Electricity & Magnetism
FE Review
Voltage
v
dw
joule
dq
coulom b
J

1 V ≡ 1 J/C
1 J of work must be done to
move a 1-C charge through
a potential difference of 1V.
 V ( volts )
C
Current
i
dq
coulom b
dt
second

C
 A ( am peres )
s
Power
p
dw
dt

dw dq

 vi
dq dt
joule
second

J
s
W
( w atts )
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
E 
 0  8.85  10
Qar
4  0 r
 12
a r  unit vector in direction of E
2
• Force on charge Q1 a distance r from Q
F  Q 1E 
Q 1Q a r
4  0 r
2
• Work to move Q1 closer to Q
r2
W  
r1
Q 1Q
4  0 r
2
F/m (perm ittivity of free space)
a r a r dr 
Q 1Q  1 1 
  
4  0  r2 r1 
Parallel Plate Capacitor
Q=charge on plate
A=area of plate
ε0=8.85x10-12 F/m
Electric Field between Plates
E 
Capacitance
Q
C 
 0A
d
Potential difference (voltage) between the plates
V  Ed 
 0A
Qd
 0A
R 
Resistivity
L
A

L
L = length
A
A = cross-sectional area
 = resistivity of m aterial
 = conductivity of m aterial
Example: Find the resistance of a 2-m copper wire if the wire has a diameter of 2 mm.
8
 C u  2  10   m
A   r    0.001 
2
R 
L
A

2
 2  10
8
  m   2m 
  0.001  m
2
2
2
 1.273  10   12.73 m 
Resistor
Power absorbed
v  Ri
p  vi  R i 
2
v
2
R
Energy dissipated
w 
Parallel Resistors
p

dt 
p dt
one period
Rp 
1
1
R1

1
R2

1
R3

1
R4
Series Resistors
R s  R1  R 2  R 3  R 4
Capacitor
iC
v
dv
Stores Energy
dt
1
1
C
t


i d  v  0  
1
C
w 
t
Cv
2
2
 i d
0
Parallel Capacitors
C p  C1  C 2  C 3  C 4
Series Capacitors
Cs 
1
1
C1

1
C2

1
C3

1
C4
Inductor
vL
Stores Energy
di
dt
i
1
L
t


v d  i  0  
1
w 
t
v

L
d
1
Li
2
2
0
Parallel Inductors
Lp 
1
1
L1

1
L2

1
L3

1
L4
Series Inductors
L s  L1  L 2  L 3  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
R1  R 2  R 3  R 4
Current Divider
i1 
vs
1
R1
1
R1

1
R2

1
R3

1
R4
is
Example: Find vx and vy and the power absorbed by the 6-Ω resistor.
vx 
vy 
2
24
6V   2V
2
22
P6  
v
v x  1V
2
2
R

vy
6

1
6
W
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
v2

4

1
4
2
v1 
v2 
20
3
4
v2  2
8  v1
3


8  10
3
v1 = 10 V
v2 = 6 V
 v 1  3v 2  8
 6V
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
i2  2A
All meshes not containing current sources are KVL equations
KVL at Mesh 1
Find the power absorbed by the 4-Ω resistor
 10  4i 1  2  i 1  i 2   0
i1 
10  2i 2
6

10  2   2 
6
 1A
P4   i R  (1) (4)  4 W
2
2
Put in some numbers
RL Circuit
R=2Ω
L = 4 mH
vs(t) = 12 V
i(0) = 0 A
di
 vs  Ri  L
di
dt

R
L
i
di
dt
1
L
0
vs
 500i  3000
dt
in (t)  Ae
 Rt /L
 Ae
 500 t
ip (t)  K
0  500 K  3000

K 
vs
R
i( t )  A e
i(0)  0
i( t ) 
vs
R
v(t)  L
di
dt
 500 t
6

1  e
0  Ae
 Rt /L
 0.004
 500 ( 0 )
  6 1  e
d
 6  6e
dt
6
 500 t
 500 t

A  6
A
  12e
 500 t
V
6
Put in some numbers
RC Circuit
R=2Ω
C = 2 mF
vs(t) = 12 V
v(0) = 5V
dv
C
dv

v  vs
dt
dv
dt

dt
0
vn (t)  Ae
R
1
RC
v
 250 v  3000
1
RC
vs
 t /RC
 Ae
 250 t
vp (t)  K
0  250 K  3000
v(t)  Ae
v (0)  5
 250 t

dv
dt
K  v s  12
 12
5  Ae
v ( t )  v s  ( v s  v 0 )e
i( t )  C

 0.002
d
 250 ( 0 )
 t /RC
 12
 12  7 e
12  7 e
dt
 250 t

 250 t
  3.5e
A  7
V
 250 t
A
DC Steady-State
0
vL
di
0
dt
Short Circuit
i = constant
0
iC
dv
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 
2
a +b
2
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 
z1  A e
jθ
z2  Be
j
 A  θ = a x  ja y
5  60 
 B    b x  jb y
z 1  z 2   a x  ja y    b x  jb y    a x  b x   j  a y  b y 
Multiplication
 AB θ   
5
  40   60  
= 4   20 
Addition
z1 z 2   A  θ   B   

20
Division
z1

z2

Aθ
B
A
B
 θ   
Phasors
A complex number representing a sinusoidal current or voltage.
V m cos   t   

Vm  
Only for:
• Sinusoidal sources
• Steady-state
Impedance
Z
V
A complex number that is the ratio of the phasor voltage and current.
units = ohms (Ω)
I
Admittance
Y 
I
V
units = Siemens (S)
Phasors
Converting from sinusoid to phasor
20 cos  40 t  15   A

100 cos 10 t  V
100  0  V
3

20 15  A
6 sin  100 t  10   A  6 cos(100 t  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
j C

1
C
  90 
1
Example: Find the steady-state output, v(t).
I
20  1.38  j3.45
5 0
1
1

 j50 20  1.38  j3.45
 4.88   24.67  A
V  Z I  20  4.88   24.67  
 97.61  24.67  V
v ( t )  97.61 cos(10 t  24.67  )V
3
10 
j4   3.71 68.20   1.38  j3.45 
Source Transformations
Voc  I sc  Z th
Thévenin Equivalent
Two special cases
Norton Equivalent
Example: Find the steady-state voltage, vout(t)
I
10  0 
Z
 j2
1
1
 j2
-j1 Ω
 5  90  A

1
 j2
  j1
V   5  90     j1   5  0  V
I
5 0
j1 Ω
Z
1
1
j1
j0.5 Ω
 5   90  A
j1

1
j1
 j0.5 
V   5   90    j0 .5   2.5  0  V
50+j0.5 Ω
No current flows through the impedance
V ou t  2.5  0  V
v out ( t )  2.5 cos(2 t )V
Thévenin Equivalent
AC Power
Complex Power
S
1
V I  P  jQ
*
units = VA (volt-amperes)
2
Average Power
P
1
2
V m I m cos 
units = W (watts)
a.k.a. “Active” or “Real” Power
Reactive Power
Power Factor
Q 
1
2
V m I m sin 
P F  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
Z

2000  0 
20  j50
Complex Power Absorbed
 37.14   68.20  A
S
1
2
Power Factor
VI 
*
1
2
 2000  0    37.14   68.20  
 37139  68.20  V A
 13793  j34483 V A
P F  cos   cos  68.20    0.371 lagging
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.
S total  S  S cap   13793  j34483    0  jQ cap 
Q cap  P tan  new  Q  13793 tan  31.79    34483   24934 V A R
Q cap   25934 V A R

S cap   j25934 V A
S for an ideal capacitor
S  j
1
2
v(t) = 2000 cos(100t) V
 C Vm
2
 j25934   j
1
2
100  C  2000 
2
C  0.13 m F
S total  S  S cap   13793  j34483    0  j25934 
 13793  j8549 V A  16227.5  31.79  V A
Without the capacitor
I  37.14   68.20  A
Current after capacitor added
S
1
2
 2  13793  j8549  
 2S 
I


  16.23   31.79  A

V
2000

0





*
VI
*

*
RMS Current & Voltage
V rm s
1

T

2
v
dt


0

T
1/ 2
I rm s
a.k.a. “Effective” current or voltage
1

T

2
i
dt
 
0

T
1/ 2
RMS value of a sinusoid
A cos   t   

A
2
P  V rm s  I rm s cos 
Balanced Three-Phase Systems
Phase Voltages
V an  V m  0 
V b n  V m   120 
V cn  V m  120 
Y-connected
source
Line Currents
Y-connected
load
Line Voltages
V ab  V an  V b n 

V b c  V L    120 
V ca  V L    120 

3  30  V an  V L  
I aA  I A N 
VAN
I bB  I BN 
VBN
I cC  I C N 
VC N
Z LN
Z LN
Z LN
 IL 
 I L    120 
 I L    120 
Balanced Three-Phase Systems
Line Voltages
V ab  V m  0 
V b c  V m   120 
V ca  V m  120 
Δ-connected
source
Δ-connected
load
Phase Currents
I AB 
VAB
I BC 
VBC
I CA 
VC A
Line Currents
I aA  I A B  I C A 

I b B  I L    120 
I cC  I L    120 

3   30  I A B  I L  
Z LL
Z LL
Z LL
 IL 
 I L    120 
 I L    120 
Currents and Voltages are specified in RMS
S
1
VI

*
S  VI
*

S  3V I
*
2
V L  line voltage  V ab
I L  line current  I aA
S for peak
voltage
& current
S for RMS
voltage
& current
S for 3-phase
voltage
& current
  im pedance angle   Z
Average Power
Complex Power
for Y-connected load
S  3VAN I AN

*
3V L I L  
Complex Power
for Δ-connected load
S  3VAB I AB

P  3V L I L cos 
*
3V L I L  
Power Factor
P F  cos  (leading or lagging)
Example:
Find the total real power supplied by the source in the balanced wye-connected circuit
Given:
V an  540  0  V
Z LN  270  j270 
I aA  I A N 
VAN
Z LN

540  0 
270  j270
 1.41  45  A
S  3 V A N I A N  3  540  0    1.41 45    2291 45  V A  1620  j1620 V A
*
P=1620 W
Ideal Operational Amplifier (Op Amp)
With negative feedback
i=0
Δv=0
Linear Amplifier
0
K V L : -v in  R 1i   v  0

i
v in
R1
K V L : -v in  R 1i  R 2 i  v out  0
v 
v 
-v in  R 1  in   R 2  in   v out  0
 R1 
 R1 
v out  
R2
R1
v in
mV or μV reading from sensor
0-5 V output to A/D converter
Magnetic Fields

B ds  0
S

H dl 
l

J dS  I
S
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 
1
2
2
  H dV
Enclosing a surface with N turns
of wire produces a voltage across
the terminals
V
v  N
d
dt
Magnetic field produces a force perpendicular to
the current direction and the magnetic field direction
F  IL  B
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
d l  H  2  r  I 0  10 A
H 
w 
10
2 r
1


2
V
for
2
3
10 m  r  10
H dV 
1
2
1 2  0.01

0 0
2
m
2
 10 

 0  2 r  rdrd  dz  18.3  0 J
0.001
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  N I0
H L  N I0
 B 

 L  N I0
 0 

I
BL
N0

 2T  1m 
1000  4   10
7

 1590 A
Questions?