Alternating Current Circuits

Chapter 33

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 πT

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

Rv Vi ωt ωtR 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

2rmsII I

maxmax0.707

2rmsVV 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 0Cqv

dtdqiC tVC cosmax

2sinmax

tVC

CVI/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

dtdqiC tVC cosmax

2sinmax

tVC

CVI/1max

max

CX C

1

CXIV maxmax

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 0dtdiLv

tdtL

ViL sinmax tL

V

cosmax

tV sinmax

LVImax

max

dtdiLv L

2sinmax

t

LV

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 tL

V

cosmax

tV sinmax

LVImax

max

dtdiLv L

2sinmax

t

LV

CXIV maxmax

Chapter 33Problem 11

Determine the maximum magnetic flux through an inductor connected to a standard electrical outlet (ΔVrms= 120 V, f = 60.0 Hz).

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 ωtπv 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 ( )

tanR L C

L C

R

V V V VV V

V

Phasor DiagramsΔVR = Imax RΔVL = Imax XL

ΔVC = Imax XC

2 2max ( )

tanR L C

L C

R

V V V VV V

V

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( )

tanL C

L C

Z R X XX X

R

Summary of Circuit Elements, Impedance and Phase Angles

Problem Solving for AC Circuits

• Calculate as many unknown quantities as possible (e.g., find XL and XC)

• Be careful with units – use F, H, Ω

• Apply Ohm’s Law to the portion of the circuit that is of interest

• Determine all the unknowns asked for in the problem

Chapter 33Problem 24

An AC source with Vmax = 150 V and f = 50.0 Hz is connected between points a and d in the figure. Calculate the maximum voltages between (a) points a and b, (b) points b and c, (c) points c and d, and (d) points b and d.

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

Chapter 33Problem 28

A series RLC circuit has a resistance of 45.0 Ω and an impedance of 75.0 Ω. What average power is delivered to this circuit when ΔVrms = 210 V?

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

LC1

0

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

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

NNV

2211 VIVI

Chapter 33Problem 43

A transmission line that has a resistance per unit length of 4.50 × 10-4 Ω/m is to be used to transmit 5.00 MW across 400 miles (6.44 × 105 m). The output voltage of the generator is 4.50 kV. (a) What is the line loss if a transformer is used to step up the voltage to 500 kV? (b) What fraction of the input power is lost to the line under these circumstances?

Answers to Even Numbered Problems

Chapter 33:

Problem 4

(a) 25.3 rad/s (b) 0.114 s

Answers to Even Numbered Problems

Chapter 33:

Problem 10

3.80 J