Download Electricity and Magnetism Review

Electricity & Magnetism
FE Review
Voltage
v
dw
dq
joule
J
  V (volts )
coulomb C
1 V ≡ 1 J/C
1 J of work must be done to
move a 1-C charge through
a potential difference of 1V.
Current
dq
i
dt
coulomb C
  A (amperes)
second
s
Power
p
dw dw dq


 vi
dt dq dt
joule
J
  W ( watts )
second s
Point 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
r1
W    F dr
0
• E field at distance r from charge Q in free space
Qa r
E
4 0 r 2
 0  8.85 1012 F/m (permittivity of free space)
a r  unit vector in direction of E
• Force on charge Q1 a distance r from Q
F  Q1E 
Q1Qa r
4 0 r 2
• Work to move Q1 closer to Q
Q1Q
Q1Q  1 1 
a
a
dr

  
r
2 r
4

r
4

r1 
0
0  r2
r1
r2
W  
Parallel Plate Capacitor
Q=charge on plate
A=area of plate
ε0=8.85x10-12 F/m
Electric Field between Plates
E
Q
 0A
C
Potential difference (voltage) between the plates
V  Ed 
Capacitance
Qd
 0A
 0A
d
L
L
R

A A
Resistivity
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.
Cu  2 108   m
A   r 2    0.001
R
L
A
2
2 10


8
  m   2m 
  0.001 m 2
2
 1.273  102   12.73 m
Resistor
Power absorbed
v  Ri
v2
p  vi  Ri 
R
2
Energy dissipated
w   p dt 
Parallel Resistors
Rp 

p dt
one period
1
1
1
1
1



R1 R 2 R 3 R 4
Series Resistors
R s  R1  R 2  R 3  R 4
Capacitor
dv
iC
dt
t
t
1
1
v   i d  v  0    i d 
C 
C0
Stores Energy
1 2
w  Cv
2
Parallel Capacitors
Cp  C1  C2  C3  C4
Series Capacitors
Cs 
1
1
1
1
1



C1 C2 C3 C4
Inductor
vL
di
dt
t
i
Stores Energy
t
1
1
v
d


i
0

   v d

L 
L0
w
1 2
Li
2
Parallel Inductors
Lp 
1
1
1
1
1



L1 L 2 L3 L 4
Series Inductors
Ls  L1  L 2  L3  L 4
KVL – Kirchhoff’s Voltage Law
The sum of the voltage drops around a closed path is zero.
KCL – Kirchhoff’s Current Law
The sum of the currents leaving a node is zero.
Voltage Divider
v1 
R1
vs
R1  R 2  R 3  R 4
Current Divider
i1 
1
R1
1
1
1
1



R1 R 2 R 3 R 4
is
Example: Find vx and vy and the power absorbed by the 6-Ω resistor.
vx 
2
 6V   2V
24
vy 
2
v x  1V
22
P6 
2
v2 v y 1


 W
R
6 6
Node 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
v 2  v1 v 2
 20
4
2
1
3
 v1  v 2  2   v1  3v 2  8
4
4
8  v1 8  10
v2 

 6V
3
3
v1 = 10 V
v2 = 6 V
Mesh 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
i 2  2A
All meshes not containing current sources are KVL equations
KVL at Mesh 1
10  4i1  2  i1  i 2   0
i1 
10  2i 2 10  2  2 

 1A
6
6
Find the power absorbed by the 4-Ω resistor
P4   i 2 R  (1) 2 (4)  4W
RL Circuit
Put in some numbers
R=2Ω
L = 4 mH
vs(t) = 12 V
i(0) = 0 A
di
0
dt
di R
1
 i  vs
dt L
L
 vs  Ri  L
di
 500i  3000
dt
i n (t)  Ae  Rt /L  Ae 500t
i p (t)  K
0  500K  3000

K
i(t)  Ae 500t  6
i(0)  0  0  Ae 500(0)  6  A  6
v
i(t)  s 1  e  Rt /L   6 1  e 500t  A
R
v(t)  L
di
d
 0.004  6  6e 500t   12e 500t V
dt
dt
vs
6
R
RC Circuit
Put in some numbers
R=2Ω
C = 2 mF
vs(t) = 12 V
v(0) = 5V
dv v  vs

0
dt
R
dv 1
1

v
vs
dt RC
RC
C
dv
 250v  3000
dt
v n (t)  Ae  t /RC  Ae 250t
v p (t)  K
0  250K  3000

v(t)  Ae 250t  12
v(0)  5

5  Ae 250(0)  12
 A  7
v(t)  vs  (vs  v 0 )e  t /RC  12  7e 250t V
i(t)  C
K  vs  12
dv
d
 0.002 12  7e 250t   3.5e 250t A
dt
dt
DC Steady-State
0
vL
di
0
dt
Short Circuit
i = constant
0
dv
iC
0
dt
v = constant
Open Circuit
Example:
Find the DC steady-state voltage, v, in the following circuit.
v  (20 )  5 A   100 V
Complex Arithmetic
Rectangular
Exponential
Polar
a+jb
Aejθ
A∠θ
tan θ 
b
a
z  a+jb  a 2 +b2
Plot z=a+jb as an ordered pair on the real
and imaginary axes
Euler’s Identity
Complex Conjugate
(a+jb)* = a-jb
(A∠θ)* = A∠-θ
ejθ = cosθ + j sinθ
Complex Arithmetic
2040 20
   40  60 
560
5
= 4  20
z1  Ae  Aθ = a x  ja y
jθ
z 2  Be j  B  bx  jby
Addition
z1  z 2   a x  ja y    b x  jb y    a x  b x   j  a y  b y 
Multiplication
z1 z 2   Aθ  B 
 AB  θ   
Division
z1 Aθ

z 2 B

A
 θ   
B
Phasors
A complex number representing a sinusoidal current or voltage.
Vm cos  t     Vm 
Only for:
• Sinusoidal sources
• Steady-state
Impedance
Z
V
I
A complex number that is the ratio of the phasor voltage and current.
units = ohms (Ω)
Admittance
Y
I
V
units = Siemens (S)
Phasors
Converting from sinusoid to phasor
20 cos  40t  15  A  2015A
100 cos 103 t  V  1000V
6sin 100t  10  A  6 cos(100t  10  90)A  6  80A
Ohm’s Law for Phasors
V  ZI
Current Divider
Ohm’s Law
KVL
KCL
Voltage Divider
Mesh Current Analysis
Node Voltage Analysis
Impedance
Z  R  R0
Z  j L   L90
Z
1
1

  90
j C  C
Example: Find the steady-state output, v(t).
1
20  1.38  j3.45
I
50
1
1

 j50 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(103 t  24.67)V
10 j4  3.7168.20  1.38  j3.45
Source Transformations
Voc  Isc  Zth
Thévenin Equivalent
Two special cases
Norton Equivalent
Example: Find the steady-state voltage, vout(t)
I
100
 590A
 j2
Z
-j1 Ω
1
1
1

 j2  j2
  j1
V   590   j1  50V
I
j1 Ω
Z
j0.5 Ω
50
 5  90A
j1
1
1 1

j1 j1
 j0.5
V   5  90  j0.5   2.50V
50+j0.5 Ω
No current flows through the impedance
Vout  2.50V
vout (t)  2.5cos(2t)V
Thévenin Equivalent
AC Power
Complex Power
1 *
S  VI  P  jQ
2
Average Power
1
P  Vm I m cos 
2
units = VA (volt-amperes)
units = W (watts)
a.k.a. “Active” or “Real” Power
Reactive Power
Power Factor
Q
1
Vm I m sin 
2
PF  cos 
units = VAR
(volt-ampere reactive)
θ = impedance angle
leading or lagging
current is leading the voltage
θ<0
current is lagging the voltage
θ>0
Example:
v(t) = 2000 cos(100t) V
V  20000 V
Z  20  j50 
 53.8568.20 
Current, I
I
V 20000

 37.14  68.20A
Z 20  j50
Power Factor
PF  cos   cos  68.20   0.371 lagging
Complex Power Absorbed
1 * 1
VI   20000  37.14  68.20 
2
2
 3713968.20 VA
 13793  j34483 VA
S
Average Power Absorbed
P=13793 W
Power Factor Correction
We want the power factor close to 1 to reduce the current.
Correct the power factor to 0.85 lagging
 new  cos 1 0.85  31.79
Total Complex Power
Add a capacitor in parallel with the load.
Stotal  S  Scap  13793  j34483   0  jQ cap 
Qcap  P tan  new  Q  13793 tan  31.79   34483  24934 VAR
Qcap  25934 VAR
 Scap   j25934 VA
S for an ideal capacitor
v(t) = 2000 cos(100t) V
1
S   j CVm2
2
 j25934   j
1
2
100
C
2000
  

2
C  0.13 mF
Stotal  S  Scap  13793  j34483   0  j25934 
 13793  j8549 VA  16227.531.79 VA
Without the capacitor
I  37.14  68.20A
Current after capacitor added
S
1 *
VI
2
 2S   2 13793  j8549  
 I  
  16.23  31.79 A
V
2000

0

  

*
*
RMS Current & Voltage
1/2
Vrms
1T 2 
   v dt 
T 0

a.k.a. “Effective” current or voltage
1/2
I rms
1T 2 
   i dt 
T 0

RMS value of a sinusoid
A cos  t   

A
2
P  Vrms  I rms cos 
Balanced Three-Phase Systems
Phase Voltages
Van  Vm 0
Vbn  Vm   120
Vcn  Vm 120
Y-connected
source
Line Currents
Y-connected
load
Line Voltages
Vab  Van  Vbn 

Vbc  VL   120
Vca  VL   120

330 Van  VL 
I aA  I AN 
VAN
 I L 
ZLN
I bB  I BN 
VBN
 I L   120
ZLN
I cC  I CN
VCN

 I L   120
ZLN
Balanced Three-Phase Systems
Line Voltages
Vab  Vm 0
Vbc  Vm   120
Vca  Vm 120
Δ-connected
source
Δ-connected
load
Phase Currents
I AB 
VAB
 I L 
ZLL
I BC 
VBC
 I L   120
ZLL
I CA 
VCA
 I L   120
ZLL
Line Currents
I aA  I AB  I CA 

I bB  I L   120
I cC  I L   120

3  30 I AB  I L 
Currents and Voltages are specified in RMS
S
1 *
VI  S  VI*  S  3VI*
2
S for peak
voltage
& current
S for RMS
voltage
& current
S for 3-phase
voltage
& current
VL  line voltage  Vab
I L  line current  I aA
  impedance angle  Z
Average Power
Complex Power
for Y-connected load
S  3VAN I AN*
 3VL I L 
Complex Power
for Δ-connected load
P  3VL I L cos 
S  3VAB I AB*
 3VL I L 
Power Factor
PF  cos  (leading or lagging)
Example:
Find the total real power supplied by the source in the balanced wye-connected circuit
Given:
Van  5400 V
ZLN  270  j270 
I aA  I AN 
VAN
5400

 1.41  45A
ZLN 270  j270
S  3VAN I AN*  3  5400 1.4145   229145VA  1620  j1620VA
P=1620 W
Ideal Operational Amplifier (Op Amp)
With negative feedback
i=0
Δv=0
Linear Amplifier
0
KVL : -vin  R1i  v  0  i 
vin
R1
KVL : -vin  R1i  R 2i  v out  0
v 
v 
-vin  R1  in   R 2  in   v out  0
 R1 
 R1 
R
v out   2 vin
R1
mV or μV reading from sensor
0-5 V output to A/D converter
Magnetic Fields

S
B ds  0
 H dl  
l
S
J dS  I
B – magnetic flux density (tesla)
H – magnetic field strength (A/m)
J - current density
Net magnetic flux through
a closed surface is zero.
B  H
Magnetic Flux φ passing through a surface
   B dS
S
Energy stored in the magnetic field
w
2
1

H
dV

2 V
Enclosing a surface with N turns
of wire produces a voltage across
the terminals
Magnetic field produces a force perpendicular to
the current direction and the magnetic field direction
F  IL  B
v  N
d
dt
Example:
A coaxial cable with an inner wire of radius 1 mm carries 10-A current. The outer
cylindrical conductor has a diameter of 10 mm and carries a 10-A uniformly
distributed current in the opposite direction. Determine the approximate magnetic
energy stored per unit length in this cable. Use μ0 for the permeablility of the
material between the wire and conductor.
 H dl  H 2 r  I
H 
10
2 r
for
0
 10 A
103 m  r  102 m
1 2 0.01
2
1
1
w    H dV  
2 V
20
2
 10 

0 0.001
 0  2 r  rdrd dz  18.30 J
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.
 H dl  NI
0
HL  NI0
2T 1m 
B

BL

 1590 A
  L  NI0  I 
7
N0 1000  4 10 
 0 
Questions?