chapter 33 Alternating Current Circuits

33.1 AC Sources33.2 Resistors in an AC Circuit33.3 Inductors in an AC Circuit33.4 Capacitors in an AC Circuit

33.5 The RLC Series Circuit33.6 Power in an AC Circuit

33.7 Resonance in a Series RLCCircuit

What is direct current?• Direct current is travels in one direction only

• Cells and batteries supply direct current

• Direct current is also known as d.c. current or d.c. supply

Direct current

• With d.c. current the voltage is constant

• The cathode ray oscilloscope (CRO) trace will show d.c. supply as a horizontal line

voltage

What is alternating current?

• Alternating current is also known as a.c. current of a.c supply

• This current is always changing direction, so the voltage goes up and down all the time too

Alternating current

• a.c. supply will produce a wave on the cathode ray oscilloscope

• The voltage with a.c. supply is constantly going up and down

AC Circuit

• An AC circuit consists of a combination of circuit elements and an AC generator or source

• The output of an AC generator is sinusoidal and varies with time according to the following equation– Δv = ΔVmax sin 2ƒt

• Δv is the instantaneous voltage• ΔVmax is the maximum voltage of the generator• ƒ is the frequency at which the voltage changes, in Hz

AC Voltage• The angular frequency is

– ƒ is the frequency of the source– T is the period of the source

• The voltage is positive during one half of the cycle and negative during the other half

• Commercial electric power plants in Canada/US use a frequency of 60 Hz– This corresponds with an angular frequency of 377 rad/s

22 ƒ

πω π

T

Resistors in an AC Circuit

• For a sinusoidal applied voltage, the current in a resistor is always in phase with the voltage across the resistor

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

• Resistors behave essentially the same way in both DC and AC circuits

RMS Current and Voltage• The average current in one cycle is zero• The rms current is the average of

importance in an AC circuit• rms stands for root mean square

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

07072max

max

II . Irms

07072max

max.rms

VV V

Resistor in an AC Circuit

• Consider a circuit consisting of an AC source and a resistor

• 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

Example:

In the U.S., standard wiring supplies 120 V at 60 Hz. Write this in sinusoidal form, assuming V(t)=0 at t=0.

This 120 V is the RMS amplitude: so Vp=Vrms = 170 V.This 60 Hz is the frequency f: so w=2 p f=377 s -1.

So V(t) = 170 sin(377t + fv).Choose fv=0 so that V(t)=0 at t=0: V(t) = 170 sin(377t).

2

More About Resistors in an AC Circuit

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

• The rate at which electrical energy is dissipated in the circuit is given by– P = i2 R

• where i is the 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

rms ValuesIt is useful to compare ac circuits and dc circuits. We

cannot use, however, the average ac current or voltage which are both zero. Instead we must use the root mean square or rms values.

The rms value of current or voltage in an ac circuit can be compared to the equivalent quantities in a simple dc circuit.

Vrms = IrmsR

Pav = Irms2R = Vrms

2/R

II

VV

rms rms max max,2 2

Root Mean Square Voltage and Current

Vrms = Square root of the mean (average) of V-squared.

2/ ,/2

1

)(sin

)sin()/(/)()(

)sin()(

22max

22max

22

max

max

Maxrmsrms VVRVR

VP

tR

VP

RIR

VPPower

tRVRtVtI

tVtV

Average AC Power consumed by a resistor independent of form of V vs t curve

RIRVP rmsrms22 /

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

maxmax

rms I707.02

II

maxmax

rms V707.02

VV

A 3.33-kW resistor is connected to a generator with a maximum voltage of 101 V. Find (a) the average and

(b) the maximum power delivered to this circuit.

WVRVP

WVRVP

VV

VV

Max

rms

rms

06.3)3330/()101(/

53.1)3330/()4.71(/

4.71414.1/101

2/

22max

22

max

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

– ΔVrms = Irms R• Also applies to the maximum values of v and i

Capacitors in an AC Circuit

• Consider a circuit containing a capacitor and an AC source

• The current starts out at a large value and charges the plates of the capacitor– There is initially no resistance to hinder the flow of the

current while the plates are not charged

• As the charge on the plates increases, the voltage across the plates increases and the current flowing in the circuit decreases

More About Capacitors in an AC Circuit

• The current reverses direction

• The voltage across the plates decreases as the plates lose the charge they had accumulated

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

Capacitors in an AC Circuit

•The circuit contains a capacitor and an AC source

•Kirchhoff’s loop rule gives: v + vc = 0 and so

v = vC = Vmax sin t– vc is the instantaneous

voltage across the capacitor

Phasor Diagram for Capacitor

•The phasor diagram shows that for a

sinusoidally applied voltage, the current

always leads the voltage across a capacitor by 90o

–This is equivalent to saying the voltage lags

the current

Capacitive Reactance

• The maximum current in the circuit occurs at cos t = 1 which gives

• The impeding effect of a capacitor on the current in an AC circuit is called the capacitive reactance and is given by

Voltage Across a Capacitor

• The instantaneous voltage across the capacitor can be written as vC = Vmax sin t = Imax XC sin t

• As the frequency of the voltage source increases, the capacitive reactance decreases and the maximum current increases

• As the frequency approaches zero, XC approaches infinity and the current approaches zero– This would act like a DC voltage and the capacitor

would act as an open circuit

Capacitive Reactance and Ohm’s Law

• The impeding effect of a capacitor on the current in an AC circuit is called the capacitive reactance and is given by

– When ƒ is in Hz and C is in F, XC will be in ohms• Ohm’s Law for a capacitor in an AC circuit

– ΔVrms = Irms XC

Cƒ2

1XC

Capacitive Reactance

The larger the capacitance, the smaller the capacitive reactance

As frequency increases, reactance decreasesDC: capacitor is an “open circuit” and

high frequency: capacitor is a “short circuit”

and

CX

0CX

fCXC 2

1

Capacitive Reactance

A particular example:

2.12 F 107.5Hz 1002

1

2

14-fC

XC

V0 = 50 V

+

- C = 750 mFf = 100 Hz

An rms voltage of 10.0 V with a frequency of 1.00 kHz is applied to a 0.395-mF capacitor. (a) What is the rms current in this circuit? (b) By what factor does the current change if the frequency of the voltage is doubled? (c) Calculate the current for a frequency of 2.00 kHz.

mAAVXVIA

V

VC

s

F

s

FHzCX

Crmsrms

C

8.240248.0)403/(0.10/

403403/

403

403)10395.0)(1000)(2(

1)/(1

6

If frequency is doubled, XC drops by factor of 2,Current is doubled: Irms = 49.6 mA @2.00 kHz

Inductors in an AC Circuit

• Consider an AC circuit with a source and an inductor

• The current in the circuit is impeded by the back emf of the inductor

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

Inductors in an AC Circuit

•Kirchhoff’s loop rule can be applied and

gives:

Current in an Inductor

• The equation obtained from Kirchhoff's loop rule can be solved for the current

• This shows that the instantaneous current iL in the inductor and the instantaneous voltage ∆vL across the inductor are out of phase by (p/2) rad = 90o

Phase Relationship of Inductors in an AC Circuit

• The current in the circuit is impeded by the back emf of the inductor

• For a sinusoidal applied voltage, the current in an inductor always lags behind the voltage across the inductor by 90° (π/2)

Phasor Diagram for an Inductor

• The phasors are at 90o with respect to each other

• This represents the phase difference between the current and voltage

• Specifically, the current lags behind the voltage by 90o

Inductive Reactance

• The factor wL has the same units as resistance and is related to current and voltage in the same way as resistance

• Because wL depends on the frequency, it reacts differently, in terms of offering resistance to current, for different frequencies

• The factor is the inductive reactance and is given by:– XL = wL

Inductive Reactance, cont.

• Current can be expressed in terms of the inductive reactance

• As the frequency increases, the inductive reactance increases– This is consistent with Faraday’s Law:

• The larger the rate of change of the current in the inductor, the larger the back emf, giving an increase in the inductance and a decrease in the current

Voltage Across the Inductor

• The instantaneous voltage across the inductor is

Inductive Reactance and Ohm’s Law

• The effective resistance of a coil in an AC circuit is called its inductive reactance and is given by– XL = 2ƒL

• When ƒ is in Hz and L is in H, XL will be in ohms

• Ohm’s Law for the inductor– ΔVrms = Irms XL

Inductive Reactance

Larger inductance: larger reactance (more induced EMF to oppose the applied AC voltage)

Higher frequency: larger impedance (higher frequency means higher time rate of change of current, which means more induced EMF to oppose the applied AC voltage)

fLX L 2

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

Current and Voltage Relationships in an RLC 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°

More About Voltage in RLC Circuits

• VR is the maximum voltage across the resistor and VR = ImaxR

• VL is the maximum voltage across the inductor and VL = ImaxXL

• VC is the maximum voltage across the capacitor and VC = ImaxXC

• In series, voltages add and the instantaneous voltage across all three elements would be v = vR + vL+ vC – Easier to use the phasor diagrams

Resulting Phasor Diagram

•The individual phasor diagrams can be

combined•Here a single phasor

Imax is used to represent the current

in each element–In series, the current is

the same in each element

Vector Addition of the Phasor Diagram

•Vector addition is used to combine the voltage

phasors•∆ VL and ∆VC are in

opposite directions, so they can be combined

•Their resultant is perpendicular to ∆VR

Total Voltage in RLC Circuits

•From the vector diagram, ∆Vmax can be calculated

Impedance

• The current in an RLC circuit is

• Z is called the impedance of the circuit and it plays the role of resistance in the circuit, where – Impedance has units of ohms– Also, ∆Vmax = ImaxZ

Impedance Triangle

•Since Imax is the same for each element, it

can be removed from each term in the phasor diagram

•The result is an impedance triangle

Impedance Triangle, cont.• The impedance triangle confirms that

• The impedance triangle can also be used to find the phase angle,

• The phase angle can be positive or negative and determines the nature of the circuit

• Also, cos =

Phasor Diagrams

• To account for the different phases of the voltage drops, vector techniques are used

• Represent the voltage across each element as a rotating vector, called a phasor

• The diagram is called a phasor diagram

Phasor Diagram for RLC Series Circuit

• The voltage across the resistor is on the +x axis since it is in phase with the current

• The voltage across the inductor is on the +y since it leads the current by 90°

• The voltage across the capacitor is on the –y axis since it lags behind the current by 90°

Phasor Diagram, cont

• 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

ΔVmax From the Phasor Diagram

• 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

R

CL

2CL

2Rmax

V

VVtan

)VV(VV

Impedance of a Circuit

• The impedance, Z, can also be represented in a phasor diagram

R

XXtan

)XX(RZ

CL

2CL

2

Impedance and Ohm’s Law

• Ohm’s Law can be applied to the impedance– ΔVmax = Imax Z

ProblemAn AC circuit with R= 2 W, C = 15 mF, and L = 30 mH is

driven by a generator with voltage V(t)=2.5 sin(8pt) Volts. Calculate the maximum current in the circuit, and the phase angle.

2 2( )L CZ R X X

2 212 (8 .030 ) 2.76

8 .015Z

Imax = 2.5/2.76 = .91 Amps

tan( ) L CX X

R

1(8 .030 )

8 .015 43.52

Imax = Vgen,max /ZL

R

C

Problem

An AC circuit with R= 2 W, C = 15 mF, and L = 30 mH is driven by a generator with voltage V(t)=2.5 sin(8pt) Volts. Calculate the maximum current in the circuit, and the phase angle.

2 2( )L CZ R X X

2 212 (8 .030 ) 2.76

8 .015Z

Imax = 2.5/2.76 = .91 Amps

tan( ) L CX X

R

1

(8 .030 )8 .015 43.5

2

Imax = Vgen,max /Z L

R

C

A 65.0-Hz generator with an rms voltage of 115 V is connected in series to a 3.35-kW resistor and a 1.50-mF capacitor. Find (a)

the rms current in the circuit and (b) the phase angle, f, between the current and the voltage.

IVR= IR

VC= I XC = I/(wC)

115 V

0.64 ),/(1/)/()(tan

109.30) 3726/(115

)]1050.1)(/65(2/[1)3350(/115/115

)()(115

3

262

2222

RCRXIRIX

AVI

FsVZVI

IZXRIIXIRV

CC

CC

Power in AC Circuits: Example• Series RLC Circuit: • R = 15W, C = 4.7 mF, L = 25mH.

The circuit is driven by a source of • Vrms = 75 V and f = 550 Hz.

• At what average rate is energy dissipated by the resistor?

29

)1

( 22

CLRZ

dd

srad

Hzd

/3457

)550)(2(

Paverage Irms2R

Erms

Z

2

R 75V

29

2

(15) 100.3W

Example: An AC power supply with Vmax=48 V is connected to a resistor with 12 W. Calculate (a) the rms current, (b) P and (c) Pmax.

(a) Irms=(0.70748 V)/12 W

Irms=2.83 A

(b) P=(0.70748 V)(2.83 A)

P=96 W

(c) Pmax=2P=192 W

Example

A coil has a resistance R = 1.00 Ω and an inductance of 0.300 H. Determine the current in the coil if (a) 120-V dc is applied to it, and (b) 120-V ac (rms) at 60.0 Hz is applied.

a. For dc current, there is no magnetic induction, so the current is determined by the resistance: I = V/R = 120 A.b. The inductive reactance is XL = 2πfL = 113 Ω. The current is Irms = Vrms/XL = 1.06 A.

Example

What is the rms current in the circuit shown if C = 1.0 μF and Vrms = 120 V? Calculate (a) for f = 60 Hz and then (b) for f = 6.0 x 105 Hz.

: a. XC = 1/2πfC = 2.7 kΩ; I = V/XC = 44 mA.b. XC = 0.27 Ω and I = 440 A.

Example

Suppose R = 25.0 Ω, L = 30.0 mH, and C = 12.0 μF, and they are connected in series to a 90.0-V ac (rms) 500-Hz source. Calculate (a) the current in the circuit, (b) the voltmeter readings (rms) across each element, (c) the phase angle , and (d) the power dissipated in the circuit.

a. XL = 2πfL = 94.2 Ω; XC = 1/(2πfC) = 26.5 Ω; so Z = 72.2 Ω and I = 1.25 A.b. The voltages are the currents multiplied by the reactances (or resistance). VL = 118 V; VC = 33.1 V; VR = 31.2 V.c. cos φ = 0.346, so φ = 69.7°.d. P = IV cos φ = 39.0 W.

Summary of Circuit Elements, Impedance and Phase Angles

Problem Solving for AC Circuits

• Calculate as many unknown quantities as possible– For example, find XL and XC

– Be careful of 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

Power in an AC Circuit

• No power losses are associated with capacitors and pure inductors in an AC circuit– In a capacitor, during one-half 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, cont

• The average power delivered by the generator is converted to internal energy in the resistor– Pav = IrmsΔVR = IrmsΔVrms 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, ƒo, where the current has its maximum value– To achieve maximum

current, the impedance must have a minimum value

– This occurs when XL = XC

LC2

1ƒo

Resonance in RLC circuit in series

22 )( CL XXRZ

R

XX CL1tan

For a fixed AC voltage source, maximum power dissipation occurs when XL = XC,

ie. When Z = R

PR = IDV = Irms2R

Where Irms = Vrms/Z

LC

1i.e.

Resonance, cont• 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• Metal Detector

– The portal is an inductor, and the frequency is set to a condition with no metal present

– When metal is present, it changes the effective inductance, which changes the current which is detected and an alarm sounds

Summary of Resonance

• At resonance– Z is minimum (=R)– Imax is maximum (=Vgen,max/R)– Vgen is in phase with I– XL = XC VL(t) = -VC(t)

• At lower frequencies– XC > XL Vgen lags I

• At higher frequencies– XC < XL Vgen lead I

Imax(XL-XC)

ImaxXL

ImaxXC

ImaxR

V gen,max

f

Power in AC circuits

• The voltage generator supplies power. – Only resistor dissipates power. – Capacitor and Inductor store and release energy.

• P(t) = I(t)VR(t) oscillates so sometimes power loss is large, sometimes small.

• Average power dissipated by resistor:P = ½ Imax VR,max

= ½ Imax Vgen,max cos(f)

= Irms Vgen,rms cos(f)

AC Summary

Resistors: VR,max=Imax R

In phase with ICapacitors: VC,max =Imax XC Xc = 1/(2pf C)

Lags IInductors: VL,max=Imax XL XL = 2pf L

Leads IGenerator: Vgen,max=Imax Z Z = √R2 +(XL -XC)2

Can lead or lag I tan(f) = (XL-XC)/R

Power is only dissipated in resistor: P = ½ImaxVgen,max cos(f)