FE Exam Review
Electrical 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%5Fmaterial
s/fe%5Fhandbook/
Electric Field
Electric field due to single charge: E = k
k=8.89x109Nm2/C2
q
r2
Uniform electric field due touniform distribution of
surface charge: Eplane  2
0
kq
V

Electric potential due to single charge
r
Potential difference in uniform electric field: ΔV = E●d
Potential energy : ΔU = qΔV
Charge in uniform electric field
qΔV= Kf -Ki
F =qE
qE = ma
Capacitance
Capacitance C = Q/ΔV, unit: farad [F]
C =εoA /d, with dielectric C=koA/d
Cseries = (1/C1+1/C2 + … + 1/Cn)-1
Cparallel = C1 + C2 + …. + Cn
U = ½CV2 εo =8.85x10-12C2/Nm2
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 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
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
1
2
Example 5
a.
b.
c.
d.
e.
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
180μC
120μC
90μC
60μC
30μC
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]
Combination of Resistors
Rseries = 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 C
I=Q/t
1A=1C/1s
b. 20 C
c. 2 C
Q = It Q = 0.1A*20s = 2C
d. 0.005 C
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?
Resistivity
a. 0.056 
b. 180 
R =r * L/A
c. 160  Resistance
R=18*10-6 Ωm*30m /
d. 90 
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 A
b. 2.0 A
P = V*I = V2/R = I2*R
c. 60 A
d. 7 200 A
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 V
b. 120 V
P = V*I = V2/R = I2*R
c. 125 V
d. 220 V
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
resistors in parallel
I1+I2 = I – Kirkchoff’s second law
I2
voltage V across resistors is the same
I2
I1
I
Current I splits but 6*I1 = 7*I2
The larger the resistor the smaller the current
I
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.
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
a.
b.
c.
d.
e.
What is the magnitude of the potential
difference across the 20-Ω resistor?
3.2 V
7.8 V
11 V
5.0 V
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)
t is the time constant
• t = 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 given
by equation:
FB  qv  B
– FB is the magnetic force
q is the charge
– v is the velocity of the moving charge
– B is the magnetic field
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:
mv
FB  qvB 
r
• Solving for r:
mv
r
qB
2
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
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 = N2 μA / ℓ
UM = ½LI2
Lparallel = (1/L1 + 1/L2 + … + 1/Ln)-1
Lseries = L1 + L2 + …. + Ln
Symbols
DC current source – keeps constant
current flowing out in the direction shown
DC voltage source – keeps constant
potential between + and – side of
battery
AC source V(t) = V0sin(wt) or I(t) = I0sin(wt)
R=0
R = infinity
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
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
Z  R   X L  XC 
2
2
Impedance has units of ohms
X – reactance of the circuit; X=ωL – 1/ωC
XL = ωL XC = 1/ωC
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 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.
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.
sin(wt-q)
sin(wt)
sin(wt+q)
Blue leads the red
or
Red lags the blue
sin(wt) sin(wt+q)
sin(wt-q) sin(wt)
blue argument is always larger than red one
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
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
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(wt) = 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
20 is the amplitude
Answer = B
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.
Sample Problem
After t = 5t,
the capacitor
acts like an
open circuit
Use Ohm’s
Law for Ix
Answer = C