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FE Exam Review Electrical Circuits

FE Exam Review Electrical Circuits.

harper-johnston
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FE Exam Review Electrical Circuits

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  1. FE Exam ReviewElectrical Circuits The FE exam consists of 180 multiple-choice questions. During the morning session, all examinees take a general exam common to all disciplines. During the afternoon session, examinees can opt to take a general exam or a discipline-specific (chemical, civil, electrical, environmental, industrial, or mechanical) exam. See exam specifications for more details.

  2. XI. Electricity and Magnetism 9% • A. Charge, energy, current, voltage, power • B. Work done in moving a charge in an electric field (relationship between voltage and work) • C. Force between charges • D. Current and voltage laws (Kirchhoff, Ohm) • E. Equivalent circuits (series, parallel) • F. Capacitance and inductance • G. Reactance and impedance, and admittance • H. AC circuits • I. Basic complex algebra

  3. Exam Strategies • Only 4 minutes per problem. • Don’t dwell on a problem. • 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)

  4. FE supplies equations • You can visit their page • To get one • http://www.ncees.org/exams/study%5Fmaterials/fe%5Fhandbook/

  5. Electric Field Electric field due to single charge: E = kk=8.89x109Nm2/C2 Uniform electric field due to uniform distribution of surface charge: Electric potential due to single charge Potential difference in uniform electric field: ΔV = E●d Potential energy : ΔU = qΔV Charge in uniform electric field F =qE qE = ma qΔV= Kf -Ki

  6. Capacitance Capacitance C = Q/ΔV, unit: farad [F] • C =εoA /d, with dielectric C=keoA/d • Cseries = (1/C1+1/C2 + … + 1/Cn)-1 • Cparallel = C1 + C2 + …. + Cn • U = ½CV2 εo =8.85x10-12C2/Nm2

  7. Example 1 Charges Q, -Q = 2 nC are placed at the vertices of an equilateral triangle with side a = 2 cm as shown. Find the magnitude of electric force on charge q = 6 nC placed at point A.

  8. Example 2,3 2. An electron with a speed of 5 x106 m/s i enters an uniform electric field E =1000 N/C i. a. How long will it take for the electron to come to stop? qe = 1.6x10-19 C me = 9.11x10-31kg 3. Find the potential difference needed for the electron to obtain a speed of 3x107 m/s.

  9. Example 4 Determine the charge on capacitor C1 when C1=10 µF, C2=12 µF, C3= 15 µF, Ceq= 4μF and V0=7 V. (Hint: If capacitors are connected in series, then charge on each capacitor is the same as that on equivalent capacitor • 0.5x10-5 C • 2.8x10-5 C • 5.2x10-5 C • 7.0x10-5 C • 1.1x10-4 C

  10. Example 5 Determine the charge stored in C2 when C1 = 15 µF, C2 = 10 µF, C3 = 20 µF, and V0 = 18 V. Hint: find the equivalent capacitor first a. 180μC b. 120μC c. 90μC d. 60μC 30μC

  11. Direct Current Current (A) – flow of charge Q in time t I= ΔQ /Δt units: ampere [A] Current density J = (ne)vd e = 1.6x10-19C Ohm’s Law: V = IR, unit: volts [V] • Resistance, unit: ohm [Ω] – opposition to flow of charge • R = ρL / A {in a conductor of length L and area A} Power: P = I.V = I2R = V2/R, unit: watt [W]

  12. Combination of Resistors Rseries = R1 + R2 + …. + Rn Rparallel = (1/R1 + 1/R2 + … + 1/Rn)-1

  13. A wire carries a steady current of 0.1 A over a period of 20 s. What total charge passes through the wire in this time interval? a. 200 C b. 20 C c. 2 C d. 0.005 C I=Q/t 1A=1C/1s Q = It Q = 0.1A*20s = 2C

  14. A metallic conductor has a resistivity of 18  106 m. What is the resistance of a piece that is 30 m long and has a uniform cross sectional area of 3.0 mm2? a. 0.056  b. 180  c. 160  d. 90  Resistivity R = r * L / A Resistance R=18*10-6 Ωm*30m / 3*10-6m2 R = 180 

  15. A 60-W light bulb is in a socket supplied with 120 V. What is the current in the bulb? a. 0.50 A b. 2.0 A c. 60 A d. 7 200 A P = V*I = V2/R = I2*R 60 = 120*I >> I = 60/120 = 0.5

  16. If a lamp has resistance of 120  when it operates at 100 W, what is the applied voltage? a. 110 V b. 120 V c. 125 V d. 220 V P = V*I = V2/R = I2*R 100 = V2 /120 V = sqrt(120*100) = 11*10 = 110

  17. 14. If R1 =R2=R3=R4 = 10Ω and R = 20Ω, what is the equivalent resistor of the circuit?

  18. R*0/(R+0) = 0

  19. Example 14 cd • Req =(1/R2+ 1/R3)-1 + R4 + R • R eg =(1/10+1/10)-1 + 10 +20 = 35Ω

  20. I1 I I2 I2 I resistors in parallel I1+I2 = I – Kirkchoff’s second law voltage V across resistors is the same Current I splits but 6*I1 = 7*I2 The larger the resistor the smaller the current

  21. Rseries = R1 + R2 + …. + Rn • Rseries = is always larger than any of the elements • if R1 and R2 are the same (R) Rseries is 2R • Current through each resistor is the same.

  22. Rparallel = (1/R1 + 1/R2 + … + 1/Rn)-1 • Rparallel = is always smaller than any of the elements

  23. Rparallel = (1/R1 + 1/R2 + … + 1/Rn)-1 • if either of R1, R2, and, … Rn is 0 (wire or closed switch) while in parallel Rparallel is 0 0

  24. if R1 and R2 are the same (R) in parallel • Rparallel is R/2 • Rparallel = (1/R1 + 1/R2 + … + 1/Rn)-1

  25. Example What is the magnitude of the potential difference across the 20-Ω resistor? a. 3.2 V b. 7.8 V c. 11 V • 5.0 V • 8.6 V

  26. Charging a Capacitor • At the instant the switch is in position a the charge on the capacitor is zero, the capacitor starts to charge. The capacitor continues to charge until it reaches its maximum charge (Q = Cε) • Once the capacitor is fully charged, the current in the circuit is zero. • Once the maximum charge is reached, the current in the circuit is zero • The potential difference across the capacitor matches that supplied by the battery • The charge on the capacitor varies with time • q(t) = Ce(1 – e-t/RC) = Q(1 – e-t/RC) t is the time constant •  = RC

  27. Discharging a Capacitor in an RC Circuit • When a switch is thrown from a to b the charged capacitor C can discharge through resistor R • q(t) = Qe-t/RC • The charge decreases exponentially

  28. Force on a Charge Moving in a Magnetic Field Force on a charge moving in a magnetic fieldis given by equation: • is the magnetic force q is the charge • is the velocity of the moving charge • is the magnetic field The magnitude of the magnetic force on a charged particle is FB = |q| v B sin θ

  29. Charged Particle in Magnetic Field • Equating the magnetic and centripetal forces: • Solving for r:

  30. Mass Spectrometer • Example: The magnetic field in the deflection chamber has a magnitude of 0.035 T. Calculate the mass of a single charged ion if the radius r of the its path in the chamber is 0.278 m and its velocity is 7.14x104m/s

  31. Inductance, Inductors Inductance, unit: henry [H] = ability to store magnetic energy A circuit element that has a large self-inductance is called an inductor. The circuit symbol is Potential across inductor: vL(t) = L diL(t) / dt L = N2μA / ℓ UM = ½LI2 Lparallel = (1/L1 + 1/L2 + … + 1/Ln)-1 Lseries = L1 + L2 + …. + Ln

  32. R = 0 R = infinity DC current source – keeps constant current flowing out in the direction shown Symbols DC voltage source – keeps constant potential between + and – side of battery AC source V(t) = V0sin(wt) or I(t) = I0sin(wt)

  33. Complex Numbers rectangular form z=a+jb, z=zcosθ+jzsinθ) phasor form z=c/θ c = (a2+b2)½ θ = tan-1(b/a) z1+z2 = (a1+a2)+j(b1+b2) z1·z2= c1·c2/(θ1+θ2)z1/z2=c1/c2/(θ1-θ2 ) AC circuits: impedance Z=R+jX In series Zeq= (R1+R2)+j(X1+X2) In parallel Zeq=[1/(R1 +jX1)+1/(R2 +jX2)]-1

  34. AC Circuits • The instantaneous voltage would be given by v = Vmax sin ωt • The instantaneous current would be given by i = Imax sin (ωt - φ) • φ is the phase angle,Imax= Vmax /Z Z is called the impedance of the circuit and it plays the role of resistance in the circuit, where Impedance has units of ohms X – reactance of the circuit; X=ωL – 1/ωC XL = ωL XC = 1/ωC

  35. AC Circuits • Root mean square value of V and I is given by expressions: • Vrms = Vmax/√2 , Irms = Imax/√2 • Z = V / I θ =tan-1(X/R) • V = Vrms sin ωt, I = Irmssin (ωt+ θ) in phasor form • V=Vrms∟0I =Irms∟θ • Impedance in rectangle form: • Z =R+jX X=XL-Xc Xc = 1/(ωC) XL = ωL

  36. AC Circuits R = Zcosθ X = Zsin θ

  37. AC Circuits Power can be expressed in rectangle form: S = P + jQ P- real power Q–reactive power P=VrmsIrmscos(θ) =I2rmsR Q = VrmsIrmssin(θ) = V2rms/X S2 = P2 + Q2 power factor PF= cos(θ)

  38. Example A series RLC circuit has R = 425 Ω, L = 1.25 H C = 3.5 μF. It is connected to and AC source with f = 60 Hz and Vmax = 150 V a. Find the impedance of the circuit. b. Find the phase angle. c. Find the current in the circuit.

  39. Example A series RLC circuit has R = 425 Ω, L = 1.25 H C = 3.5 μF. It is connected to and AC source with f = 60 Hz and Vmax = 150 V Calculate the average real and reactive power delivered to the circuit.

  40. sin(wt-q) sin(wt) sin(wt+q)

  41. Blue leads the red • or • Red lags the blue sin(wt) sin(wt+q) blue argument is always larger than red one sin(wt-q) sin(wt)

  42. Sample Problem Read from the plot: Amplitude of i(t) Io= 50 A Irms=50*0.71=35 v(t) = sin(wt) i(t) = sin(wt-90) current lags voltage by 900 Answer = B

  43. Sample Problem Magnitude = 5 from Pythagoram principle Angle (phase) from tan(q)=4/3 q = tan-1(4/3) Tan>1 so angle > 450 Answer = D

  44. Sample Problem Information 10 kV power line is useless. It is not the potential difference between two ends of the wire. You must use P = I2R to calculate the power dissipated. Answer = C

  45. Sample Problem • For AC circuit with Vrms=115V, Irms = 20.1A and phase constant θ=320, find the average real power and average reactive power drawn by the circuit. P = 115V*20.1Acos320 = 1965 W Q = 115V*20.1Asin 320 = 1217 kVAR ( kilovolt-amps reactive)

  46. Sample Problem Answer = A

  47. Sample Problem Time constant of the circuit is t = RC = 15 ms. Time constant is the time to charge capacitor to 63%. [1- e-1]. To charge more (80%) you need more time. Answer = D

  48. Sample Problem Inductances are like resistors in series and in parallel. Lseries = L1 + L2 Energy stored in an inductor: W[J] = 0.5L*I2 IL = 10 A from current source • Answer = B

  49. Sample Problem Average of any sin(wt) = 0 so ignore the AC Source For DC current inductor resistance is zero (made of copper wire) the battery and 10 W resistor are shorted by the 2 H inductance. The current is Iavg = 12/10 Answer = C

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