chapter 28: alternating current - stony brook...

Chapter 28: Alternating Current

Phasors and Alternating CurrentsAlternating 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 sourcemeans either a source of alternating current or voltage.

amplitudecurrent I current, galternatinfor sinamplitude voltageV voltage,galternatinfor sin

====

tIitV

ωωυ

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

kin textboo sin,sinsin,sin :Note tIItVVtIitV PP ωωωωυ ==→==

Phasors and Alternating Currents

Phasorsω

O

ωt

IP

phasor• A convenient way to express a quantityvarying sinusoidally with time is bya phasor in phasor diagram as shown.

I=I P

sin

ωt

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(

21sinsin

222222 P

PPPIItItIItII =→−==→= ωωω

time averaged

2P

rmsII =

• Root-mean-square voltage of a sinusoidal voltage

2P

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

Reluctance

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

∼ε

IR

R

⇒sinR R mV RI tε ω= =

0

0 t

IRR

mε

Rmε

−

0

0

VR

t

mε

mε−

Voltage across R in phase with current through R

s i nmRI t

Rε ω=

IVR

IRωt

tm ωεε sin=Given:

υR

At time t

I=εm/R

Reluctance

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

⇒I L

∼

εL


tdtdILV m

LL ωε sin== tdt

LdI m

L ωε sin=

c o smL LI d I t

Lε ωω

= = −∫ ( )2/sin πωωε

−= tLm⇒

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

t

0

0

mε

mε−

VL

t

0

0

Lm

ωε

Lm

ωε

−

IL I

VL

IL

ωtε εm

I=εm/(ωL)At time t

Reluctance

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

∼ε

C IC


tCQV mC ωε sin== ⇒ tCQ m ωε sin=

)2

sin(cos πωεωωεω +=== tCtCdtdQI mmC⇒

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

0

0 t

mε

mε−

VC

t

0

0

mCεω

mCεω−

IC

I

VC

εm

ICωt

At time t

ωtI=ωCεm

ε

Reluctance

LRC series circuit and reluctanceLRC circuit summary


Assume the solution for current: )sin()( φω −= tItI m

)cos(

)cos(1)sin(

φωω

φωω

φω

−=

−−=

−=

tLIV

tIC

V

tRIV

mL

mC

mR

(See derivation later)

R mV I R∝ 1C mV I

Cω∝ L mV I Lω∝

XCXL

reactance

amplitude

Reluctance

LRC series circuit and reluctance (cont’d)

You can think of it as a frequency-dependent resistance.What is reactance? f=ω/2π

1CX

Cω=

For high ω, χC~0- Capacitor looks like a wire (“short”)

For low ω, χC ∞- Capacitor looks like a break

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 Lω=

( " " )RX R=

LRC Circuits

LC

∼ε

R

φ

ω

φ

φ

Im R

Imω L

I mω C

εm

⇒

)sin( φω −== tRIRIV mR

)cos( φωω −== tLIdtdILV mL

)cos(1 φωω

−−== tICC

QV mC

• Assume:• Given: tm ωεε sin=

)sin( φω −= tII m ⇒)cos( φω

ω−−= tIQ m

)cos( φωω −= tIdtdI

m

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

Im

)cos( )2/sin( : Note

φωπφω

−=+−

tt

LRC Circuits

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

(No L or C → f = 0)

LC

∼ε

R

0at )sin( )sin(

=−=


−=ti

tii

m

mR

φφω −φ

φ

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

LRC Circuits


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.

-φVR = I R

VL= I XL

VC = I XC

−φ

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.

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

Phasor diagrams for LRC circuits: Example

y

xε

VC

IR( )2 2 2( )CIR IX ε+ =

2 2 2 2( )CI R X ε+ =

2 2C

IR X

ε=

+

~

Vout

y

xε

LRC Circuits

amplitude of current

LRC CircuitsFilters : Example

~Vout

22out

C

RV IRR X

ε= =+

( ) ( )02 22 1

1

1

out

C

V R

R ωω ω

ε= =

+ +Ex.: C = 1 µF, R = 1Ω

01

RCω =

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 ωπ

=

LRC CircuitsFilters

outVε

0ω ω

High-pass filterω=0 No current

Vout ≈ 0

ω=∞ Capacitor ~ wireVout ≈ ε

~Vout

outVε

0ω ω

~Vout

ω = ∞ No currentVout ≈ 0

ω = 0 Inductor ~ wireVout ≈ ε

~

ω = 0 No current because of capacitor

ω = ∞ No current because of inductor

Low-pass filter

Band-pass filter

outVε

0ω(Conceptual sketch only)

LRC CircuitsPhasor diagrams for LRC circuits: Example 2

⇒Im R

εm

φ

Im (XL -X C)

φφ

φ

Im R

ε m

Im XC

Im XL

LX L ω≡

CX C ω

1≡

( )22CL XXRZ −+≡

RXX CL −

=φtanReluctance for inductor

Reluctance for capacitor

⇓( )( )2222

CLmm XXRI −+=ε

( ) ZXXRI m

CL

mm

εε=

−+=

22

amplitude

Impedance Z

LRC CircuitsPhasor 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

εm

“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

εm

Im XC

ImX

Ly

xf f

f

CLm XXI −ZIm

RIm

“Impedance Triangle”

f| |

φ

φ

φ

φ

φ

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.

( )22CL

mmm

XXRZI

−+==

εεi.e.:

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

⇒C

Lo

o ωω 1

=LC

fLCo π

ω2

1;10 ==

LC

~ε

R

resonance frequency

• 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 CLφ


Resonance (cont’d)

Im

002ωoω

0Rmε

R=Ro

R=2Ro

R

XL

XC

φ

Z

XL - XC

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

RXX CL −

=φtan

cos φR

I mm

ε=

ZI m

mε

=φcos

RZ =


Resonance (cont’d)

On Resonance:LC

~ε

R

RV IR ε= = IRε

=

φ=0 and Z=R

LL L

XV IX QR

ε ε= = = CC C

XV IX QR

ε ε= = =

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

Power in Alternating Current Circuits

Power

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

( )( ))sin(sin)()()( φωωεε −== tIttIttP mm

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

⟩−⟨=⟩⟨ )sin(sin)( φωωε ttItP mm

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


Power

sinωtcosωt

ω t0 2π

0

+1

-1

(Product of even and odd function = 0)

sin2ωt

ω t0 2π

0

+1

-1

⟩⟨−⟩⟨=⟩⟨ tttItP mm ωωφωφε cossinsinsincos)( 2

φε cos21)( mm ItP =⟩⟨

01/2

• Expanding,

• Taking the averages,

)sincoscos(sinsin)sin(sin φωφωωφωω ttttt −=−

0cossin =⟩⟨ tt ωω

• Generally:

∫ ==⟩⟨π

π

2

0

22

21sin

21sin xdxx

• Putting it all back together again,


Power This result is often rewritten in terms of rms values:

mrms εε2

1≡ mrms II

21

≡ φε cos)( rmsrmsItP =⟩⟨

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

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

Therefore...

φε cos)( rmsrms ItP =⟩⟨


Power Power, as well as current, peaks at ω = ω0. The sharpness of the resonance

depends on the values of the components.

Recall:

φε cosR

I mm =

22 2

rms( ) cosrmsP t I RR

ε φ⟨ ⟩ = =

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

2222

22

)1()(

−+=⟩⟨

xQxx

RtP rmsε

…introducing the curious factors Q and x...


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

UUQ∆

≡ max2π

2maxmax 2

1 LIU =For RLC circuit, Umax is

Losses only come from R:⎟⎟⎠

⎞⎜⎜⎝

⎛==∆

res

RIRTIUω

π221

21 2

max2max

periodR

LQ resω≡This gives

res

xω

ω≡And for completeness, note


Power and resonance

FWHMQ resω

≈For Q > few,

<P>

002ωoω

0

2

Rrmsε

R=Ro

R=2Ro

Q=3

FWHMFWHM

Full Width at Half Maximum

QQuality of the peakHigher Q = sharper peak = better quality

TransformersTransformers

• 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

21(primary) (secondary)

~ε

NN

iron

V2V1

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

• 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.

TransformersIdeal transformer without a load

Nothing connected on secondaryNo resistance losses All flux contained in iron

1

1

NV

dtd turn =

φ

∼ε

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 secondarycoil 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

NN

dtdNV turn ==

φ• 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!

TransformersIdeal transformer with a load

RVI 2

2 =

∼ε


iron

V2V1 R

What happens when we connect a resistive loadto 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 N1I1which exactly cancels the flux produced by I2.

21

21 I

NNI =

TransformersIdeal transformer with a load (cont’d)

Power is dissipated only in the load resistor R.∼ε


iron

V2V1 R2

2 2dissipated 2 2 2

VP I R V IR

= = =

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

∴ 1 1 2 2V I V I=generated 1 1P V I=

1 2

2 1

I VI V

=2

1 1 2

1 1

NN V NV N

==

21 2

1

NI IN

=2

2 2 1 2

1 1

V N V NR N R N

⎛ ⎞= ⎜ ⎟

⎝ ⎠=The primary circuit has to drive the resistance R

of the secondary.

ExercisesExercise 1

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

XL = ωL = 6.28 ×1000 × 0.00422H = 26.5Ω

Z = 102 + (26.5)2 = 28.3Ω

A. 53.33.28

100===

ZI mε

V. 3.3553.310 =×== RIVR

22 )( LRZ ω+=

V. 5.9353.35.26 =×== IXV LL

ExercisesExercise 2: Calculate power lost in R in Exercise 1

Pavg = Irms2 R

Irms =I2

=3.53A1.414

= 2.50A

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 = εrmsIrms cosφ

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

chapter 28: alternating current - stony brook...

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