Electricity & Magnetism€¦ · RMS value of a sinusoid A A cos t 2 ZM o rms rms P V I cos T....

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?