electricity & magnetism€¦ · rms value of a sinusoid a a cos t 2 zm o rms rms p v i cos t....
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Electricity & Magnetism
FE Review
Voltage
dwv
dq joule
( )coulom b
JV volts
C
Current
dqi
dt
coulom b ( )
second
CA am peres
s
Power
dw dw dqp vi
dt dq dt
joule ( )
second
JW w atts
s
1 V ≡ 1 J/C 1 J of work must be done to move a 1-C charge through a potential difference of 1V.
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
• 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
r
2
0
Q
4 r
aE
12
0
r
8.85 10 F/m (perm ittivity of free space)
unit vector in direction of
a E
1 r
1 2
0
Q QQ
4 r
aF E
2
1
r
1 1
r r2
0 0 2 1r
Q Q Q Q 1 1W dr
4 r 4 r r
a a
Parallel Plate Capacitor
0
QE
A
Electric Field between Plates
Q=charge on plate A=area of plate ε0=8.85x10-12 F/m
0
Q dV Ed
A
0A
Cd
Potential difference (voltage) between the plates
Capacitance
Resistivity L L
RA A
L = length
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 u2 10 m
22
A r 0.001
8
2
2 2
2 10 m 2mLR 1.273 10 12.73 m
A 0.001 m
Resistor v R iPower absorbed
2
2 vp vi R i
R
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
Capacitor
t t
0
dvi C
dt
1 1v i d v 0 i d
C C
Stores Energy
21w C v
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
Inductor
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 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
Current Divider
1
1 s
1 2 3 4
Rv v
R R R R
1
1 s
1 2 3 4
1
Ri i
1 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
22
y
6
vv 1P 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
2 1 2v v v
2 04 2
1 2 1 2
1
2
1 3v v 2 v 3v 8
4 4
8 v 8 10v 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
2i 2A
All meshes not containing current sources are KVL equations
KVL at Mesh 1
1 1 210 4i 2 i i 0
2
1
10 2 210 2ii 1A
6 6
Find the power absorbed by the 4-Ω resistor
2 2
4P i R (1) (4) 4 W
RL Circuit
s
div R i L 0
dt
s
di R 1i v
dt L L
Put in some numbers
R = 2 Ω L = 4 mH
vs(t) = 12 V i(0) = 0 A
R t / L 500 t
n
p
di500i 3000
dt
i ( t ) Ae Ae
i ( t ) K
sv
0 500K 3000 K 6R
500 t
500 ( 0 )
R t / L 500 ts
i( t ) Ae 6
i(0) 0 0 Ae 6 A 6
vi( t ) 1 e 6 1 e A
R
500 t 500 tdi dv( t ) L 0.004 6 6e 12e V
dt dt
RC Circuit
s
s
v vdvC 0
dt R
dv 1 1v v
dt R C R C
Put in some numbers
R = 2 Ω C = 2 mF
vs(t) = 12 V v(0) = 5V
t / R C 250 t
n
p
dv250 v 3000
dt
v ( t ) Ae Ae
v ( t ) K
s0 250K 3000 K v 12
250 t
250 ( 0 )
t / R C 250 t
s s 0
v( t ) Ae 12
v(0) 5 5 Ae 12 A 7
v( t ) v (v v )e 12 7e V
250 t 250 tdv di( t ) C 0.002 12 7e 3.5e A
dt dt
DC 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 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 Arithmetic
jθ
1
j
2
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 20
40 605 60 5
= 4 20
Phasors A complex number representing a sinusoidal current or voltage.
m mV cos t V
Only for: • Sinusoidal sources • Steady-state
Impedance A complex number that is the ratio of the phasor voltage and current.
Z V
Iunits = ohms (Ω)
Admittance
Y I
Vunits = Siemens (S)
Phasors
Converting from sinusoid to phasor
3
20 cos 40 t 15 A 20 15 A
100 cos 10 t V 100 0 V
6 sin 100 t 10 A 6 cos(100 t 10 90 )A 6 80 A
Ohm’s Law for Phasors
ZV IOhm’s Law
KVL KCL
Current Divider
Voltage Divider
Mesh Current Analysis
Node Voltage Analysis
Impedance
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
1
20 1.38 j3.45I 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 ) 97.61 cos(10 t 24.67 )V
Source Transformations
Thévenin Equivalent Norton Equivalent
th
Z oc sc
V I
Two special cases
Example: Find the steady-state voltage, vout(t)
-j1 Ω
10 0I 5 90 A
j2
1Z j1
1 1
j2 j2
V 5 90 j1 5 0 V
j1 Ω
5 0I 5 90 A
j1
j0.5 Ω
1Z j0.5
1 1
j1 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 out
V
outv ( t ) 2.5 cos(2 t )V
AC Power
Complex Power *1
S P jQ2
V I 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 Factor PF cos θ = impedance angle
leading or lagging
current is leading the voltage θ<0
current is lagging the voltage θ>0
v(t) = 2000 cos(100t) V 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 1
S 2000 0 37.14 68.202 2
37139 68.20 V A
13793 j34483 V A
V I
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
1
newcos 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 13793 tan 31.79 34483 24934 V AR
cap capQ 25934 V AR S j25934 V A
S for an ideal capacitor
2
m
1S j C V
2
21
j25934 j 100 C 20002
C 0.13 m F
v(t) = 2000 cos(100t) V
total capS S S 13793 j34483 0 j25934
13793 j8549 V A 16227.5 31.79 V A
**
*2 13793 j85491 2S
S 16.23 31.79 A2 2000 0
V I IV
Current after capacitor added
I 37.14 68.20 A
Without the capacitor
RMS Current & Voltage a.k.a. “Effective” current or voltage
1/ 2 1/ 2T T
2 2
rm s rm s
0 0
1 1V v dt I i dt
T T
RMS value of a sinusoid
A
A cos t 2
rm s rm sP V I cos
Balanced Three-Phase Systems
Y-connected source
Y-connected load
m
m
m
V 0
V 120
V 120
an
bn
cn
V
V
V
Phase Voltages
Line Currents
L
LN
L
LN
L
LN
IZ
I 120Z
I 120Z
A N
aA A N
B N
bB B N
C N
cC C N
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 Systems
Phase Currents
Line Currents
Line Voltages
m
m
m
V 0
V 120
V 120
ab
bc
ca
V
V
V
L
LL
L
LL
L
LL
IZ
I 120Z
I 120Z
A B
A B
B C
B C
C A
C A
VI
VI
VI
L
L
L
3 30 I
I 120
I 120
aA A B C A A B
bB
cC
I I I I
I
I
Δ-connected source
Δ-connected load
Currents and Voltages are specified in RMS
* * *1S S S 3
2 V I V I V I
S for peak voltage
& current
S for RMS voltage
& current
S for 3-phase voltage
& current
*
L L
S 3
3V I
A N A NV I *
L L
S 3
3V I
A B A BV I
Complex Power for Y-connected load
Complex Power for Δ-connected load
L
L
V line voltage
I line current
im pedance 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 V A 1620 j1620V A
A N
aA A N
A N A N
VI I
V I
P=1620 W
Ideal Operational Amplifier (Op Amp)
With negative feedback i=0
Δv=0
Linear Amplifier
in
in 1
1
in 1 2 out
in in
in 1 2 out
1 1
2
out in
1
vK V L : -v R i v 0 i
R
K V L : -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 Fields S S
d 0 d d I B s H J Sl
l
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
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 turns of wire produces a voltage across the 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 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.
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 7
0 0
d N I
H L N I
2T 1mB BLL N I I 1590 A
N 1000 4 10
H l
Questions?
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