FE Exam Review Electrical Circuits

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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.
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• 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

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Exam Strategies

• Only 4 minutes per problem.
• Dont dwell on a problem.
• 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)

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FE supplies equations You can visit their page To
get one http//www.ncees.org/exams/study5Fmateria
ls/fe5Fhandbook/
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Electric Field
Electric field due to single charge E k
k8.89x109Nm2/C2 Uniform electric field due
to uniform distribution of surface charge
Electric potential due to single charge
Potential difference in uniform electric field
?V E?d Potential energy ?U q?V Charge in
uniform electric field F qE qE ma q?V
Kf -Ki
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Capacitance
Capacitance C Q/?V, unit farad F
C eoA /d, with dielectric CkeoA/d Cseries
(1/C11/C2 1/Cn)-1 Cparallel C1 C2 .
Cn U ½CV2 eo 8.85x10-12C2/Nm2
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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.

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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.

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Example 4

• Determine the charge on capacitor C1 when C110
  µF,
• C212 µF, C3 15 µF, Ceq 4µF and V07 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

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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

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Direct Current
Current (A) flow of charge Q in time t I ?Q
/?t units ampere A Current density J
(ne)vd e 1.6x10-19C
Ohms Law V IR, unit volts V
Resistance, unit ohm O opposition to flow of
charge R ?L / A in a conductor of length L and
area A
Power P I.V I2R V2/R, unit watt W
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Combination of Resistors

• Rseries R1 R2 . Rn
• Rparallel (1/R1 1/R2 1/Rn)-1

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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
IQ/t 1A1C/1s
Q It Q 0.1A20s 2C
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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
R1810-6 Om30m / 310-6m2
R 180 ?
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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 VI V2/R I2R
60 120I gtgt I 60/120 0.5
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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 VI V2/R I2R
100 V2 /120
V sqrt(120100) 1110 110
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14. If R1 R2R3R4 10O and R 20 O, what is
the equivalent resistor of the circuit?
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(No Transcript)
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R0/(R0) 0
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Example 14 cd

• Req (1/R2 1/R3)-1 R4 R
• R eg (1/101/10)-1 10 20 35O

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resistors in parallel I1I2 I Kirkchoffs
second law
voltage V across resistors is the same
Current I splits but 6I1 7I2
The larger the resistor the smaller the current
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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.
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Rparallel (1/R1 1/R2 1/Rn)-1
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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
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Rparallel (1/R1 1/R2 1/Rn)-1
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Example

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

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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 Ce)
• 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

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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

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Force on a Charge Moving in a Magnetic Field

• Force on a charge moving in a magnetic field is
  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 ?
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Charged Particle in Magnetic Field

• Equating the magnetic
• and centripetal forces
• Solving for r

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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

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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
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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)
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Complex Numbers

• rectangular form zajb, zzcos?jzsin?)
• phasor form zc/? c (a2b2)½ ?
  tan-1(b/a)
• z1z2 (a1a2)j(b1b2)
• z1z2 c1c2/(?1?2) z1/z2c1/c2/(?1-?2 )
• AC circuits impedance ZRjX
• In series Zeq (R1R2)j(X1X2)
• In parallel Zeq1/(R1 jX1)1/(R2 jX2)-1

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AC Circuits

• The instantaneous voltage would be given by v
  Vmax sin ?t
• The instantaneous current would be given by i
  Imax sin (?t - f)
• f 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?L 1/?C
• XL ?L XC 1/?C

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AC Circuits
Root mean square value of V and I is given by
expressions Vrms Vmax/v2 , Irms Imax /v2
Z V / I ? tan-1(X/R) V
Vrms sin ?t, I Irmssin (?t ?) in phasor
form VVrms?0 I Irms?? Impedance in
rectangle form Z RjX XXL-Xc
Xc 1/(?C) XL ?L
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AC Circuits

• R Zcos?
• X Zsin ?

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AC Circuits

• Power can be expressed in rectangle form
• S P jQ
• P- real power Qreactive power
• PVrmsIrmscos(?) I2rmsR
• Q VrmsIrmssin(?) V2rms/X S2 P2 Q2
• power factor PF cos(?)

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Example

• A series RLC circuit has
• R 425 O, 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.

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Example

• A series RLC circuit has
• R 425 O, 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.

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sin(wt)
sin(wtq)
sin(wt-q)
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• Blue leads the red
• or
• Red lags the blue

sin(wt) sin(wtq)
blue argument is always larger than red one
sin(wt-q) sin(wt)
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Sample Problem
Read from the plot Amplitude of i(t) Io 50 A
Irms500.7135
v(t) sin(wt) i(t) sin(wt-90)
current lags voltage by 900
Answer B
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Sample Problem
Magnitude 5 from Pythagoram principle
Angle (phase) from tan(q)4/3 q tan-1(4/3)
Tangt1 so angle gt 450
Answer D
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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
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Sample Problem

• For AC circuit with Vrms115V, Irms 20.1A and
  phase constant ?320, find the average real power
  and average reactive power drawn by the circuit.
• P 115V20.1Acos320 1965 W
• Q 115V20.1Asin 320 1217 kVAR ( kilovolt-amps
  reactive)

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Sample Problem
Answer A
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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
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Sample Problem
Inductances are like resistors in series and in
parallel.
Lseries L1 L2
Energy stored in an inductor WJ 0.5LI2
IL 10 A from current source

• Answer B

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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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Sample Problem
Two 4? resistors are in parallel 2 ? .
Then 2 ? and 2 ? resistors Are in series. I
40/4
Answer C
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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 200.7 14. PI2R 14250 10kW.
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Sample Problem
After t 5t, the capacitor acts like an open
circuit
Use Ohms Law for Ix
Answer C