PHYS 2421 - Fields and Waves - Academics Portal Index 2421/class material...Horizontal projection...

PHYS 2421 - Fields and Waves

cosi I t

Idea: how to deal with alternating currents

Edison wanted direct current, Tesla alternating current, who won?

Mathematically:

Instantaneous current:

Instantaneous voltage:

w is the angular frequency

w=2pf, f = 60 Hertz (50 Hertz in Europe)

V=120 volts (220 in Europe, South America)

Symbol in circuits:

cosv V t

Average I ?: 0Avei

How to measure a changing current?

There are several ways.

cosi I tHow to characterize alternating currents?

Average of rectified I ?: 2 / / 2

0 / 2

2

0

22

0.6372

cos cos

rav

t t

t t

t

t

dt I dI

I

dt

I t

I

Method 1: Rectified average

iUse average of


2 2 /

2

0 0

1

2cos2

2

Ave

t t

t t

I dt dt

i

t

2

2

0

1

2 2rmst

t

II

dt

I

Method 2: Root mean square average

2iUse average of

Average of ?: 2i

2 2 2 2 1

2cos 1 cos2i I t I t

2

rms AveiI

0.7071rms I I 0.637rav II Compare to


Method 3: Phasor

• Take instantaneous current as component of a vector

• Let vector rotate with angular speed .

• Useful for comparing times

Let us see some examples

Not a Mexican Wrestler!

0Avea) I

2.7rms

Given:

AI

22 2.7 7.3i iI I 2 2 2

rms Ave rmsAve

A Ab)

2.7 2 2 2.7 3.82

rms rms A A Ac) I

I I I

Hmwk: Probls 11th ed.: 31.1 and 2 (soln.: 2.97 A and 1.89 A)

or 12th Ed.: 31.3 and 2

Summary of Section 31.1

cosi I tSymbol in circuits:

cosv V t

Average I ?: 0Avei

2rav II

2rms

II

Phasor

Hmwk Sect. 31.1: Probls 11th ed.: 31.1 and 2

or 12th Ed.: 31.3 and 2

cos

cos

R

R

v iR

IR t

V t

Phasor diagram

Resistor in an ac circuit

Voltage drop across R?

cosi I t

cosi I t

cos

sin

cos 90

cos 90

L

o

o

L

d Idi

dt dt

tv L L

I L t

I L t

V t

Phasor diagram

Inductor in an ac circuit

L

L

is like a resistance,

call it "Reactance"

Inductive reactance

V I L L

X L

Voltage drop across L?

Reactance

a) Want XL to have i=250 mA when V=3.6 V610/ 3.6 /250 14.4L L L L A V kV IX X V I

Now want L

3 610 1.6 10 1.43

2

/ 2 14.4 / 2

L

L mH

X L fL

L X f

b) Want I for 16 MHz and 160 kHz / / 2L L L

As increases by 10, decreases by 10 I=25 A

As decreases by 10, increases by 10 I=2500 A

I V X V fL

f I

f I

m

m

Hmwk: Probls 31.3 and 5 (11th Ed) or 5 and 9 (12th Ed.)

cos

sin

sin cos 90o

C C

dqI

dt

I

I I

i t

q t

q Cv v t tC C

Capacitor in an ac circuit

1

IV IX

C

XC

L C

C

Take

is capacitive reactance

Phasor diagram

cosi I t

Voltage across C?

Reactance

Hmwk: Probls

11th Ed.: 31.7 & 10 Soln: a) 1736 W, vR=1.10cos [120t]

12th Ed.: 31.11 & 12 Soln: a) 1736 , vR=1.10cos [120t]

1.2 cos 25000.006cos 2500

200

Aa) A

tvi t

R

1 1

8062500 5 10

C rad/s F

b) XC

/ 2 0.48 2500 / 2cos cosC C V rad/sc) t tv V


cosR Rv V tR in an ac circuit

cosi I t

cos 90o

L Lv V t

Reactance

L in an ac circuit

LX L

Reactance

C in an ac circuit

cos 90o

C

Iv t

C

1CX

C


Homework for section 31.2:

11th Ed.: Probls 3, 5, 7 and 10

12th Ed.: Probls 5, 9, 11 and 12

cosR Rv V t

cosi I t

cos 90o

L Lv V t

cos 90o

C

Iv t

C

i and vR are in phase

vL is ahead 90o

vC is behind 90o

At this point phasors become useful

vL is ahead 90o

vC is behind 90o

vR is in phase with i

v across source?

v v v v R L C

cos cos 90 cos 90I

v V t V t tC

o o

R L

Horizontal projection Horizontal projectionHorizontal projection

Sum of horizontal projections of vectors

= horizontal projection of sum of vectors

Sum voltage vectors:

Magnitude of sum :

2 2 2 22 2

2 22 2; 1/ impedance " " is

R L C L C L C

L C

V V V IR IX IX R X X

IZ Z R X X R L C

V I

Phase angle : tan L C L C L C

R

V V IX IX X X

V IR R

In conclusion, if the current is :

The source voltage is :

cosi I t

cosv IZ t

If R is missing, set R=0

Let us see an example!

2 22 2 1/L CZ R X X R L C

If L is missing, set L=0

If C is missing, set C

Formalism is still valid if an element is missing

10,000 60 600L rad/s mH LX

61/ 1/ 10,000 0.5 10 200C rad/s F CX

2 22 2300 600 200 500 L CZ R X X

500.1

500

V A

VI

Z

1 1 0600 200tan tan 53

300

L CX X

R

0.1 300 30 0.1 600 60

0.1 200 20

A V; A V

A V

R L L

C C

IR IX

IX

V V

V

Hmwk: Problem 31.12 (11th Ed.) or 14 (12th Ed.)

cos 30 cos 10,000R V rads/siR IR t tv

Hmwk: Problem 31.13 (11th Ed.) or 16 (12th Ed.)

0cos 90 sin 60 sin 10,000L V rads/sL L LiX V t V t tv

0cos 90 sin 20 sin 10,000C V rads/sC C CiX V t V t tv

0

0cos 50 cos 10,000 53

180 V rads/sV t tv

Instantaneous source voltage

cos 0.1 cos 10,000i I t t A rads/s


Hmwk Sect. 31.3: Probls 31.12 & 31.13 (11th Ed.) or 14 & 16 (12th Ed.)

2 22 2 1

1/ ,L C L CZ R X X R L C X L XC

I is mpedance

Phase angle : tan L C L C L C

R

V V IX IX X X

V IR R

If the current is :

The source voltage is :

cosi I t cosv IZ t

cosR IR tv 0 0cos 90 cos 90L LV t IX tv L

0 0cos 90 cos 90C CV t IX tv C

cosV tv Instantaneous source voltage

Voltages across elements:

Idea: vary to increase current

22

L C

V VI

Z R X X

I increases as Z decreases

Z varies with

0

1/

1

L CX X L C

LC

Minimum of Z when XL=XC

“Resonanting frequency”

22 1/Z R L C

Max. I at 0

XL & XC vary with

6 35 10 0.4 10 2000b) rad/s H LX L

6 121/ 1/ 5 10 100 10 2000 rad/s F CX C

2 22 2 2000 2000 500 L CZ R X X R R

12

500

Vc) mA

rmsrms

Vi

Z

0.002 500 1d) A Vrms R rmsV I R

4 Vrms L rms LV I X 4 Vrms C rms CV I X

0

3 12

1 15,000,000

0.4 10 100 10LC

a) rad/s

H F


Hwk for sect. 31.5: Probs 27 a)& b) and 32 (11th Ed.)or 31a)& b) and 36 (12th Ed.)

Solution of 32/36: 945 rad/s, 70.6 , 450 V

I output is max at resonance0

1

LC Max. I

at 0L CX X Z R

rmsrms

Vi

Z

rms R rmsV I R

rms L rms LV I X

rms C rms CV I X

Idea: Use an ac circuit at a given v and i to induce a different i and v in another circuit

Add a magnetic core

Add a secondary

loop

Start with a circuit with

i and v

1 1 1 2 2 2I B B I

A changing current I1 induces a changing current I2

Relations between currents and voltages

1 1Bd

Ndt

Notice that the frequency is the same on both circuits

2 2Bd

Ndt

where is the flux per turn (equal on both sides of core)

B

1 2 22 1

1 2 1

Bd NV V

dt N N N

Assuming no resistance in wires

Power-in in circuit 1 = power-out in circuit 2

1 11 1 1 2 2 2 2 1 2 1

2 2

or V N

P I V P I V I I I IV N

2 2

1 1

2402

120

Va) Use a tranformer with

V

N V

N V

12 1 1

2

b) Current: cannot use as is not knownV

I I IV

1 1 2 2

2 1

2 1

960 9604 8

240 120

Must use

W W A or A

V V

P I V I V

P PI I

V V

2 2

2 2/ 240 / 4 60

c) Resistance: take

V A

V RI

R V I

Hmwk: Probl 33 (11th Ed.) or 37 (12th Ed.)


Hmwk Section 31.6: Probls 32 and 33 (11th Ed.)

or 36 and 37 (12th Ed.)

Frequency is the same on both circuits

22 1

1

NV V

N

Power-in in circuit 1 = power-out in circuit 2

1 1 1 2 2 2P I V P I V

1 12 1 2 1

2 2

or V N

I I I IV N

PHYS 2421 - Fields and Waves

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