electricity & magnetism
DESCRIPTION
Electricity & Magnetism. FE Review. Voltage. 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. Power. Point Charge, Q. Experiences a force F =Q E in the presence of electric field E ( E is a vector with units volts/meter) - PowerPoint PPT PresentationTRANSCRIPT
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Electricity & Magnetism
FE Review
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Voltage
dwvdq
joule ( )coulomb
J V voltsC
Currentdqidt
coulomb ( )second
C A amperess
Powerdw dw dqp vidt dq dt
joule ( )second
J W wattss
1 V ≡ 1 J/C1 J of work must be done to move a 1-C charge througha potential difference of 1V.
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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
r2
0
Q4 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 QQ4 r
aF E
2
1
r1 1
r r20 0 2 1r
Q Q Q Q 1 1W dr4 r 4 r r
a a
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Parallel Plate Capacitor
0
QEA
Electric Field between Plates
Q=charge on plateA=area of plateε0=8.85x10-12 F/m
0
QdV EdA
0ACd
Potential difference (voltage) between the plates
Capacitance
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ResistivityL LR
A 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
82
2 2
2 10 m 2mLR 1.273 10 12.73 mA 0.001 m
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Resistor v RiPower absorbed
22 vp vi Ri
R
Energy dissipated
one period
w p dt p dt Parallel Resistors
Series Resistors
p
1 2 3 4
1R 1 1 1 1R R R R
s 1 2 3 4R R R R R
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Capacitor
t t
0
dvi Cdt
1 1v i d v 0 i dC C
Stores Energy
21w Cv2
Parallel Capacitors
Series Capacitors
p 1 2 3 4C C C C C
s
1 2 3 4
1C 1 1 1 1C C C C
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Inductor
t t
0
div Ldt
1 1i v d i 0 v dL L
Stores Energy
21w Li2
Parallel Inductors
Series Inductors
p
1 2 3 4
1L 1 1 1 1L L L L
s 1 2 3 4L L L L L
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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
11 s
1 2 3 4
Rv vR R R R
11 s
1 2 3 4
1Ri i1 1 1 1
R R R R
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Example: Find vx and vy and the power absorbed by the 6-Ω resistor.
x2v 6V 2V
2 4
y x2v v 1V
2 2
22y
6
vv 1P WR 6 6
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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
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KCL at Node 2
2 1 2v v v 2 04 2
1 2 1 2
12
1 3v v 2 v 3v 84 4
8 v 8 10v 6V3 3
v1 = 10 Vv2 = 6 V
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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
21
10 2 210 2ii 1A6 6
Find the power absorbed by the 4-Ω resistor
2 24P i R (1) (4) 4W
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RL Circuit
sdiv Ri L 0dt
sdi R 1i vdt L L
Put in some numbers
R = 2 ΩL = 4 mH
vs(t) = 12 Vi(0) = 0 A
Rt/L 500tn
p
di 500i 3000dti (t) Ae Aei (t) K
sv0 500K 3000 K 6R
500t
500(0)
Rt /L 500ts
i(t) Ae 6
i(0) 0 0 Ae 6 A 6v
i(t) 1 e 6 1 e AR
500t 500tdi dv(t) L 0.004 6 6e 12e Vdt dt
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RC Circuit
s
s
v vdvC 0dt R
dv 1 1v vdt RC RC
Put in some numbers
R = 2 ΩC = 2 mF
vs(t) = 12 Vv(0) = 5V
t /RC 250tn
p
dv 250v 3000dtv (t) Ae Aev (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 Adt dt
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DC Steady-State
div L 0dt
0
Short Circuit
i = constant
dvi C 0dt
0
Open Circuit
v = constant
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Example: Find the DC steady-state voltage, v, in the following circuit.
v (20 ) 5 A 100 V
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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 θa
2 2z a+jb a +b
Euler’s Identity
ejθ = cosθ + j sinθ
Complex Conjugate
(a+jb)* = a-jb (A∠θ)* = A∠-θ
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Complex Arithmetic
jθ1
j2
z Ae A θ = a
z Be B
x y
x y
ja
b jb
1
2
z A θz B
A θB
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
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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 VI units = ohms (Ω)
Admittance
Y IV units = Siemens (S)
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Phasors
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
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Impedance
Z R R 0
Z j L L 90
1 1Z 90j C C
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Example: Find the steady-state output, v(t).
10 j4 3.71 68.20 1.38 j3.45
3
120 1.38 j3.45I 5 01 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(10 t 24.67 )V
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Source Transformations
Thévenin Equivalent Norton Equivalent
thZ oc scV I
Two special cases
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Example: Find the steady-state voltage, vout(t)
-j1 Ω
10 0I 5 90 Aj2
1Z j11 1j2 j2
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V 5 90 j1 5 0 V
j1 Ω
5 0I 5 90 Aj1
j0.5 Ω
1Z j0.51 1j1 j1
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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
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AC Power
Complex Power *1S P jQ2
VI units = VA (volt-amperes)
Average Power
a.k.a. “Active” or “Real” Power
m m1P V I cos2
units = W (watts)
Reactive Power m m1Q V I sin2
units = VAR (volt-ampere reactive)
Power Factor PF cos θ = impedance angle
leading or lagging
current is leading the voltageθ<0
current is lagging the voltageθ>0
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v(t) = 2000 cos(100t) VExample:
Z 20 j50 53.85 68.20
2000 0 V V
Current, I
V 2000 0I 37.14 68.20 AZ 20 j50
Power Factor
PF cos cos 68.20 0.371 lagging
Complex Power Absorbed
*1 1S 2000 0 37.14 68.202 2
37139 68.20 VA 13793 j34483 VA
VI
Average Power Absorbed
P=13793 W
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Power Factor Correction
We want the power factor close to 1 to reduce the current.
Correct the power factor to 0.85 lagging1
new 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
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cap capQ 25934 VAR S j25934 VA
S for an ideal capacitor
2m
1S j CV2
21j25934 j 100 C 20002
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 2SS 16.23 31.79 A
2 2000 0
VI IV
Current after capacitor added
I 37.14 68.20 A Without the capacitor
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RMS Current & Voltage a.k.a. “Effective” current or voltage
1/2 1/2T T2 2
rms rms0 0
1 1V v dt I i dtT T
RMS value of a sinusoid
AA cos t 2
rms rmsP V I cos
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Balanced Three-Phase Systems
Y-connectedsource
Y-connectedload
m
m
m
V 0V 120V 120
an
bn
cn
VVV
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 120V 120
ab an bn an
bc
ca
V V V V
VV
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Balanced Three-Phase Systems
Phase Currents
Line Currents
Line Voltages
m
m
m
V 0V 120V 120
ab
bc
ca
VVV
LLL
LLL
LLL
IZ
I 120Z
I 120Z
ABAB
BCBC
CACA
VI
VI
VI L
L
L
3 30 I
I 120I 120
aA AB CA AB
bB
cC
I I I I
II
Δ-connectedsource
Δ-connectedload
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Currents and Voltages are specified in RMS
* * *1S S S 32
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)
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Example:
Find the total real power supplied by the source in the balanced wye-connected circuit
LN
540 0 VZ 270 j270
anVGiven:
LN
*
540 0 1.41 45 AZ 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
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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 iR
KVL : -v R i R i v 0
v v -v R R v 0R R
R v vR
0
mV or μV reading from sensor
0-5 V output to A/D converter
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Magnetic Fields S Sd 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 Ndt
2
V
1w H dV2
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
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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 m2 r
H l
21 2 0.012
0 0V 0 0 0.001
1 1 10w H dV rdrd dz 18.3 J2 2 2 r
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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 AN 1000 4 10
H l
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Questions?