28. Alternating Current Circuits

1. Alternating Current 2. Current Elements in AC Circuits3. LC Circuits4. Driven RLC Circuits & Resonance5. Power in AC Circuits6. Transformers & Power Supplies

Why does alternating current facilitate the transmission and distribution of electric power?

EM induction allows voltage transformation.

28.1. Alternating Current

Reminder: All waves can be analyzed in terms of sinusoidal waves (Fourier analysis Chap 14).

Sinusoidal wave (Chap 13) :

2p

rms

VV

2p

rms

II

Angular frequency : 2 f [] = rad/s

sinp VV V t sinp II I t

= phase

= / 6

Vp sin

2 22 2

0 0

1 1 1sin cos2 2 2

x dx x dx

Example 28.1. Characterizing Household Voltage

Standard household wiring supplies 110 V rms at 60 Hz.

Express this mathematically, assuming the voltage is rising through 0 at t = 0.

2p rmsV V 156V

1377s2 f 2 60 Hz

sinpV V t

0 0V t 0 sinpV

0 sinpV V t 156 sin 377 Vt

28.2. Current Elements in AC Circuits

Resistors

Capacitors

Inductors

Phasor Diagrams

Capacitors & Inductors: A Comparison

Displacing Functions

f x xg

xx

fg g is moved to the right (forward) by to give f.

sin cos

cosn2

si x x

Displacement: sin is cos moved forward by /2.Phase: sin lags cos by /2.

sin sincos2

x x x

Derivative: moves sinusoidal functions backward by /2.

phase is increased by /2.

cossin2

sinx dx x x Integral: moves sinusoidal functions forward by /2.

phase is decreased by /2.

Resistors

V tI

R sinpV

tR

I & V in phasesinpI I t pp

VI

R

rmsrms

VIR

+VR

I

+

sin 0pV t I R 0RV t V When V(t) > 0 :

Capacitors

q C V t sinpCV t

d qId t

cospCV t

sin2pI C V t

I leads V by 90

p pI C V p

C

VX

1CX

C Capacitive reactance

CX

0CX as DC: open ckt.

0CX as HF: short ckt.

I peaks ¼ cycle before V

d VCd t

+VC

I

+

0CV t V sin 0pqV tC

When V(t) > 0 :

Inductors

sinpVd I td t L

sinpVt d t

L cospV

tL

sin2

pVI t

L

I trails V by 90

pp

VI

L p

L

VX

LX L Inductive reactance

LX

0 0LX as DC: short ckt.

LX as HF: open ckt.

I peaks ¼ cycle after V

1I V t d tL

I

+

0LV t E

sin 0pd IV t Ld t

When V(t) > 0 :

L

+

Table 28.1. Amplitude & Phase in Circuit Elements

Circuit Element Peak Current vs Voltage Phase Relation

Resistor

Capacitor

Inductor

pp

VI

R

pp

C

VI

X

1 /pVC

pp

L

VI

X pV

L

V & I in phase

V lags I 90

V leads I 90

GOT IT? 28.1.A capacitor and an inductor are connected across separate but identical electric generators,and the same current flows in each.

If the frequency of the generators is doubled, which will carry more current?

p C pI V C

pp L

VI

L

Ans. is capacitor

2p C pI V C

2p

p L

VI

L

2 p CI

12 p LI

Example 28.2. Equal Currents?

A capacitor is connected across a 60-Hz, 120-V rms power line,

and an rms current of 200 mA flows.

(a) Find the capacitance.

(b) What inductance, connected across the same powerline,

would result in the same current?

(c) How would the phases of the inductor & capacitor currents compare?

(a) rms

rms

ICV

3200 10

120 2 60A

V Hz

64.42 10 F 4.42 F

(b) rms

rms

VLI

3

120200 10 2 60

VA Hz

1.59 H

(c)

rmsrms

IVC

rms rmsV L I

Capacitor: IC leads V by 90.Inductor: V leads IL by 90.

IC leads IL by 180.

Phasor DiagramsPhasor = Arrow (vector) in complex plane. Length = mag. Angle = phase.

V leads I by 90. V leads I by 90.( V lags I by 90 )

V leads I by 0.( same phase )

V I X iV V e i tV e

Capacitors Revisitedq C V i t

pCV e

d qId t

i tpCV i e

I leads V by 90

dVCdt

+VC

Vp e i t

Taking the real part as physical

cospV V t

sinpI CV t cos2pCV t

Taking the imaginary part as physical

sinpV V t

cospI CV t sin2pCV t

I i C V 2i

CV e

V I Z

1Z iC

Impedance

I

Inductors Revisited

I lags V by 90

Vp e i t

Taking the real part as physical

cospV V t

sinpVI t

L

cos

2pV

tL

Taking the imaginary part as physical

sinpV V t

cospVI t

L

sin

2pV

tL

VIi L

2iV e

L

I

L

+

i tp

d IL V ed t

p i tVI e d t

L

LZ L i

0LV t E

p i tVe

i L

Capacitors & Inductors: A Comparison

C L translator:

E B

q B

V I

Z Y

Table 28.2. Capacitors & InductorsCapacitor Inductor

Behavior in low freq limit

Defining relation; differential form

Opposes change in

Energy storage

Defining relation

Behavior in high freq limit

Reactance

Phase

qCV

BLI

dVI Cd t

d IV Ld t

V I

212EU CV 21

2BU L I

Open circuit Short circuit

Open circuitShort circuit

I leads by 90 V leads by 90

1/ CX C LX L

Admittance / Impedance 1/C CY i C Z LZ i L

Application: Loudspeaker Systems

Loudspeaker

Loudspeaker system with high & low frequency filters.

C passes High freq

L passes low freq

CQV I RC

1

CR Ii C

LL

d IV L I Rd t

Li L R I

28.3. LC CircuitsI

V+

Analyzing the LC Circuit

B EU U U 2 21 12 2L I C V

0d Ud t

d I d VL I C Vd t d t

qVC

1d V d q

d t C d t

d qId t

2

2

d I d qd t d t

2

2

10 d q d q q d qL Cd t d t C C d t

2

2 0d q qLd t C

cospq q t1L C

I

V+ 0q d IL

C d t

GOT IT? 28.2

You have an LC circuit that oscillates at a typical AM radio frequency of 1 MHz.

You want to change the capacitor so it oscillates at a typical FM frequency , 100 MHz.

Should you make the capacitor

(a) larger or

(b) smaller ?

By what factor ? 104

1L C

2

1CL

Example 28.3. Tuning a Piano

You wish to make an LC circuit oscillate at 440 Hz ( A above middle C ) to use in tuning pianos.

You have a 20-mH inductor.

(a)What value of capacitance should you use ?

(b) If you charge the capacitor to 5.0 V, what will be the peak current in the circuit ?

2

1CL

23

120 10 2 440H Hz

66.54 10 F 6.54 F(a)

(b) B p E pU U 2 21 12 2p pL I C V

p pCI VL

6

3

6.54 10 5.020 10

F VH

0.09 A 90 mA

Resistance in LC Circuits – Damping

2 21 12 2L CU L I C V

2dU I Rd t

CC

d I dVL I C Vdt d t

CqVC

CdV Id t C

2 0d I IL I q I Rd t C

0d I qL I Rd t C

2

2 0d q dq qL Rd t d t C

/ 2 cosR t Lpq t q e t (see next page)

I

VC

+

L

+

+ VR

dqId t

Resistance in LC Circuits – Damping

0q d II R LC d t

dqId t

2

2 0d q dq qL Rd t d t C

/ 2 cosR t Lpq t q e t

I

(see next page)

VC

+

L

+

+ VR

Solutions to Damped Oscillator

2

2 0d q dq qL Rdt d t C

0a tq q e 2

01 0a tL a R a q eC

Ansatz:

2 1 0L a R aC

21 42

La R RL C

/2R t L i t i tq e c e c e 21 4

2L R

L C

212R

LC L

220 2

RL

/ 2 cos sinR t Le A t B t

/20 cosR t Lq e t

24L R C

01LC

2R iL

28.4. Driven RLC Circuits & Resonance

Driven damped oscillator :

Long time: oscillates with frequency d.

Resonance if d = 0.

I

VC +

L

+

+ VR

+

0d I qV I R Ld t C

0dd

IV I R i L Ii C

1d

d

V I R i LC

2021dd

Z R i L

Resonance in the RLC Circuit

1/C p

C pd

VI

C L p

L pd

VI

L L p L pV V if

1d

d

LC

2 20d i.e.,

VC & VL are 180 out of phase.

2021dd

V I R i L

L dV i L I

202C Ld

V V

L CV V V I R

GOT IT? 28.3

You measure the capacitor & inductor voltage in adriven RLC circuit,

and find 10 V for the rms capacitor voltage

and 15 V for the rms inductor voltage.

Is the driving frequency

(a) above or

(b) below resonance ?

VCp < VLp 1 /

C pC p

d

VI

C L p

L pd

VI

L

1d

d

LC

21dLC

Frequency Response of the RLC Circuit

Series circuit same I phasor for all.

VR in phase with I.

VC lags I by 90.

VL leads I by 90.

Q d II R L VC d t

22 1Z R L

C

1I R i Li C

p pV I Z

High Q

Low Q

See Prob 71 for definition of Q.

tan L p C p

R p

V VV

L CX XR

1LC

R

At resonance, = 0.

1Z R i LC

2021L

R

Example 28.4. Designing a Loud Speaker System

Current flows to the midrange speaker in a loudspeaker system through a 2.2-mH inductor in series with a capacitor.

(a)What should the capacitance be so that a given voltage produces the greatest current at 1 kHz ?

(b) If the same voltage produces half this current at 618 Hz,what is the speaker’s resistance ?

(c) If the peak output voltage of the amplifier is 20 V, what will be the peak capacitor voltage be at 1 kHz ?

(a) Greatest I is at resonance: 20

1LC

20

1CL

23 3

1

2.2 10 2 10H Hz

611.5 10 F 11.5 F

(b) If the same voltage produces half this current at 618 Hz, what is the speaker’s resistance ?

22 1Z R L

C

p pV I Z At resonance: Z R

2p

p

IZ I R

22 1 2R L R

C

1 13

R LC

36

1 12 618 2.2 102 618 11.5 103

Hz HHz F

8.0

(c) If the peak output voltage of the amplifier is 20 V,

what will be the peak capacitor voltage be at 1 kHz ?

p pV I RPeak voltage is at resonance (1 kHz).

C p p CV I X 1pVR C

3 6

20 18.0 2 10 11.5 10

VHz F

35V

28.5. Power in AC Circuits

P I V

sin sinp pP I t V t

sin sinp pI t V t

sin sinp pI V t t

sin cos cos sin sinp pI V t t t

2sin cos cos sin sinp pI V t t t

1 cos2 p pP I V cosrms rmsI V

Power factor

Capacitor: I leads V by 90 , P = 0

Resistor: I & V in phase , P > 0

I & V out of phase , P

Dissipative power = I2 R large power factor reduces I & hence heat loss.

Conceptual Example 28.1. Managing Power Factor

You’re chief engineer of a power company.

Should you strive for a high or a low power factor on your lines?

cosrms rmsP I V

Power factor

Generator : fixed Vrms .

To maintain fixed <P>, Irms cos = const.

Smaller power factor higher Irms .

higher power loss.

Ans.: keep power factor close to 1.

Making the Connection

Transmission losses on a well-managed electric grid average about 8% of the total power delivered.

How does this figure change if the power factor drops from 1 to 0.71?

To deliver the same power

cosrms rmsP I V

0.71newII 1.4 I

Transmission losses: 2L rmsP I R 2

, 1.4L new LP P 2 LP

( doubles to 16% )

28.6. Transformers & Power Supplies

Transformer: pair of coils wound on the same (iron) core.

sec secondary ondary

primary primary

V NV N

Works only for AC.

Direct-Current Power SuppliesDiode passes + half of each cycle

Diode cuts off half of each cycle

Diode

RC (low freq) filter