Chapter 28: Alternating Current Phasors and Alternating Currents Alternating current (AC current) Current which varies sinusoidally in time is called

Chapter 28: Alternating Current

Phasors and Alternating Currents Alternating current (AC current)

• Current which varies sinusoidally in time is called alternating current (AC) as opposed to direct current (DC). One example of AC current source is a coil of wire rotating with constant angular velocity in a magnetic field.

The symbol ~ is used to denote an AC source. In general a source

means either a source of alternating current or voltage.

• In the U.S. and Canada, commercial electric-power distribution system uses a frequency of f = 60 Hz, corresponding to = 377 rad/s. In much of the rest of the world uses f = 50 Hz. In Japan, however, the country is divided in two regions with f = 50 Hz and 60 Hz.

amplitudecurrent I current, galternatinfor sin

amplitude voltageV voltage,galternatinfor sin

tIi

tV

kin textboo sin,sinsin,sin :Note tIItVVtIitV PP

Phasors and Alternating Currents

Phasors

O

t

IP

I=I P

sin

t

• A convenient way to express a quantity varying sinusoidally with time is by a phasor in phasor diagram as shown.

phasor

Rectifier and rectified current

+ -- +


Rectifier and rectified current (cont’d)


Root-mean-square current and voltage• Root-mean-square current of a sinusoidal current

2)2cos1(

2

1sinsin

222222 P

PPP

IItItIItII

time averaged

2P

rms

II

• Root-mean-square voltage of a sinusoidal voltage

2P

rms

VV For 120-volt AC, V=170 V.

Reluctance

Resistance, inductance, capacitance and reactance• Resistor in an AC circuit

IR

R

s inR R mV RI t

0

0t

IRRm

Rm

0

0

VR

t

m

m

Voltage across R in phase with current through R

sinmRI t

R

I

VR

R

IR

t

tm sinGiven:

At time t

I=m/R

Resistance, inductance, capacitance and reactance• Inductor in an AC circuit

Voltage across L leads current through L by one-quarter cycle (90°).

I L

L

cosmL LI dI t

L

2/sin

tLm

t

0

0

m

m

VL

t

0

0

Lm

Lm

IL I

VL

IL

t

tm sinGiven:

Reluctance

At time t

I=m/(L)

m

tdt

dILV m

LL sin tdt

LdI m

L sin

Resistance, inductance, capacitance and reactance• Capacitor in an AC circuit

C IC

Voltage across C lags current through C by one-quarter cycle (90°).

tC

QV mC sin tCQ m sin

0

0 t

m

m

VC

t

0

0

mC

mC

IC

I

VC

m

IC

t

tm sinGiven:

Reluctance

At time t

)2

sin(cos tCtC

dt

dQI mmC

t

I=Cm

LRC series circuit and reluctance

R mV I R 1C mV I

C L mV I L

XCXL

reactance

tm sinGiven:

Assume the solution for current: )sin()( tItI m

LRC circuit summary

)cos(

)cos(1

)sin(

tLIV

tIC

V

tRIV

mL

mC

mR

amplitude

Reluctance

(See derivation later)

For high ω, χC~0

- Capacitor looks like a wire (“short”)

For low ω, χC∞

- Capacitor looks like a break

1CX

C

For low ω, χL~0

- Inductor looks like a wire (“short”)

For high ω, χL∞

- Inductor looks like a break(inductors resist change in current)

LX L

( " " )RX R

You can think of it as a frequency-dependent resistance.

What is reactance?

LRC series circuit and reluctance (cont’d)

Reluctance

f

LC

R

ImR

Im L

ImC

m

)sin( tRIRIV mR

)cos( tLIdt

dILV mL

)cos(1

tI

CC

QV mC

• Assume:

• Given: tm sin

)sin( tII m )cos(

t

IQ m

)cos( tIdt

dIm

This picture corresponds to a snapshot at t=0.

The projections of these phasors along the vertical axis are the actual values of the voltages at the given time.

LRC series circuit (cont’d)

amplitude

LRC Circuits

Im

)cos(

)2/sin( : Note

t

t

Problem: Given Vdrive = εm sin ωt,find VR, VL, VC, IR, IL, IC

Strategy:1. Draw Vdrive phasor at t=0

2. Guess iR phasor

3. Since VR = iR R, this is also the direction for the VR phasor.

-φ

4. Realize that due to Kirchhoff’s current law, iL = iC = iR (i.e., the same current flows through each).

(No L or C → f = 0)

LC

R


0at )sin(

)sin(

ti

tii

m

mR

LRC Circuits

-φ

VR = I R

VL= I XL

VC = I XC

5. The inductor current IL always lags VL draw VL 90˚ further counterclockwise.

6. The capacitor voltage VC always lags IC draw VC 90˚ further clockwise.

The lengths of the phasors depend on R, L, C, and ω. The relative orientation of the VR, VL, and VC phasors is always the way we have drawn it.

is determined such that VR + VL + VC = ε (Kirchhoff’s voltage rule)These are added like vectors.


LRC Circuits

Phasor diagrams for LRC circuits: Example

y

xε

VC

IR 2 2 2( )CIR IX

2 2 2 2( )CI R X

2 2C

IR X

~

Vout

y

xε

LRC Circuits

amplitude of current

Filters : Example

22out

C

RV IR

R X

02 22 1

1

1

out

C

V R

R

0

1

RC

Ex.: C = 1 μF, R = 1Ω

High-pass filter

High-pass filter

0

0.2

0.4

0.6

0.8

1

0.E+00 1.E+06 2.E+06 3.E+06 4.E+06 5.E+06 6.E+06

(Angular) frequency, omega

"tra

nsm

issi

on"

Note: this is ω,2

f

~Vout

LRC Circuits

Filters

~Vout

~

ω=0 No currentVout ≈ 0

ω=∞ Capacitor ~ wireVout ≈ ε

~Vout

ω = ∞ No currentVout ≈ 0

ω = 0 Inductor ~ wireVout ≈ ε

ω = 0 No current because of capacitor

ω = ∞ No current because of inductor

outV

0

outV

0

(Conceptual sketch only)

High-pass filter

Low-pass filter

Band-pass filter

outV

0

LRC Circuits

Phasor diagrams for LRC circuits: Example 2

ImR

m

Im(XL -XC)

ImR

m

ImXC

ImXL

LX L

CXC

1

22CL XXRZ

R

XX CL tan

2222CLmm XXRI

ZXXRI m

CL

mm

22

Impedance Z

amplitude

LRC Circuits

Reluctance for inductor

Reluctance for capacitor

LRC Circuits Phasor diagrams for LRC circuits: Tips

• This phasor diagram was drawn as a snapshot of time t=0 with the voltages being given as the projections along the y-axis.

From this diagram, we can also create a triangle which allows us to calculate the impedance Z:f

ImR

ImXL

ImXC

m

“Full Phasor Diagram”

• Sometimes, in working problems, it is easier to draw the diagram at a time when the current is along the x-axis (when I=0).

Im

R

m

Im

XC

Im

XL

y

xf f

f

CLm XXI ZIm

RIm“Impedance Triangle”

f| |

Resonance in Alternating Current Circuits

Resonance For fixed R, C, L the current Im will be a maximum at the resonant

frequency w0 which makes the impedance Z purely resistive.

• Note that this resonant frequency is identical to the natural frequency of the LC circuit by itself!

• At this frequency, the current and the driving voltage are in phase!

0tan

R

XX CL

22CL

mmm

XXRZI

i.e.:

reaches a maximum when: X XC=LThis condition is obtained when:

C

Lo

o 1

LC

fLC

o

2

1;

10

LC

~

R

resonance frequency


Resonance (cont’d)

Im

00

o

0Rm

R=Ro

R=2Ro

R

XL

XC

Z

XL - XC

Plot the current versus , the frequency of the voltage source: →

R

XX CL tan

cos R

I mm

ZI mm

cos

RZ


Resonance (cont’d)

On Resonance:LC

~

R

RV IR IR

LL L

XV IX Q

R

CC C

XV IX Q

R

On resonance, the voltage across the reactive elements is amplified by Q!Necessary to pick up weak radio signals, cell phone transmissions, etc.

and Z=R

Power in Alternating Current Circuits

Power

• The instantaneous power (for some frequency, w) delivered at time t is given by:

• The most useful quantity to consider here is not the instantaneous power but rather the average power delivered in a cycle.

• To evaluate the average on the right, we first expand the sin(t-) term.

)sin(sin)()()( tIttIttP mm

)sin(sin)( ttItP mm


Power

sintcost

t0

0

+1

-1

(Product of even and odd function = 0)

sin2t

t0

0

+1

-1

tttItP mm cossinsinsincos)( 2

cos2

1)( mmItP

01/2

• Expanding,

• Taking the averages,

)sincoscos(sinsin)sin(sin ttttt

0cossin tt

• Generally:

2

0

22

2

1sin

2

1sin xdxx

• Putting it all back together again,


Power

mrms 2

1 mrms II

2

1 cos)( rmsrmsItP

Power delivered depends on the phase, f, the “power factor”

Phase depends on the values of L, C, R, and

Therefore...

This result is often rewritten in terms of rms values:

cos)( rmsrms ItP


Power

cos

RI mm

22 2

rms( ) cosrms

P t I RR

We can write this in the following manner (which we won’t try to prove):

2222

22

)1()(

xQx

x

RtP

rms

…introducing the curious factors Q and x...

Power, as well as current, peaks at = 0. The sharpness of the resonance

depends on the values of the components.

Recall:


Power and resonance

where Umax is max energy stored in the system and U is the energy dissipated in one cycle

A parameter “Q” is often defined to describe the sharpness of resonance peaks in both mechanical and electrical oscillating systems. “Q” is defined as

U

UQ

max2

For RLC circuit, Umax is 2maxmax 2

1LIU

Losses only come from R:

res

RIRTIU

2

2

1

2

1 2max

2max

This gives R

LQ res

And for completeness, note res

x

period


Power and resonance

<P>

00

o

0

2

Rrms

R=Ro

R=2Ro

Q=3

FWHM

For Q > few, FWHM

Q res

FWHM

Full Width at Half Maximum

Q

Quality of the peak

Higher Q = sharper peak = better quality

Transformers Transformers

• AC voltages can be stepped up or stepped down by the use of transformers.

The AC current in the primary circuitcreates a time-varying magnetic field in the iron

• The iron is used to maximize the mutual inductance. We assume that the entire flux produced by each turn of the primary is trapped in the iron.

21(primary) (secondary)

~

NN

iron

V2V1

This induces an emf on the secondarywindings due to the mutual inductance ofthe two sets of coils.

Transformers Ideal transformer without a load

1

1

N

V

dt

d turn

No resistance losses All flux contained in iron Nothing connected on secondary

N2N1(primary) (secondary)

iron

V2V1

The primary circuit is just an AC voltagesource in series with an inductor. Thechange in flux produced in each turn is given by:

• The change in flux per turn in the secondary coil is the same as the change in flux per turn in the primary coil (ideal case). The induced voltage appearing across the secondary coil is given by:

11

222 V

N

N

dt

dNV turn

• Therefore,

• N2 > N1 secondary V2 is larger than primary V1 (step-up) • N1 > N2 secondary V2 is smaller than primary V1 (step-down)

• Note: “no load” means no current in secondary. The primary current, termed “the magnetizing current” is small!

Transformers Ideal transformer with a load

R

VI 2

2

21

21 I

N

NI

N2N1

(primary) (secondary)

iron

V2V1 R

What happens when we connect a resistive load to the secondary coil?

Changing flux produced by primary coil inducesan emf in secondary which produces current I2

This current produces a flux in the secondary coilµ N2I2, which opposes the change in the originalflux -- Lenz’s law

This induced changing flux appears in the primarycircuit as well; the sense of it is to reduce the emf inthe primary, to “fight” the voltage source. However, V1 is assumed to be a voltage source. Therefore, theremust be an increased current I1 (supplied by the voltagesource) in the primary which produces a flux µ N1I1 which exactly cancels the flux produced by I2.

Transformers Ideal transformer with a load (cont’d)

Power is dissipated only in the load resistor R.

N2N1

(primary) (secondary)

iron

V2V1 R

22 2

dissipated 2 2 2

VP I R V I

R

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

1 2

2 1

I V

I V

2

1 1 2

1 1

NN V N

V N=

21 2

1

NI I

N

2

2 2 1 2

1 1

V N V N

R N R N

=

The primary circuit has to drive the resistance R

of the secondary.

Where did this power come from?It could come only from the voltage source in the

primary:

Exercises Exercise 1

Suppose m = 100 volts, f=1000 Hz, R=10 Ohms, L=4.22 mH,Find XL, Z, I, VR, and Vl.

XL L 6.28 10000.00422H 26.5

Z 102 (26.5)2 28.3

A. 53.33.28

100

ZI m

V. 3.3553.310 RIVR

V. 5.9353.35.26 IXV LL

22 )( LRZ

Exercises Exercise 2: Calculate power lost in R in Exercise 1

Pavg Irms2 R

Irms I

2

3.53A

1.4142.50A

Pavg (2.50A)210 62.5Watts

To calculate power produced by the generator you need to take account of the phase difference between the voltage and the current. In general you can write:

Pavg rmsIrms cos

For an inductor P = 0 because the phase difference between current through the inductor and voltage across the inductor is 90 degrees

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Chapter 28: Alternating Current Phasors and Alternating Currents Alternating current (AC current) Current which varies sinusoidally in time is called