Chapter 32A – AC Circuits A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University A PowerPoint Presentation.

Chapter 32A – AC Chapter 32A – AC CircuitsCircuits

A PowerPoint Presentation byA PowerPoint Presentation by

Paul E. Tippens, Professor of Paul E. Tippens, Professor of PhysicsPhysics

Southern Polytechnic State Southern Polytechnic State UniversityUniversity

A PowerPoint Presentation byA PowerPoint Presentation by

Paul E. Tippens, Professor of Paul E. Tippens, Professor of PhysicsPhysics

Southern Polytechnic State Southern Polytechnic State UniversityUniversity© 2007

Objectives: Objectives: After completing After completing this module, you should be this module, you should be

able to:able to:

• Write and apply equations for Write and apply equations for calculating the calculating the inductive and inductive and capacitive reactancescapacitive reactances for inductors for inductors and capacitors in an ac circuit.and capacitors in an ac circuit.• Describe, with diagrams and equations, Describe, with diagrams and equations, thethe phase relationshipsphase relationships for circuits for circuits containing containing resistance,resistance, capacitance,capacitance, and and inductanceinductance..

• Describe the sinusoidal variation in Describe the sinusoidal variation in ac ac current and voltagecurrent and voltage, and calculate , and calculate their their effectiveeffective values. values.

Objectives (Cont.)Objectives (Cont.)

• Write and apply equations for Write and apply equations for calculating the calculating the impedanceimpedance, the , the phase phase angleangle, the , the effective currenteffective current, the , the average average powerpower, and the , and the resonant frequencyresonant frequency for a for a series ac circuit.series ac circuit.

• Describe the basic operation of a Describe the basic operation of a step-upstep-up and a and a step-down step-down transformertransformer..

• Write and apply the Write and apply the transformer transformer equationequation and determine the and determine the efficiencyefficiency of a transformer. of a transformer.

Alternating CurrentsAlternating Currents

An An alternating currentalternating current such as that such as that produced by a generator has no produced by a generator has no direction in the sense that direct current direction in the sense that direct current has. The magnitudes vary has. The magnitudes vary sinusoidallysinusoidally with time as given by:with time as given by:

An An alternating currentalternating current such as that such as that produced by a generator has no produced by a generator has no direction in the sense that direct current direction in the sense that direct current has. The magnitudes vary has. The magnitudes vary sinusoidallysinusoidally with time as given by:with time as given by:

Emax

iimaxmax

time, t

E = Emax sin i = imax sin

AC-voltage and current

450 900 1350

1800 2700 3600

E

R = Emax

E = Emax sin

Rotating Vector Rotating Vector DescriptionDescription

The coordinate of the emf at any instant is the value of Emax sin Observe for

incremental angles in steps of 450. Same is true for i.

The coordinate of the emf at any instant is the value of Emax sin Observe for

incremental angles in steps of 450. Same is true for i.

450 900 1350

1800 2700 3600

E

Radius = Emax

E = Emax sin

Effective AC CurrentEffective AC Currentiimaxmax

The average The average current in a cycle current in a cycle is zero—half + and is zero—half + and half -.half -.But energy is But energy is expended, regardless expended, regardless of direction. So the of direction. So the “root-mean-square”“root-mean-square” value is useful.value is useful.

2

2 0.707rms

I II

2

2 0.707rms

I II

I = imax

The The rmsrms value value IIrmsrms is is sometimes called sometimes called the the effectiveeffective current current IIeffeff::

The effective ac current:ieff = 0.707 imax

AC DefinitionsAC DefinitionsOne One effective ampereeffective ampere is that ac is that ac current for which the power is the current for which the power is the same as for one ampere of dc current.same as for one ampere of dc current.

One One effective volteffective volt is that ac voltage is that ac voltage that gives an effective ampere that gives an effective ampere through a resistance of one ohm.through a resistance of one ohm.

Effective current: ieff = 0.707 imax

Effective current: ieff = 0.707 imax

Effective voltage: Veff = 0.707 Vmax

Effective voltage: Veff = 0.707 Vmax

Example 1:Example 1: For a particular device, the For a particular device, the house ac voltage is house ac voltage is 120-V120-V and the ac and the ac current is current is 10 A10 A. What are their . What are their maximummaximum values?values?

ieff = 0.707 imax

ieff = 0.707 imax

Veff = 0.707 Vmax

Veff = 0.707 Vmax

max

10 A

0.707 0.707effii max

120V

0.707 0.707effVV

imax = 14.14 Aimax = 14.14 A Vmax = 170 V

Vmax = 170 V

The ac voltage actually varies from The ac voltage actually varies from +170 +170 V to -170 VV to -170 V and the current from and the current from 14.1 14.1 A to –14.1 AA to –14.1 A..

The ac voltage actually varies from The ac voltage actually varies from +170 +170 V to -170 VV to -170 V and the current from and the current from 14.1 14.1 A to –14.1 AA to –14.1 A..

Pure Resistance in AC Pure Resistance in AC CircuitsCircuits

A

a.c. Source

R

V

Voltage and current are in phase, and Voltage and current are in phase, and Ohm’s law applies for effective currents Ohm’s law applies for effective currents

and voltages.and voltages.

Voltage and current are in phase, and Voltage and current are in phase, and Ohm’s law applies for effective currents Ohm’s law applies for effective currents

and voltages.and voltages.

Ohm’s law: Veff = ieffR

Vmax

iimaxmax

Voltage

Current

AC and InductorsAC and Inductors

Time, t

I i

Current Current RiseRise

0.63I

Inductor

The voltage The voltage V V peaks first, causing rapid rise peaks first, causing rapid rise in in i i current which then peaks as the emf current which then peaks as the emf goes to zero. Voltage goes to zero. Voltage leadsleads ( (peaks beforepeaks before) ) the current by 90the current by 9000. . Voltage and current are Voltage and current are out of phaseout of phase..

Time, t

I i

Current Current DecayDecay

0.37I

Inductor

A Pure Inductor in AC A Pure Inductor in AC CircuitCircuit

A

L

V

a.c.

Vmax

iimaxmax

Voltage

Current

The voltage peaks 90The voltage peaks 900 0 before the current before the current peaks. One builds as the other falls and peaks. One builds as the other falls and

vice versa.vice versa.

The voltage peaks 90The voltage peaks 900 0 before the current before the current peaks. One builds as the other falls and peaks. One builds as the other falls and

vice versa.vice versa.The The reactance reactance may be defined as the may be defined as the nonresistive oppositionnonresistive opposition to the flow of ac to the flow of ac current.current.

Inductive ReactanceInductive Reactance

A

L

V

a.c.

The The backback emfemf induced by a induced by a changing current changing current provides opposition provides opposition to current, called to current, called inductive reactance inductive reactance XXLL..Such losses are Such losses are temporarytemporary, however, since , however, since the current the current changes directionchanges direction, periodically , periodically re-supplying energy so that no net power is re-supplying energy so that no net power is lost in one cycle.lost in one cycle.Inductive reactanceInductive reactance XXLL is a function of is a function of both the both the inductanceinductance and the and the frequencyfrequency of of the ac current.the ac current.

Calculating Inductive Calculating Inductive ReactanceReactance

A

L

V

a.c.

Inductive Reactance:2 Unit is the LX fL

Ohm's law: L LV iX

The The voltagevoltage reading reading VV in the above circuit in the above circuit at the instant the at the instant the acac current is current is ii can be can be found from the found from the inductanceinductance in in HH and the and the frequencyfrequency in in HzHz..

(2 )LV i fL (2 )LV i fL Ohm’s law: VL = ieffXL

Example 2:Example 2: A coil having an A coil having an inductance of inductance of 0.6 H0.6 H is connected to a is connected to a 120-V120-V, , 60 Hz60 Hz ac source. Neglecting ac source. Neglecting resistance, what is the effective current resistance, what is the effective current through the coil?through the coil?

A

L = 0.6 H

V

120 V, 60 Hz

Reactance: XReactance: XLL = = 22fLfLXXLL = = 22(60 Hz)(0.6 (60 Hz)(0.6

H)H)XXLL = 226 = 226

120V

226 eff

effL

Vi

X

ieff = 0.531 Aieff = 0.531 A

Show that the peak current is Show that the peak current is IImaxmax = = 0.7500.750 AA

AC and AC and CapacitanceCapacitance

Time, t

Qmaxq

Rise in Rise in ChargeCharge

Capacitor

0.63 I

Time, t

I i

Current Current DecayDecay

Capacitor

0.37 I

The voltage The voltage VV peaks ¼ of a cycle after the peaks ¼ of a cycle after the current current ii reaches its maximum. The voltage reaches its maximum. The voltage lagslags the current. the current. Current Current ii and V out of and V out of phasephase..

A Pure Capacitor in AC A Pure Capacitor in AC CircuitCircuit

Vmax

iimaxmax

Voltage

CurrentA V

a.c.

C

The voltage peaks 90The voltage peaks 900 0 after after the current the current peaks. One builds as the other falls and peaks. One builds as the other falls and

vice versa.vice versa.

The voltage peaks 90The voltage peaks 900 0 after after the current the current peaks. One builds as the other falls and peaks. One builds as the other falls and

vice versa.vice versa.The diminishing current The diminishing current ii builds charge on builds charge on

CC which increases the which increases the back emfback emf of of VVCC..

The diminishing current The diminishing current ii builds charge on builds charge on CC which increases the which increases the back emfback emf of of VVCC..

Capacitive ReactanceCapacitive Reactance

No No net powernet power is lost in a complete cycle, is lost in a complete cycle, even though the capacitor does provide even though the capacitor does provide nonresistive opposition (nonresistive opposition (reactancereactance) to the ) to the flow of ac current.flow of ac current.Capacitive reactanceCapacitive reactance XXCC is affected by both is affected by both the the capacitancecapacitance and the and the frequencyfrequency of the of the ac current.ac current.

A V

a.c.

CEnergyEnergy gains and gains and losses are also losses are also temporarytemporary for for capacitors due to the capacitors due to the constantly changing constantly changing ac current.ac current.

Calculating Inductive Calculating Inductive ReactanceReactance

Capacitive Reactance:1

Unit is the 2CX fC

Ohm's law: VC CiX

The The voltagevoltage reading reading VV in the above circuit in the above circuit at the instant the at the instant the acac current is current is ii can be can be found from the found from the inductanceinductance in in FF and the and the frequencyfrequency in in HzHz..

2L

iV

fL

2L

iV

fL

A V

a.c.

C

Ohm’s law: VC = ieffXC

Example 3:Example 3: A 2- A 2-F capacitor is F capacitor is connected to a 120-V, 60 Hz ac source. connected to a 120-V, 60 Hz ac source. Neglecting resistance, what is the Neglecting resistance, what is the effective current through the coil?effective current through the coil?

ReactanceReactance::

XXCC = 1330 = 1330

120V

1330 eff

effC

Vi

X

ieff = 90.5 mAieff = 90.5 mA

Show that the peak current is Show that the peak current is iimaxmax = = 128 128 mAmA

A V

C = 2 F

120 V, 60 Hz

1

2CX fC

-6

1

2 (60Hz)(2 x 10 F)CX

Memory Aid for AC Memory Aid for AC ElementsElementsAn An oldold, but very , but very effective, way to effective, way to

remember the remember the phase phase differencesdifferences for for inductorsinductors and and capacitorscapacitors is : is :

““E E L I” the “L I” the “i i C C EE” ” ManMan

Emf Emf EE is is beforebefore current current ii in in inductors inductors LL; Emf ; Emf EE is is after after current current ii

in capacitors in capacitors C. C.

Emf Emf EE is is beforebefore current current ii in in inductors inductors LL; Emf ; Emf EE is is after after current current ii

in capacitors in capacitors C. C.

“E E LL i”

“I C EE”man

the

Frequency and AC CircuitsFrequency and AC Circuits

ff

R, XR, X

1

2CX fC

1

2CX fC2LX fL

ResistanceResistance R R is constant and not affected is constant and not affected by by f.f.

Inductive reactanceInductive reactance XXLL

varies directly with varies directly with frequency as expected frequency as expected since since EE i/i/tt..

Capacitive reactanceCapacitive reactance XXCC

variesvaries inversely inversely with with ff since rapid ac allows little since rapid ac allows little time for charge to build up time for charge to build up on capacitors.on capacitors.

RR

XXLLXXCC

Series LRC CircuitsSeries LRC Circuits

L

VR VC

CRa.c.

VL

VT

A

Series ac circuit

Consider an Consider an inductorinductor LL,, a a capacitorcapacitor CC,, and a and a resistorresistor RR all connected in all connected in seriesseries with with an ac sourcean ac source. The . The instantaneous current and voltages instantaneous current and voltages can be measured with meters.can be measured with meters.

Consider an Consider an inductorinductor LL,, a a capacitorcapacitor CC,, and a and a resistorresistor RR all connected in all connected in seriesseries with with an ac sourcean ac source. The . The instantaneous current and voltages instantaneous current and voltages can be measured with meters.can be measured with meters.

Phase in a Series AC Phase in a Series AC CircuitCircuit

The voltage The voltage leadsleads current in an inductor current in an inductor and and lagslags current in a capacitor. current in a capacitor. In phaseIn phase for for

resistance resistance RR..

450 900 1350

1800 2700 3600

V V = Vmax sin

VRVC

VL

Rotating Rotating phasor diagramphasor diagram generates voltage generates voltage waves for each element waves for each element RR, , LL, and , and C C showing showing phase relations. Current phase relations. Current i i is always is always in in phasephase with with VVR.R.

Phasors and VoltagePhasors and VoltageAt time t = 0, suppose we read At time t = 0, suppose we read VVLL, , VVRR and and VVCC

for an ac series circuit. What is the source for an ac series circuit. What is the source voltage voltage VVTT??

We handle phase differences by finding We handle phase differences by finding the the vector sumvector sum of these readings. of these readings. VVTT = = VVii. . The angle The angle is the is the phase anglephase angle for for the ac circuit.the ac circuit.

VR

VL - VCVVTT

Source voltageSource voltage

VRVC

VL

Phasor Phasor DiagraDiagra

mm

Calculating Total Source Calculating Total Source VoltageVoltage

VR

VL - VCVVTT

Source voltageSource voltage Treating as vectors, we Treating as vectors, we find:find:

2 2( )T R L CV V V V 2 2( )T R L CV V V V

tan L C

R

V V

V

tan L C

R

V V

V

Now recall that:Now recall that: VVRR = iR = iR; ; V VLL = iX = iXLL;; andand V VCC = iV = iVCC

Substitution into the above voltage equation Substitution into the above voltage equation gives:gives:

2 2( )T L CV i R X X 2 2( )T L CV i R X X

Impedance in an AC Impedance in an AC CircuitCircuit

R

XL - XCZZ

ImpedanceImpedance2 2( )T L CV i R X X

2 2( )T L CV i R X X

ImpedanceImpedance Z Z is is defined:defined:

2 2( )L CZ R X X 2 2( )L CZ R X X

Ohm’s law for ac Ohm’s law for ac current and current and impedance:impedance:

or TT

VV iZ i

Z or T

T

VV iZ i

Z

The impedance is the combined opposition to ac current consisting of both resistance and reactance.

The impedance is the combined opposition to ac current consisting of both resistance and reactance.

Example 3:Example 3: A A 60-60- resistor, a resistor, a 0.5 H0.5 H inductor, and an inductor, and an 8-8-FF capacitor are capacitor are connected in series with a connected in series with a 120-V, 60 Hz120-V, 60 Hz ac ac source. Calculate the impedance for this source. Calculate the impedance for this circuit.circuit.

A

60 Hz

0.5 H

60

120 V8 F

12 and

2L CX fL XfC

2 (60Hz)(0.6 H) = 226LX

-6

1332

2 (60Hz)(8 x 10 F)CX

2 2 2 2( ) (60 ) (226 332 )L CZ R X X

Thus, the impedance Thus, the impedance is:is:

Z = 122

Example 4:Example 4: Find the effective current and Find the effective current and the phase angle for the previous example.the phase angle for the previous example.

A

60 Hz

0.5 H

60

120 V8 F

XXLL = = 226 226 XXCC = = 332 332

R = R = 60 60 Z = Z = 122 122 120 V

122 T

eff

Vi

Z

ieff = 0.985 Aieff = 0.985 A

Next we find the Next we find the phase phase angleangle::

R

XL - XCZZ

ImpedanceImpedanceXXL L – X– XC C = 226 – 332 = -106 = 226 – 332 = -106

R = 60 R = 60 tan L CX X

R

tan L CX X

R

Continued . . .Continued . . .

Example 4 (Cont.):Example 4 (Cont.): Find the Find the phase anglephase angle for the previous example.for the previous example.

-106

60

ZZ

XXL L – X– XC C = 226 – 332 = -106 = 226 – 332 = -106

R = 60 R = 60 tan L CX X

R

tan L CX X

R

106tan

60

= -

60.50

= -60.50

The The negativenegative phase angle means that phase angle means that the ac voltage the ac voltage lags lags the current by 60.5the current by 60.500. .

This is known as a This is known as a capacitivecapacitive circuit. circuit.

The The negativenegative phase angle means that phase angle means that the ac voltage the ac voltage lags lags the current by 60.5the current by 60.500. .

This is known as a This is known as a capacitivecapacitive circuit. circuit.

Resonant FrequencyResonant Frequency

BecauseBecause inductance inductance causes the voltage to causes the voltage to leadlead the current and the current and capacitance capacitance causes it causes it to to laglag the current, they tend to the current, they tend to cancelcancel each each other out.other out.

BecauseBecause inductance inductance causes the voltage to causes the voltage to leadlead the current and the current and capacitance capacitance causes it causes it to to laglag the current, they tend to the current, they tend to cancelcancel each each other out.other out.

ResonanceResonance (Maximum (Maximum Power) occurs when XPower) occurs when XL L = =

XXCCRXC

XL XXLL = = XXCC

2 2( )L CZ R X X R 2 2( )L CZ R X X R

12

2fL

fC

1

2rf

LC

1

2rf

LCResonantResonant

ffrr X XLL = X = XC C

Example 5:Example 5: Find the resonant frequency Find the resonant frequency for the previous circuit example: L = .5 H, for the previous circuit example: L = .5 H, C = 8 C = 8 FF

1

2rf

LC

1

2rf

LC

-6

1

2 (0.5H)(8 x 10 Ff

Resonant fr = 79.6 Hz

Resonant fr = 79.6 Hz

At resonant frequency, there is zero reactance At resonant frequency, there is zero reactance ((only resistanceonly resistance) and the circuit has a phase ) and the circuit has a phase

angle of zero.angle of zero.

At resonant frequency, there is zero reactance At resonant frequency, there is zero reactance ((only resistanceonly resistance) and the circuit has a phase ) and the circuit has a phase

angle of zero.angle of zero.

A

? Hz

0.5 H

60

120 V8 F

Resonance XL = XC

Power in an AC CircuitPower in an AC Circuit

No power is consumed by inductance or No power is consumed by inductance or capacitance. Thus power is a function of capacitance. Thus power is a function of the component of the impedance along the component of the impedance along

resistance:resistance:

No power is consumed by inductance or No power is consumed by inductance or capacitance. Thus power is a function of capacitance. Thus power is a function of the component of the impedance along the component of the impedance along

resistance:resistance:In terms of ac In terms of ac

voltage:voltage:P = iV cos P = iV cos

In terms of the resistance In terms of the resistance R:R:

P = i2RP = i2R

R

XL - XCZZ

ImpedanceImpedance

P P lost in lost in RR only only

The fraction The fraction Cos Cos is known as the is known as the power power factor.factor.

Example 6:Example 6: What is the average power What is the average power loss for the previous example: loss for the previous example: VV = 120 V, = 120 V, = -60.5 = -60.500, , ii = 90.5 A, and R = 60 = 90.5 A, and R = 60 . .

The The higher higher the power factor, the more the power factor, the more efficientefficient is the circuit in its use of ac is the circuit in its use of ac

power. power.

The The higher higher the power factor, the more the power factor, the more efficientefficient is the circuit in its use of ac is the circuit in its use of ac

power. power.

A

? Hz

0.5 H

60

120 V8 F

Resonance XL = XC

P = iP = i22RR = (0.0905 A) = (0.0905 A)22(60 (60

Average P = 0.491 W

Average P = 0.491 W

The power factor is: The power factor is: Cos Cos 60.560.500

Cos = 0.492 or 49.2%Cos = 0.492 or 49.2%

The TransformerThe TransformerA A transformertransformer is a device that uses is a device that uses induction and ac current to step voltages induction and ac current to step voltages up or down.up or down.

R

a.c.

Np Ns

Transformer

P PN t

EP PN t

E

S SN t

ES SN t

EInduced

emf’s are:Induced

emf’s are:

An ac source of emf An ac source of emf EEpp is connected to is connected to primary coil with primary coil with NNpp

turns. Secondary turns. Secondary has has NNss turns and turns and emf of emf of EEss..

An ac source of emf An ac source of emf EEpp is connected to is connected to primary coil with primary coil with NNpp

turns. Secondary turns. Secondary has has NNss turns and turns and emf of emf of EEss..

Transformers (Continued):Transformers (Continued):

R

a.c. Np Ns

TransformerP PN t

EP PN t

E

S SN t

ES SN t

E

Recognizing that Recognizing that //tt is the same in each is the same in each coil, we divide first relation by second and coil, we divide first relation by second and obtain:obtain:

The transformer equation:

The transformer equation:

P P

S S

N

N

EEP P

S S

N

N

EE

Example 7:Example 7: A generator produces 10 A A generator produces 10 A at 600 V. The primary coil in a at 600 V. The primary coil in a transformer has 20 turns. How many transformer has 20 turns. How many secondary turns are needed to step up secondary turns are needed to step up the voltage to 2400 V?the voltage to 2400 V?

R

a.c. Np Ns

I = 10 A; Vp = 600 V

20 turns

P P

S S

V N

V NP P

S S

V N

V N

Applying the Applying the transformer transformer equation:equation:

(20)(2400V)

600VP S

SP

N VN

V NS = 80 turnsNS = 80 turns

This is a This is a step-up transformerstep-up transformer; reversing ; reversing coils will make it a step-down coils will make it a step-down

transformer.transformer.

This is a This is a step-up transformerstep-up transformer; reversing ; reversing coils will make it a step-down coils will make it a step-down

transformer.transformer.

Transformer EfficiencyTransformer EfficiencyThere is no power gain in stepping up the There is no power gain in stepping up the voltage since voltage is increased by voltage since voltage is increased by reducing current. In an ideal transformer reducing current. In an ideal transformer with no internal losses:with no internal losses:

or SPP P S S

s P

ii i

i

EE E

E or SP

P P S Ss P

ii i

i

EE E

E

An ideal An ideal transformer:transformer:

R

a.c. Np Ns

Ideal Transformer

The above equation assumes no internal The above equation assumes no internal energy losses due to heat or flux changes. energy losses due to heat or flux changes. Actual efficienciesActual efficiencies are usually between are usually between 90 90 and 100%.and 100%.

The above equation assumes no internal The above equation assumes no internal energy losses due to heat or flux changes. energy losses due to heat or flux changes. Actual efficienciesActual efficiencies are usually between are usually between 90 90 and 100%.and 100%.

Example 7:Example 7: The transformer in The transformer in Ex. 6Ex. 6 is is connected to a power line whose connected to a power line whose resistance is resistance is 12 12 . How much of the . How much of the power is lost in the transmission line?power is lost in the transmission line?

VVSS = 2400 = 2400 VV

R

a.c. Np Ns

I = 10 A; Vp = 600 V

20 turns

12 P P

P P S S SS

ii i i

EE E

E(600V)(10A)

2.50 A2400VSi

PPlostlost = i = i22RR = (2.50 A) = (2.50 A)22(12 (12 ))

PPlostlost = 75.0 = 75.0 WWPPinin = (600 V)(10 A) = 6000 = (600 V)(10 A) = 6000

WW%Power Lost = (75 W/6000 W)(100%) = %Power Lost = (75 W/6000 W)(100%) = 1.25%1.25%

%Power Lost = (75 W/6000 W)(100%) = %Power Lost = (75 W/6000 W)(100%) = 1.25%1.25%

SummarySummary

Effective current: ieff = 0.707 imax

Effective current: ieff = 0.707 imax

Effective voltage: Veff = 0.707 Vmax

Effective voltage: Veff = 0.707 Vmax

Inductive Reactance:2 Unit is the LX fL

Ohm's law: L LV iX

Capacitive Reactance:1

Unit is the 2CX fC

Ohm's law: VC CiX

Summary (Cont.)Summary (Cont.)

2 2( )T R L CV V V V 2 2( )T R L CV V V V tan L C

R

V V

V

tan L C

R

V V

V

2 2( )L CZ R X X 2 2( )L CZ R X X

or TT

VV iZ i

Z or T

T

VV iZ i

Z

tan L CX X

R

tan L CX X

R

1

2rf

LC

1

2rf

LC

Summary (Cont.)Summary (Cont.)

In terms of ac In terms of ac voltage:voltage:

P = iV cos P = iV cos

In terms of the resistance In terms of the resistance R:R:

P = i2RP = i2R

Power in AC Circuits:Power in AC Circuits:

P P

S S

N

N

EE P P S Si iE E

Transformers:Transformers:

CONCLUSION: Chapter 32ACONCLUSION: Chapter 32AAC CircuitsAC Circuits