Ac circuits 15 april 2013(1)
Upcoming SlideShare
Loading in...5
×
 

Ac circuits 15 april 2013(1)

on

  • 880 views

 

Statistics

Views

Total Views
880
Views on SlideShare
880
Embed Views
0

Actions

Likes
0
Downloads
55
Comments
0

0 Embeds 0

No embeds

Accessibility

Categories

Upload Details

Uploaded via as Microsoft PowerPoint

Usage Rights

© All Rights Reserved

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Processing…
Post Comment
Edit your comment

Ac circuits 15 april 2013(1) Ac circuits 15 april 2013(1) Presentation Transcript

  • AC Circuits15 April 2013
  • What we are going to cover20.1 Resistors in an AC Circuit21.2 Capacitors in an AC Circuit21.3 Inductors in an AC Circuit21.4 The RLC Series Circuit21.5 Power in an AC Circuit21.6 Resonance in a Series RLC Circuit21.7 The Transformer
  • INTRODUCTION• Every time we turn on a television set, a stereo system,or any of a multitude of other electric appliances, wecall on alternating currents (AC) to provide the powerto operate them.• We begin our study of AC circuits by examining thecharacteristics of a circuit containing a source of emfand one other circuit element: a resistor, a capacitor, oran inductor.• Then we examine what happens when these elementsare connected in combination with each other.• Our discussion is limited to simple series configurationsof the three kinds of elements
  • INTRODUCTIONIn an AC circuit, the charge flow reverses direction periodically.In circuits that contain only resistance, the current reverses direction each timethe polarity of the generator reverses.The output of an AC generator issinusoidal and varies with time accordingto the equation:
  • 21.1 RESISTORS IN AN AC CIRCUITAn AC circuit consists of combinations of circuitelements and an AC generator or an AC source,which provides the alternating current.Previously we have seen that The output of anAC generator is sinusoidal and varies with timeaccording to the equation:
  • 21.1 RESISTORS IN AN AC CIRCUIT• Let us consider a simple circuit consisting of a resistor and anAC source (designated by the symbol . T .• The current and the voltage across the resistor are shown inActive Figure 21.2.
  • • To explain the concept of alternatingcurrent, we begin by discussing thecurrent-versus-time curve in Active Figure21.2.• At point a on the curve, the current has amaximum value in one direction, arbitrarilycalled the positive direction.• Between points a and b, the current isdecreasing in magnitude but is still in thepositive direction. At point b, the current ismomentarily zero; it then begins toincrease in the opposite (negative)direction between points b and c.• At point c, the current has reached itsmaximum value in the negative direction.
  • • The current and voltage are in step with eachother because they vary identically with time.• Because the current and the voltage reachtheir maximum values at the same time, theyare said to be in phase.• Notice that the average value of the currentover one cycle is zero. This is because thecurrent is maintained in one direction (thepositive direction) for the same amount oftime and at the same magnitude as it is in theopposite direction (the negative direction).However, the direction of the current has no effect on the behavior of the resistorin the circuit: the collisions between electrons and the fixed atoms of the resistorresult in an increase in the resistor’s temperature regardless of the direction of thecurrent
  • Power• We can quantify this discussion by recalling that the rate atwhich electrical energy is dissipated in a resistor, the poweris• where i is the instantaneous current in the resistor.• Because the heating effect of a current is proportional tothe square of the current, it makes no difference whetherthe sign associated with the current is positive or negative.• However, the heating effect produced by an alternatingcurrent with a maximum value of Imax is not the same asthat produced by a direct current of the same value.• The reason is that the alternating current has thismaximum value for only an instant of time during a cycle.
  • RMSThe important quantity in an AC circuit is a special kind of average value of current,called the rms current—the direct current that dissipates the same amount of energyin a resistor that is dissipated by the actual alternating current.To find the rms current, we first square the current, Then find its average value, andfinally take the square root of this average value. Hence, the rms current is thesquare root of the average (mean) of the square of the current.Therefore, the rms current I rms is related to the maximum value of thealternating current Imax by• This equation says that an alternating current with a maximum value of 3 A producesthe same heating effect in a resistor as a direct current of (3/2 ) A.• We can therefore say that the average power dissipated in a resistor that carriesalternating current I is
  • The mathematical part
  • • When we speak of measuring an AC voltage of 120 V from an electric outlet, wereally mean an rms voltage of 120 V.• A quick calculation using Equation 21.3 shows that such an AC voltage actuallyhas a peak value of about 170 V. In this chapter we use rms values whendiscussing alternating currents and voltages.• One reason is that AC ammeters and voltmeters are designed to read rmsvalues. Further, if we use rms values, many of the equations for alternatingcurrent will have the same form as those used in the study of direct-current(DC) circuits.Consider the series circuit in Figure 21.1, consisting of a resistor connected to an ACgenerator. A resistor impedes the current in an AC circuit, just as it does in a DCcircuit. Ohm’s law is therefore valid for an AC circuit, and we have
  • ProblemSolutionObtain the maximum voltage bycomparison of the given expression for theoutput with the general expression:Next, substitute into Equation 21.3 tofind the rms voltage of the source:Substitute this result into Ohm’s law tofind the rms current:
  • 21.2 CAPACITORS IN AN AC CIRCUIT• In a purely resistive circuit, thecurrent and voltage are always instep with each other. This isn’t thecase when a capacitor is in thecircuit.• In Figure 21.5, when an alternatingvoltage is applied across a capacitor,the voltage reaches its maximumvalue one-quarter of a cycle afterthe current reaches its maximumvalue.• We say that the voltage across acapacitor always lags the current by90°.
  • Capacitive reactance
  • NoteSubstitute the values of f and CSolve Equation 21.6 for the current, andsubstitute Xc and the rms voltage to find therms current:Capacitors do not behave the same as resistors. Whereas resistors allow a flow ofelectrons through them directly proportional to the voltage drop, capacitors opposechanges in voltage by drawing or supplying current as they charge or discharge to thenew voltage level. The flow of electrons “through” a capacitor is directly proportionalto the rate of change of voltage across the capacitor. This opposition to voltagechange is another form of reactance, but one that is precisely opposite to the kindexhibited by inductors.
  • 21.3 INDUCTORS IN AN AC CIRCUIT• Now consider an AC circuit consisting only of an inductorconnected to the terminals of an AC source, as in Active Figure21.6. (In any real circuit, there is some resistance in the wireforming the inductive coil, but we ignore this for now.)• The changing current output of the generator produces a backemf that impedes the current in the circuit. The magnitude ofthis back emf is• The effective resistance of the coil in an AC circuit is measuredby a quantity called the inductive reactance,• When f is in hertz and L is in henries, the unit of XL is the ohm.The inductive reactance increases with increasing frequency and increasing inductance.Inductors do not behave the same as resistors. Whereas resistors simply oppose the flow ofelectrons through them (by dropping a voltage directly proportional to the current),inductors oppose changes in current through them, by dropping a voltage directlyproportional to the rate of change of current
  • Phase diagrams
  • Problem
  • Review of R, X, and ZResistance (R):• This is essentially friction against the motion of electrons. It is present in allconductors to some extent (except superconductors!), most notably inresistors.• When alternating current goes through a resistance, a voltage drop isproduced that is in-phase with the current.Reactance (X):• It is essentially inertia against the motion of electrons. It is present anywhereelectric or magnetic fields are developed in proportion to applied voltage orcurrent, respectively; but most notably in capacitors and inductors.• When alternating current goes through a pure reactance, a voltage drop isproduced that is 900 out of phase with the current.Impedance (Z)• This is a comprehensive expression of any and all forms ofopposition to electron flow, including both resistance andreactance.• It is present in all circuits, and in all components. When alternatingcurrent goes through an impedance, a voltage drop is producedthat is somewhere between 0o and 90o out of phase with thecurrent.
  • Phasor Diagrams: Conventions
  • Phasor Diagrams(a) A phasor diagramfor the RLC circuit(b) Addition of the phasorsas vectors gives(c) The reactancetriangle that givesthe impedance relation
  • Summary
  • RESONANCE IN A SERIES RLC CIRCUIT
  • QuestionsThe diagram is given, thencalculate the following :(a) the impedance,(b)The maximum currentin the circuit,(c) the phase angle, and(d)the maximum voltagesacross the elements.
  • About average PowerCheck RCL Summary as done in Example 4 :Voltages and power in a series RCL Circuitpage 720 on the 9th Edition
  • Chapter 23:Tutorial ExercisesProblem 1. A 63.0 µF capacitor is connected to a generator operating at a lowfrequency. The rms voltage of the generator is 4.00 V and is constant. A fuse in serieswith the capacitor has negligible resistance and will burn out when the rms currentreaches 15.0 A. As the generator frequency is increased, at what frequency will thefuse burn out?As the frequency f of the generator increases, the capacitive reactance XC of the capacitordecreases, according to C12Xf CIn terms of the rms current and voltage, Equation 23.1 gives the capacitive reactance asrmsCrmsVXI . Substituting this relation into Equation (1) yields
  • Problem 17: A series RCL circuit includes a resistance of 275 Ω an inductive reactanceof 648 Ω and a capacitive reactance of 415 Ω. The current in the circuit is 0.233 A.What is the voltage of the generator?17. REASONING The voltage supplied by the generator can be found from Equation 23.6,V I Zrms rms . The value of Irmsis given in the problem statement, so we must obtain theimpedance of the circuit.Z R X X     2 2 2275 3 60 10( – ) .L C2 2( ) (648 – 415 )   The rms voltage of the generator is V I Zrms rmsA ) = 83.9 V  ( . )( .0 233 3 60 102
  • The phase angle is given by tan  = (XL – XC)/R (Equation 23.8). When a series circuitcontains only a resistor and a capacitor, the inductive reactance XL is zero, and the phaseangle is negative, signifying that the current leads the voltage of the generator. Theimpedance is given by Equation 23.7 with XL = 0 , or Z R X 2C2. Since the phaseangle  and the impedance Z are given, we can use these relations to find the resistanceR and XC.Since the phase angle is negative, wecan conclude that only a resistor and acapacitor are present. Using Equations23.8, then, we have CCtan or tan 75.0 3.73XX R RR     According to Equation 23.7, the impedance is2 2C150Z R X   Substituting XC = 3.73R into thisexpression for Z gives    2 22 2 22150 3.73 = 14.9 R or 150 14.9 R15038.914.9R RR      XC = 3.73 (38.9 ) = 145 hence
  • Since, on the average, only the resistor consumes power, the average power dissipated inthe circuit is 2rmsP I R (Equation 20.15b), where Irms is the rms current and R is theresistance. The current is related to the voltage V of the generator and the impedance Z of thecircuit by rms rms /I V Z (Equation 23.6). Thus, the average power can be written asThe impedance of the circuit is  22L CZ R X X   2 2CR X Therefore, the expression for theaverage power becomes2 22 2 2CV R V RPZ R X   C 61 124002 2 60.0 Hz 1.1 10 FXf C     ButHence       222 2 2 2C120 V 27003.0 W2700 2400V RPR X     
  • TutorialsProblem 52. When a resistor is connected by itself to an ac generator, the averagepower delivered to the resistor is 1.000 W. When a capacitor is added in series withthe resistor, the power delivered is 0.500 W. When an inductor is added in serieswith the resistor (without the capacitor), the power delivered is 0.250 W. Determinethe power delivered when both the capacitor and the inductor are added in serieswith the resistor.