Electricity & Magnetism

FE Review

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.

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

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

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

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

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

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

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

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

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

12

1 3v v 2 v 3v 84 4

8 v 8 10v 6V3 3

v1 = 10 Vv2 = 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

21

10 2 210 2ii 1A6 6

Find the power absorbed by the 4-Ω resistor

2 24P i R (1) (4) 4W

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

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

DC Steady-State

div L 0dt

0

Short Circuit

i = constant

dvi C 0dt

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 θa

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

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

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)

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

Impedance

Z R R 0

Z j L L 90

1 1Z 90j C C

Example: Find the steady-state output, v(t).

10 j4 3.71 68.20 1.38 j3.45

3

120 1.38 j3.45I 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

Source Transformations

Thévenin Equivalent Norton Equivalent

thZ oc scV I

Two special cases

Example: Find the steady-state voltage, vout(t)

-j1 Ω

10 0I 5 90 Aj2

1Z j11 1j2 j2

V 5 90 j1 5 0 V

j1 Ω

5 0I 5 90 Aj1

j0.5 Ω

1Z j0.51 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 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

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

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

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

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

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

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

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)

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

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

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

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

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

Questions?