposition (for multiple sources)– Thevenin Circuits & Max Power to load

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Important Concepts• AC Circuits

v(t) = Vp sin(ωt+θ) ω = 2πf f = 1/T

Vrms (root mean square) = Vp/√2

Z = V / I V=Vrms∟θv I=Irms∟θi

Zc = 1/(jωC) ZL = jωL

P = VrmsIrmscos(θv - θi) PF = cos(θv - θi)

Q = VrmsIrmssin(θv - θi)

S = P + jQ S2 = P2 + Q2Q

P

S

θv - θi

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Important Concepts

• Three-phase power– Wye-wye system

Iphase = Iline Vline = (√3)Vphase Ineutral = 0

– Delta-delta system

Vphase = Vline Iline = (√3)Iphase

• Op-AmpsI- = I+ = 0 V+ = V-

Vout = -Vin(Rf/Rin) or –Rf(Va/Ra + Vb/Rb +…)

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Sample ProblemWhat to consider• Capacitors in series

are mathematically like resistors in parallel

• 10 pF in series with 10 pF = 5 pF

• 10 µF in series with 10 µF = 5µF

• 5 pF in series with 5 µF = 5pF Answer = A

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Sample ProblemWhat to consider• 16 resistor is in parallel

with the 60V source When shorting the source

for RTH, 16 is also shorted

RTH = 4 | | 12 V16 = 60V

V12 which is in series with 4 VTH = 60V[12 /(12 +4)]

= 45VAnswer = A

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Sample ProblemWhat to consider• Magnitude of a

phasor is always the RMS value Irms = .7071 x Ipeak

• The peak for the current occurs after the peak of the voltage I lags V If Vphase = 0, then Iphase

must be negativeAnswer = A

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Sample ProblemWhat to consider• Does the summing-

point constraint apply? Yes I- = I+ = 0 A

V- = 0 V

• I1k = I10k = 2V/1k Vo = -(10k)(I1k)Answer = B

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Sample ProblemWhat to consider• Magnitude: square root

of the sum of the squares

(32 + 42)0.5 = 5• Phase angle: inverse

tangent of imaginary component over real component tan-1(4/3) > 45 angle is positive because imaginary component is positive

Answer = D

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Sample ProblemWhat to consider• Magnitudes:

multiple or divide• Angles: add or

subtract• Note: if you

calculate |z1|/|z2|, you get 1.49 so (b) is the answer with no further work

Answer = B

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Sample ProblemWhat to consider• After t = 5Τ, the

capacitor acts like an open circuit

the right 10V source and the 2 resistor are not factors to Vx

• Use Ohm’s Law for Ix

• Use voltage divider for Vx

Answer = C

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Sample ProblemWhat to consider• Total power

dissipated is due to the voltage, current, and resistance of the power line only.The power lines rating

(10kV) and length(5 km) are extra info.

• Use Joule’s Law P = I2R

Answer = C

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Sample ProblemWhat to consider• With terminals

shorted, the only resistance in the circuit is the internal resistance of the battery.

• Use Ohm’s Law

R = V / I

Answer = A

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Sample ProblemWhat to consider• Recall it takes ~ 5T for

a capacitor to fully charge.

• T = RC = 15 ms

5T = 75 ms• Only 24 ms > T so it must

be the answer

• vc(t) = Vf – (Vi-Vf)e-t/T

• 1-e-5 = 0.993• 1-e-x = 0.8 x = 1.6TAnswer = D

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Sample ProblemWhat to consider• Inductances are

mathematic like resistors in series and in parallel.

• Leq = L1 + L2

• IL = 10 A from current source

• Energy stored in an inductor:W(Joules) = 0.5(L*I2)

• Recall energy stored in a capacitor:W(Joules) = 0.5(C*V2)

Answer = B

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Sample ProblemWhat to consider

• Iavg(AC) = 0 so ignore the AC Source (short it)

• A fully-stored inductor has current but no voltage the 5 resistor is shorted out.

• Use Ohm’s Law:

I = V/R = 12V/10

Answer = C

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Sample ProblemWhat to consider

• Find Req

= 2 + 4 ||4• Use Ohm’s Law:

I = V/Req

Answer = C

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Sample ProblemWhat to consider

• Pavg = I2rmsR

= I2R/2

= (20 A)2 * 50/2= 10000 W

= 10 kW

Answer = B

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Sample ProblemWhat to consider

• Iline = Vline / (√3) Xline

• Xline = √(3002+4002)

• In a balance wye-connected circuit,

Iline = Iphase

• In a balance three=phase system, Ineutral ≡ 0

Answer = B

Answer = B

Answer = A

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Sample ProblemWhat to consider• The input current into an

op-amp is so small that it is negligible 0 A

• if = v1/R1 + v2/R2 = 0.875A

• vo = - if Rf

Answer = A

Answer = C