Alternating Current Circuits

Chapter 28

AC Circuit• An AC circuit consists of a combination of circuit

elements and an AC generator or source

• The output of an AC power source is sinusoidal and varies with time according to the following equation

Δv = ΔVmax sin ωt

• Δv: instantaneous voltage

• ΔVmax is the maximum voltage (amplitude) of the

generator

• ω is the angular frequency of the AC voltage

22 ƒ

πω π

T

Resistors in an AC Circuit• Consider a circuit consisting of

an AC source and a resistor

ΔvR = ΔVmax sin ωt

• ΔvR is the instantaneous voltage across the resistor

• The instantaneous current in the resistor is

• The instantaneous voltage across the resistor is also given as ΔvR = ImaxR sin ωt

sin sin maxmaxIR

R

v Vi ωt ωt

R R

Resistors in an AC Circuit• The graph shows the current

through and the voltage across the resistor

• The current and the voltage reach their maximum values at the same time

• The current and the voltage are said to be in phase

• The direction of the current has no effect on the behavior of the resistor

Resistors in an AC Circuit• The rate at which electrical

energy is dissipated in the circuit is given by

• i: instantaneous current

• The heating effect produced by an AC current with a maximum value of Imax is not the same as that of a DC current of the same value

• The maximum current occurs for a small amount of time

2i R

rms Current and Voltage• The rms current is the direct current that would

dissipate the same amount of energy in a resistor as is actually dissipated by the AC current

• Alternating voltages can also be discussed in terms of rms values

• The average power dissipated in resistor in an AC circuit carrying a current I is

maxmax0.707

2rms

II I

maxmax0.707

2rms

VV V

2av rmsI R

Ohm’s Law in an AC Circuit• rms values will be used when discussing AC currents

and voltages

• AC ammeters and voltmeters are designed to read rms values

• Many of the equations will be in the same form as in DC circuits

• Ohm’s Law for a resistor, R, in an AC circuit

ΔVR,rms = Irms R

• The same formula applies to the maximum values of v and i

Capacitors in an AC Circuit• Consider a circuit containing a

capacitor and an AC source

• Kirchhoff’s loop rule gives:

• ΔvC: instantaneous voltage across the capacitor

tVCtq sin)( max

0 Cvv 0C

qv

dt

dqiC tVC cosmax

2sinmax

tVC

C

VI

/1max

max

Capacitors in an AC Circuit• The voltage across the capacitor

lags behind the current by 90°

• The impeding effect of a capacitor on the current in an AC circuit is called the capacitive reactance (measured in ohms):

tVCtq sin)( max

dt

dqiC tVC cosmax

2sinmax

tVC

C

VI

/1max

max

CX C

1

CXIV maxmax

Chapter 28Problem 22

A capacitor and a 1.8-kΩ resistor pass the same current when connected across 60-Hz power. Find the capacitance.

Inductors in an AC Circuit• Consider an AC circuit with a

source and an inductor

• Kirchhoff’s loop rule gives:

• ΔvL: instantaneous voltage across the inductor

0 Lvv 0dt

diLv

tdtL

ViL sinmax t

L

V

cosmax

tV sinmax

L

VI

max

max

dt

diLv L

2sinmax

t

L

V

Inductors in an AC Circuit• The voltage across the inductor

always leads the current by 90°

• The effective resistance of a coil in an AC circuit is called its inductive reactance (measured in ohms): LX L

tdtL

ViL sinmax t

L

V

cosmax

tV sinmax

L

VI

max

max

dt

diLv L

2sinmax

t

L

V

LXIV maxmax

LC Circuit• A capacitor is connected to an inductor in an LC

circuit

• Assume the capacitor is initially charged and then the switch is closed

• Assume no resistance and no energy losses to radiation

• The current in the circuit and thecharge on the capacitor oscillatebetween maximum positive andnegative values

LC Circuit• With zero resistance, no energy is transformed into

internal energy

• Ideally, the oscillations in the circuit persist indefinitely (assuming no resistance and no radiation)

• The capacitor is fully charged and the energy in the circuit is stored in the electric field of the capacitor

Q2max / 2C

• No energy is stored in the inductor

• The current in the circuit is zero

LC Circuit• The switch is then closed

• The current is equal to the rate at which the charge changes on the capacitor

• As the capacitor discharges, the energy stored in the electric field decreases

• Since there is now a current, someenergy is stored in the magneticfield of the inductor

• Energy is transferred from theelectric field to the magnetic field

LC Circuit• Eventually, the capacitor becomes fully discharged

and it stores no energy

• All of the energy is stored in the magnetic field of the inductor and the current reaches its maximum value

• The current now decreases in magnitude, recharging the capacitor with its plates having opposite their initial polarity

• The capacitor becomes fullycharged and the cycle repeats

• The energy continues to oscillatebetween the inductor and the capacitor

LC Circuit• The total energy stored in the LC circuit remains

constant in time

• Solution:

22

22 LI

C

QU 0

22

22

LI

dt

d

C

Q

dt

d

dt

dU

0dt

dILI

dt

dQ

C

Q0

2

2

dt

QdL

C

Q

LC

Q

dt

Qd

2

2

LC

tQtQ cos)( max

LC Circuit• The angular frequency, ω, of the circuit depends on

the inductance and the capacitance

• It is the natural frequency of oscillation of the circuit

• The current can be expressed as a function of time:

LC

tQtQ cos)( max tQtQ cos)( max

1ωLC

tQdt

d

dt

dQtI cos)( max tQ sinmax

tItI sin)( max

LC Circuit• Q and I are 90° out of phase with each other, so when

Q is a maximum, I is zero, etc.

tQtQ cos)( max

tItI sin)( max

Energy in LC Circuits• The total energy can be expressed as a function of

time

• The energy continually oscillatesbetween the energy stored in theelectric and magnetic fields

• When the total energy is stored inone field, the energy stored in theother field is zero

22

22 LI

C

QU t

LIt

C

Q 22

max22

max sin2

cos2

Energy in LC Circuits

• In actual circuits, there is always some resistance

• Therefore, there is some energy transformed to internal energy

• Radiation is also inevitable in this type of circuit

• The total energy in the circuit continuously decreases as a result of these processes

Chapter 28Problem 27

An LC circuit with a 20-µF capacitor oscillates with period 5.0 ms. The peak current is 25 mA. Find (a) the inductance and (b) the peak voltage.

The RLC Series Circuit• The resistor, inductor, and capacitor

can be combined in a circuit

• The current in the circuit is the same at any time and varies sinusoidally with time

The RLC Series Circuit• The instantaneous voltage across the

resistor is in phase with the current

• The instantaneous voltage across the inductor leads the current by 90°

• The instantaneous voltage across the capacitor lags the current by 90°

max

max

max

sin sin

sin cos 2

sin cos 2

R R

L L L

C C C

v I R ωt V ωt

πv I X ωt V ωt

πv I X ωt V ωt

Phasor Diagrams• Because of the different phase

relationships with the current, the voltagescannot be added directly

• To simplify the analysis of AC circuits, a graphical constructor called a phasor diagram can be used

• A phasor is a vector rotating CCW; its length is proportional to the maximum value of the variable it represents

• The vector rotates at an angular speed equal to the angular frequency associated with the variable, and the projection of the phasor onto the vertical axis represents the instantaneous value of the quantity

Phasor Diagrams• The voltage across the resistor is in phase with the

current

• The voltage across the inductor leads the current by 90°

• The voltage across the capacitor lags behind the current by 90°

Phasor Diagrams• The phasors are added as vectors

to account for the phase differences in the voltages

• ΔVL and ΔVC are on the same line and so the net y component is ΔVL - ΔVC

Phasor Diagrams• The voltages are not in phase, so

they cannot simply be added to get the voltage across the combination of the elements or the voltage source

is the phase angle between the current and the maximum voltage

• The equations also apply to rms values

2 2max ( )

tan

R L C

L C

R

V V V V

V VV

Phasor Diagrams

ΔVR = Imax RΔVL = Imax XL

ΔVC = Imax XC

2 2max ( )

tan

R L C

L C

R

V V V V

V VV

22maxmax )( CL XXRIV

Impedance of a Circuit• The impedance, Z, can also be represented in a phasor

diagram

• φ: phase angle

• Ohm’s Law can be applied to the impedance

ΔVmax = Imax Z

• This can be regarded as a generalized form of Ohm’s Law applied to a series AC circuit

2 2( )

tan

L C

L C

Z R X X

X XR

Summary of Circuit Elements, Impedance and Phase Angles

Chapter 28Problem 30

Find the impedance at 10 kHz of a circuit consisting of a 1.5-kΩ resistor, 5.0-µF capacitor, and 50-mH inductor in series.

Power in an AC Circuit

• No power losses are associated with pure capacitors and pure inductors in an AC circuit

• In a capacitor, during 1/2 of a cycle energy is stored and during the other half the energy is returned to the circuit

• In an inductor, the source does work against the back emf of the inductor and energy is stored in the inductor, but when the current begins to decrease in the circuit, the energy is returned to the circuit

Power in an AC Circuit

• The average power delivered by the generator is converted to internal energy in the resistor

Pav = Irms ΔVR,rms

ΔVR, rms = ΔVrms cos

Pav = Irms ΔVrms cos

• cos is called the power factor of the circuit

• Phase shifts can be used to maximize power outputs

Resonance in an AC Circuit• Resonance occurs at the frequency,

ω0, where the current has its maximum

value

• To achieve maximum current, the impedance must have a minimum value

• This occurs when XL = XC and

22 )( CL

rmsrms

XXR

VI

CL

00

1

LC

10

Resonance in an AC Circuit• Theoretically, if R = 0 the current

would be infinite at resonance

• Real circuits always have some resistance

• Tuning a radio: a varying capacitor changes the resonance frequency of the tuning circuit in your radio to match the station to be received

Chapter 28Problem 29

A series RLC circuit has R = 75 Ω, L = 20 mH, and resonates at 4.0 kHz. (a) What’s the capacitance? (b) Find the circuit’s impedance at resonance and (c) at 3.0 kHz.

Damped LC Oscillations• The total energy is not constant, since there is a

transformation to internal energy in the resistor at the rate of dU/dt = – i2R

• Radiation losses are still ignored

• The circuit’s operation can be expressed as:

0 iRdt

diL

C

q

02

2

Rdt

dq

dt

qdL

C

q

Damped LC Oscillations

• Solution:

• Analogous to a damped harmonic oscillator

• When R = 0, the circuit reduces to an LC circuit (no damping in an oscillator)

teQtq dL

Rt

cos)( 2max

2

2

1

L

R

LCd

02

2

Rdt

dq

dt

qdL

C

q

Transformers• An AC transformer consists of two coils of wire wound

around a core of soft iron

• The side connected to the input AC voltage source is called the primary and has N1 turns

• The other side, called the secondary, is connected to a resistor and has N2

turns

• The core is used to increase the magnetic flux and to provide a medium for the flux to pass from one coil to the other

Transformers

• The rate of change of the flux is the same for both coils, so the voltages are related by

• When N2 > N1, the transformer is referred to as a step

up transformer and when N2 < N1, the transformer is

referred to as a step down transformer• The power input into the primary

equals the power output at the secondary

tNV B

11 tNV B

22

11

22 V

N

NV

2211 VIVI

Chapter 28Problem 37

You’re planning to study in Europe, and you want a transformer designed to step 230-V European power down to 120 V needed to operate your stereo. (a) If the transformer’s primary has 460 turns, how many should the secondary have? (b) You can save money with a transformer whose maximum primary current is 1.5 A. If your stereo draws a maximum of 3.3 A, will this transformer work?

Answers to Even Numbered Problems

Chapter 28:

Problem 14

V = (325 V) sin[(314 s−1)t]

Answers to Even Numbered Problems

Chapter 28:

Problem 26

22 to 190 pF