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.

•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

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)

FE supplies equations

You can visit their page

To get one

http://www.ncees.org/exams/study%5Fmaterials/fe%5

Fhandbook

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Electric FieldElectric field due to single charge: E = k k=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

2r

q

0plane 2

E

r

kqV

Capacitance

C =εoA /d, with dielectric C=A/d

Cseries = (1/C1+1/C2 + … + 1/Cn)-1

Cparallel = C1 + C2 + …. + Cn

U = ½CV2 εo =8.85x10-12C2/Nm2

Capacitance C = Q/ΔV, unit: farad [F]

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.

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.

Example 4Determine 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 eachcapacitor is the same as that on equivalent capacitor

a. 0.5x10-5 C

b. 2.8x10-5 C

c. 5.2x10-5 C

d. 7.0x10-5 C

e. 1.1x10-4 C

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. 120μCc. 90μC d. 60μC e. 30μC

21

Direct Current

Ohm’s Law: V = IR, unit: volts [V]

Power: P = I.V = I2R = V2/R, unit: watt [W]

Current (A) – flow of charge Q in time t

I = ΔQ /Δt units: ampere [A]

Current density J = (ne)vd e = 1.6x10-19C

Resistance, unit: ohm [Ω] – opposition to flow of charge

R = ρL / A {in a conductor of length L and area A}

Combination of ResistorsRseries = R1 + R2 + …. + Rn

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

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 Cb. 20 Cc. 2 Cd. 0.005 C

I=Q/t 1A=1C/1s

Q = It Q = 0.1A*20s = 2C

A metallic conductor has a resistivity of 18 106 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

R = * L / A

Resistance

Resistivity

R=18*10-6 Ωm*30m / 3*10-6m2

R = 180

A 60-W light bulb is in a socket supplied with 120 V.

What is the current in the bulb?a. 0.50 Ab. 2.0 Ac. 60 Ad. 7 200 A

P = V*I = V2/R = I2*R

60 = 120*I >> I = 60/120 = 0.5

If a lamp has resistance of 120 when it operates at

100 W, what is the applied voltage?a. 110 Vb. 120 Vc. 125 Vd. 220 V

P = V*I = V2/R = I2*R

100 = V2 /120

V = sqrt(120*100) = 11*10 = 110

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

circuit?

R*0/(R+0) = 0

Example 14 cd

• Req =(1/R2+ 1/R3)-1 + R4 + R

• R eg =(1/10+1/10)-1 + 10 +20 = 35Ω

voltage V across resistors is the same

resistors in parallelI1+I2 = I – Kirkchoff’s second law

Current I splits but 6*I1 = 7*I2 I

I2

I1

The larger the resistor the smaller the current

I

I 2

Rseries = R1 + R2 + …. + Rn

Rseries = is always larger than any of the elements

if R1 and R2 are the same (R) Rseries is 2RCurrent through each resistor is the same.

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

Rparallel = is always smaller than any of the elements

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

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

if R1 and R2 are the same (R) in parallel Rparallel is R/2

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

d. 5.0 V

e. 8.6 V

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) = C(1 – e-t/RC)

= Q(1 – e-t/RC)

is the time constant• = RC

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

Force on a Charge Moving in a Magnetic Field

Force on a charge moving in a magnetic field is givenby equation:

– is the magnetic force q is the charge– is the velocity of the moving charge– is the magnetic field

B q F v B

BF

v

B

The magnitude of the magnetic force on a charged particle is FB = |q| v B sin θ

Charged Particle in Magnetic Field

• Equating the magnetic and centripetal forces:

• Solving for r:

2

B

mvF qvB

r

mvr

qB

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

Inductance, Inductors

Lseries = L1 + L2 + …. + Ln

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

Potential across inductor: vL(t) = L diL(t) / dt

Inductance, unit: henry [H] = ability to store magnetic energy

L = N2 μA / ℓ UM = ½LI2

A circuit element that has a large self-inductance is called an inductor. The circuit symbol is

DC current source – keeps constant current flowing out in the direction shown

DC voltage source – keeps constant potential between + and – side of battery

Symbols

AC source V(t) = V0sin(t) or I(t) = I0sin(t)

R = infinity R = 0

Complex Numbersrectangular 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

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

22L CZ R X X

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∟0 I = Irms∟θ

Impedance in rectangle form:

Z =R+jX X=XL-Xc Xc = 1/(ωC) XL = ωL

AC Circuits

R = Zcosθ

X = Zsin θ

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)

Example A series RLC circuit hasR = 425 Ω, L = 1.25 HC = 3.5 μF. It is connectedto and AC source withf = 60 Hz and Vmax = 150 Va. Find the impedance ofthe circuit.b. Find the phase angle.c. Find the current in the circuit.

Example

A series RLC circuit hasR = 425 Ω, L = 1.25 HC = 3.5 μF. It is connectedto and AC source withf = 60 Hz and Vmax = 150 VCalculate the average real and reactive powerdelivered to the circuit.

sin(t)sin(t+)sin(t-)

sin(t) sin(t+)

Blue leads the red

or

Red lags the bluesin(t-) sin(t)blue argument is always larger than red one

Sample Problem

Read from the plot:

Amplitude of i(t) Io= 50 A

Answer = B

Irms=50*0.71=35

v(t) = sin(t)i(t) = sin(t-90)

current lags voltage by 900

Sample Problem

Angle (phase) from tan()=4/3

= tan-1(4/3)

Answer = D

Magnitude = 5 from Pythagoram principle

Tan>1 so angle > 450

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

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)

Sample Problem

Answer = A

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

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

Sample Problem

Average of any sin(t) = 0 so ignore the AC Source

Answer = C

For DC current inductor resistance is zero (made of copper wire)the battery and 10 resistor are shorted by the 2 H inductance. The

current is Iavg = 12/10

Sample Problem

+ Two 4 resistors are in parallel = 2 . Then 2 and 2 resistors Are in series.I = 40/4

Answer = C

Sample Problem

Power in AC circuits is calculated using rms values (this is why the rms was introduced)

rms value is 20*0.7 = 14. P=I2R = 14250 = 10kW.

Answer = B

20 is the amplitude

Sample Problem

Answer = C

After t = 5, the capacitor acts like an open circuit

Use Ohm’s Law for Ix