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.
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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.
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}
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