Exam FE Review Electrical 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.
Dont 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 dont 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 9s Casio 115 MS and 115 MS Plus Texas Instruments TI 30X IIS and TI 30X IIB Texas Instruments TI 36X
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Important Concepts
Voltage (V) electric potential
V = d {Electric field, distance moved between 2 points}
Current (A) flow or movement of charge
I = dQ / dt
Resistance () opposition to flow of charge
R = L / A {in a conductor} Rseries = R1 + R2 + . + Rn Rparallel = (1/R1 + 1/R2 + + 1/Rn)-1
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Important Concepts
Ohms Law: V = IR Joules Law: P = IV = I2R = V2/R Capacitance (F) = ability to store electric charge
C = A / d Cseries = (1/C1 + 1/C2 + + 1/Cn)-1 Cparallel = C1 + C2 + . + Cn ic(t) = C dvc(t) / dt W = CV2
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Important Concepts
Inductance (H) = ability to store magnetic energy
L = N2 A / Lseries = L1 + L2 + . + Ln Lparallel = (1/L1 + 1/L2 + + 1/Ln)-1 vL(t) = L diL(t) / dt W = LI2
DC Transients
T = 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. 2F || 2F = 1F) 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(t+) = 2f f = 1/T Vrms (root mean square) = Vp/2 Z=V/I I=Irmsi V=Vrmsv ZL = jL
Zc = 1/(jC) Q = VrmsIrmssin(v - i) P
v - i
Q S P = VrmsIrmscos(v - i) PF = cos(v - i)
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Important Concepts
Three-phase power
Wye-wye system
Iphase = Iline
Vline = (3)Vphase
Ineutral = 0
Delta-delta system
Vphase = Vline Op-Amps I- = I+ = 0
Iline = (3)Iphase V+ = V9
Vout = -Vin(Rf/Rin) or Rf(Va/Ra + Vb/Rb + )
Sample Problem
What to consider Capacitors in series are mathematically like resistors in parallel 10 pF in series with 10 pF = 5 pF 10 F in series with 10 F = 5F 5 pF in series with 5 F = 5pF
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Answer = A
Sample Problem
What to consider 16F resistor is in parallel with the 60V source
When shorting the source for RTH, 16 F is also shorted RTH = 4 F | | 12 F V16F = 60V V12F which is in series with 4F VTH = 60V[12F /(12F +4F)] = 45V
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Answer = A
Sample to Problem
What 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
Answer = A If Vphase = 0F, then Iphase must be negative
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Sample Problem
What to consider Does the summingpoint constraint apply? Yes
I- = I + = 0 A V- = 0 V
Answer = B
I1kF = I10kF = 2V/1kF Vo = -(10kF)(I1kF)
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Sample Problem
What 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) > 45F angle is positive because imaginary component is positive
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Answer = D
Sample Problem
What 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
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Answer = B
Sample Problem
What to consider After t = 5, the capacitor acts like an open circuit the right 10V source and the 2F resistor are not factors to Vx Use Ohms Law for Ix Use voltage divider for Vx Answer = C
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Sample Problem
What 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. Answer = C
Use Joules Law P = I2R
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Sample Problem
What to consider With terminals shorted, the only resistance in the circuit is the internal resistance of the battery. Use Ohms Law R=V/I
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Answer = A
Sample Problem
What 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
Answer = D
1-e-5 = 0....