Electricity & MagnetismElectricity & Magnetism
FE Review
VoltageVoltage
dwv
dq joule
( )coulomb
JV volts
C
CurrentCurrentdq
idt
coulomb ( )
second
CA amperes
s
PowerPowerdw dw dq
p vidt dq dt
joule ( )
second
JW watts
s
1 V ≡ 1 J/C1 J of work must be done to move a 1-C charge througha potential difference of 1V.
Point Charge, QPoint Charge, Q
• Experiences a force F=QE in the presence of electric field E(E is a vector with units volts/meter)
• Work done in moving Q within the E field
• E field at distance r from charge Q in free space
• Force on charge Q1 a distance r from Q
• Work to move Q1 closer to Q
1r
0
W d F r
r2
0
Q
4 r
aE
120
r
8.85 10 F/m (permittivity of free space)
unit vector in direction of
a E
1 r1 2
0
Q QQ
4 r
aF E
2
1
r
1 1r r2
0 0 2 1r
Q Q Q Q 1 1W dr
4 r 4 r r
a a
Parallel Plate CapacitorParallel Plate Capacitor
0
QE
A
Electric Field between Plates
Q=charge on plateA=area of plateε0=8.85x10-12 F/m
0
QdV Ed
A
0ACd
Potential difference (voltage) between the plates
Capacitance
ResistivityResistivityL L
RA A
L = length
A = cross-sectional area
= resistivity of material
= conductivity of material
Example: Find the resistance of a 2-m copper wire if the wire has a diameter of 2 mm.
8Cu 2 10 m
22A r 0.001
8
22 2
2 10 m 2mLR 1.273 10 12.73 m
A 0.001 m
ResistorResistor v RiPower absorbed
22 v
p vi RiR
Energy dissipated
one period
w p dt p dt Parallel Resistors
Series Resistors
p
1 2 3 4
1R
1 1 1 1
R R R R
s 1 2 3 4R R R R R
CapacitorCapacitor
t t
0
dvi C
dt
1 1v i d v 0 i d
C C
Stores Energy
21w Cv
2
Parallel Capacitors
Series Capacitors
p 1 2 3 4C C C C C
s
1 2 3 4
1C
1 1 1 1
C C C C
InductorInductor
t t
0
div L
dt
1 1i v d i 0 v d
L L
Stores Energy
21w Li
2
Parallel Inductors
Series Inductors
p
1 2 3 4
1L
1 1 1 1
L L L L
s 1 2 3 4L L L L L
KVL – Kirchhoff’s Voltage LawKVL – Kirchhoff’s Voltage Law
The sum of the voltage drops around a closed path is zero.
KCL – Kirchhoff’s Current LawKCL – Kirchhoff’s Current Law
The sum of the currents leaving a node is zero.
Voltage DividerVoltage Divider
Current DividerCurrent Divider
11 s
1 2 3 4
Rv v
R R R R
11 s
1 2 3 4
1R
i i1 1 1 1
R R R R
Example: Find vx and vy and the power absorbed by the 6-Ω resistor.
x
2v 6V 2V
2 4
y x
2v v 1V
2 2
22y
6
vv 1P W
R 6 6
Node Voltage AnalysisNode Voltage Analysis
Find the node voltages, v1 and v2 Look at voltage sources first
Node 1 is connected directly to ground by a voltage source → v1 = 10 V
All nodes not connected to a voltage source are KCL equations
↓
Node 2 is a KCL equation
KCL at Node 2
2 1 2v v v2 0
4 2
1 2 1 2
12
1 3v v 2 v 3v 8
4 48 v 8 10
v 6V3 3
v1 = 10 Vv2 = 6 V
Mesh Current AnalysisMesh Current Analysis
Find the mesh currents i1 and i2 in the circuit
Look at current sources first
Mesh 2 has a current source in its outer branch
2i 2A
All meshes not containing current sources are KVL equations
KVL at Mesh 1
1 1 210 4i 2 i i 0
21
10 2 210 2ii 1A
6 6
Find the power absorbed by the 4-Ω resistor
2 24P i R (1) (4) 4W
RL CircuitRL Circuit
s
div Ri L 0
dt
s
di R 1i v
dt L L
Put in some numbers
R = 2 ΩL = 4 mH
vs(t) = 12 Vi(0) = 0 A
Rt/L 500tn
p
di500i 3000
dt
i (t) Ae Ae
i (t) K
sv0 500K 3000 K 6
R
500t
500(0)
Rt /L 500ts
i(t) Ae 6
i(0) 0 0 Ae 6 A 6
vi(t) 1 e 6 1 e A
R
500t 500tdi dv(t) L 0.004 6 6e 12e V
dt dt
RC CircuitRC Circuit
s
s
v vdvC 0
dt Rdv 1 1
v vdt RC RC
Put in some numbers
R = 2 ΩC = 2 mF
vs(t) = 12 Vv(0) = 5V
t /RC 250tn
p
dv250v 3000
dt
v (t) Ae Ae
v (t) K
s0 250K 3000 K v 12 250t
250(0)
t /RC 250ts s 0
v(t) Ae 12
v(0) 5 5 Ae 12 A 7
v(t) v (v v )e 12 7e V
250t 250tdv di(t) C 0.002 12 7e 3.5e A
dt dt
DC Steady-StateDC Steady-State
div L 0
dt
0
Short Circuit
i = constant
dvi C 0
dt
0
Open Circuit
v = constant
Example: Find the DC steady-state voltage, v, in the following circuit.
v (20 ) 5 A 100 V
Complex ArithmeticComplex Arithmetic
Rectangular Exponential Polar
a+jb Aejθ A∠θ
Plot z=a+jb as an ordered pair on the real and imaginary axes
btan θ
a
2 2z a+jb a +b
Euler’s Identity
ejθ = cosθ + j sinθ
Complex Conjugate
(a+jb)* = a-jb (A∠θ)* = A∠-θ
Complex ArithmeticComplex Arithmetic
jθ1
j2
z Ae A θ = a
z Be B
x y
x y
ja
b jb
1
2
z A θ
z B
Aθ
B
1 2z z A θ B
AB θ
1 2 x y x y x x y yz z a ja b jb a b j a b
Addition
Multiplication Division
20 40 2040 60
5 60 5 = 4 20
PhasorsPhasors A complex number representing a sinusoidal current or voltage.
m mV cos t V
Only for:• Sinusoidal sources• Steady-state
ImpedanceImpedance A complex number that is the ratio of the phasor voltage and current.
Z V
Iunits = ohms (Ω)
AdmittanceAdmittance
Y I
Vunits = Siemens (S)
PhasorsPhasors
Converting from sinusoid to phasor
3
20cos 40t 15 A 20 15 A
100cos 10 t V 100 0 V
6sin 100t 10 A 6cos(100t 10 90 )A 6 80 A
Ohm’s Law for Phasors
ZV IOhm’s Law
KVLKCL
Current Divider
Voltage Divider
Mesh Current Analysis
Node Voltage Analysis
ImpedanceImpedance
Z R R 0
Z j L L 90
1 1Z 90
j C C
Example: Find the steady-state output, v(t).
10 j4 3.71 68.20 1.38 j3.45
3
120 1.38 j3.45
I 5 01 1j50 20 1.38 j3.45
4.88 24.67 A
V ZI 20 4.88 24.67
97.61 24.67 V
v(t) 97.61cos(10 t 24.67 )V
Source TransformationsSource Transformations
Thévenin Equivalent Norton Equivalent
thZ oc scV I
Two special casesTwo special cases
Example: Find the steady-state voltage, vout(t)
-j1 Ω
10 0I 5 90 A
j2
1Z j1
1 1j2 j2
V 5 90 j1 5 0 V
j1 Ω
5 0I 5 90 A
j1
j0.5 Ω
1Z j0.5
1 1j1 j1
V 5 90 j0.5 2.5 0 V
50+j0.5 Ω
Thévenin Equivalent
No current flows through the impedance
2.5 0 V outV
outv (t) 2.5cos(2t)V
AC PowerAC Power
Complex Power *1S P jQ
2 VI units = VA (volt-amperes)
Average Power
a.k.a. “Active” or “Real” Power
m m
1P V I cos
2 units = W (watts)
Reactive Power m m
1Q V I sin
2 units = VAR
(volt-ampere reactive)
Power FactorPower Factor PF cos θ = impedance angle
leading or lagging
current is leading the voltageθ<0
current is lagging the voltageθ>0
v(t) = 2000 cos(100t) VExample:Example:
Z 20 j50
53.85 68.20
2000 0 V V
Current, I
V 2000 0I 37.14 68.20 A
Z 20 j50
Power Factor
PF cos cos 68.20 0.371 lagging
Complex Power Absorbed
*1 1S 2000 0 37.14 68.20
2 2 37139 68.20 VA
13793 j34483 VA
VI
Average Power Absorbed
P=13793 W
Power Factor CorrectionPower Factor Correction
We want the power factor close to 1 to reduce the current.
Correct the power factor to 0.85 lagging
1new cos 0.85 31.79
Add a capacitor in parallel with the load.
total cap capS S S 13793 j34483 0 jQ
Total Complex Power
cap newQ P tan Q 13793tan 31.79 34483 24934 VAR
cap capQ 25934 VAR S j25934 VA
S for an ideal capacitor
2m
1S j CV
2
21j25934 j 100 C 2000
2
C 0.13 mF
v(t) = 2000 cos(100t) V
total capS S S 13793 j34483 0 j25934
13793 j8549 VA 16227.5 31.79 VA
*** 2 13793 j85491 2S
S 16.23 31.79 A2 2000 0
VI I
V
Current after capacitor added
I 37.14 68.20 A Without the capacitor
RMS Current & VoltageRMS Current & Voltage a.k.a. “Effective” current or voltage
1/2 1/2T T2 2
rms rms
0 0
1 1V v dt I i dt
T T
RMS value of a sinusoid
AA cos t
2
rms rmsP V I cos
Balanced Three-Phase SystemsBalanced Three-Phase Systems
Y-connectedsource
Y-connectedload
m
m
m
V 0
V 120
V 120
an
bn
cn
V
V
V
Phase Voltages
Line Currents
LLN
LLN
LLN
IZ
I 120Z
I 120Z
ANaA AN
BNbB BN
CNcC CN
VI I
VI I
VI I
Line Voltages
L
L
L
3 30 V
V 120
V 120
ab an bn an
bc
ca
V V V V
V
V
Balanced Three-Phase SystemsBalanced Three-Phase Systems
Phase Currents
Line Currents
Line Voltages
m
m
m
V 0
V 120
V 120
ab
bc
ca
V
V
V
LLL
LLL
LLL
IZ
I 120Z
I 120Z
ABAB
BCBC
CACA
VI
VI
VI
L
L
L
3 30 I
I 120
I 120
aA AB CA AB
bB
cC
I I I I
I
I
Δ-connectedsource
Δ-connectedload
Currents and Voltages are specified in RMS
* * *1S S S 3
2 VI VI VI
S for peakvoltage
& current
S for RMSvoltage
& current
S for 3-phasevoltage
& current
*
L L
S 3
3V I
AN ANV I *
L L
S 3
3V I
AB ABV I
Complex Powerfor Y-connected load
Complex Powerfor Δ-connected load
L
L
V line voltage
I line current
impedance angle Z
ab
aA
V
I
Average Power
L LP 3V I cos
Power Factor
PF cos (leading or lagging)
Example:
Find the total real power supplied by the source in the balanced wye-connected circuit
LN
540 0 V
Z 270 j270
anV
Given:
LN
*
540 01.41 45 A
Z 270 j270
S 3 3 540 0 1.41 45 2291 45 VA 1620 j1620VA
ANaA AN
AN AN
VI I
V I
P=1620 W
Ideal Operational Amplifier (Op Amp)Ideal Operational Amplifier (Op Amp)
With negative feedbacki=0
Δv=0
Linear Amplifier
inin 1
1
in 1 2 out
in inin 1 2 out
1 1
2out in
1
vKVL : -v R i v 0 i
R
KVL : -v R i R i v 0
v v -v R R v 0
R R
R v v
R
0
mV or μV reading from sensor
0-5 V output to A/D converter
Magnetic FieldsMagnetic FieldsS S
d 0 d d I B s H J S l l
B – magnetic flux density (tesla)H – magnetic field strength (A/m)J - current densityNet magnetic flux through
a closed surface is zero.B H
Sd B S
Magnetic Flux φ passing through a surface
dv N
dt
2
V
1w H dV
2
Energy stored in the magnetic field Enclosing a surface with N turnsof wire produces a voltage acrossthe terminals
I F L B
Magnetic field produces a force perpendicular to the current direction and the magnetic field direction
Example:
A coaxial cable with an inner wire of radius 1 mm carries 10-A current. The outercylindrical conductor has a diameter of 10 mm and carries a 10-A uniformly distributed current in the opposite direction. Determine the approximate magneticenergy stored per unit length in this cable. Use μ0 for the permeablility of the material between the wire and conductor.
0
3 2
d H 2 r I 10 A
10H for 10 m r 10 m
2 r
H l
21 2 0.012
0 0
V 0 0 0.001
1 1 10w H dV rdrd dz 18.3 J
2 2 2 r
Example:
A cylindrical coil of wire has an air core and 1000 turns. It is 1 m long with a diameter of 2 mm so has a relatively uniform field. Find the current necessary to achieve a magnetic flux density of 2 T.
0
0
0 70 0
d NI
HL NI
2T 1mB BLL NI I 1590 A
N 1000 4 10
H l
Questions?