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Alternating Current Circuits

By Lewis Stevens,2014-04-04 11:39
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Alternating Current Circuits

     p. 65

    Alternating Current Circuits

Review of rms values. rms values are root-mean-square values of quantities (such as voltage and current) that

    vary periodically with time. In AC circuits voltage and current vary sinusoidally with time:

     vVtiIt!?!??,sin, sin;;;;

where V and I are the voltage and current amplitudes, respectively. ? is the angular frequency (? = 2f, where f

    is the frequency) and is a phase constant that we will discuss later. The rms values of voltage and current are

    defined to be

    2222 VVtIIt!?!??,sin, sin;;;;rmsrms

     2Where the overbar indicates the average value of the function over one cycle. Since the average value of sin~

    over one cycle is ?, we get

    VIVI!! and rmsrms22

Note that these formulas are valid only if the voltage varies sinusoidally with time.

What we will study in this chapter is what happens to the current and power in an AC series circuit if a resistor,

    a capacitor and an inductor are present in the circuit.

Resistors and Resistance

If just a resistor of resistance R is connected across an AC generator the generator is said to have a purely

    resistive load. The phase constant is zero and we write

    . vVtiItVIR!?!?!sin, sin and ;;;;rmsrms

Since the angle for v and i is the same, the instantaneous voltage and current are said to be in phase. Note that

    Vrms RIrms

    is a constant independent of the frequency f of the AC generator. We assume that the resistor maintains its resistance regardless of how fast or slow the generator’s armature is turning. R, of course, is measured in ohms.

For a purely resistive load the average power delivered to the circuit by the generator is given by

    2 PIVPIR!! or rmsrmsrms

    which are analogous to the familiar formulas for DC circuits. P, as usual, is measured in watts.

    Rev. 1/23/08

     p. 66

    Capacitors and Capacitive Reactance

Now let us connect just a capacitor of capacitance C across an AC generator. In this case the generator is said

    to have a purely capacitive load. The phase constant is and we write 2

    ?: vVtiItVIX!?!??!sin, sin and ;;rmsrmsC,,2??

where X is called the capacitive reactance. Capacitive reactance, like resistance, is measured in ohms. C

Since the angle for the instantaneous current is greater than the angle for the instantaneous voltage by /2

    radians or 90?, the current is said to lead the voltage by 90? or lead the voltage by a quarter cycle. (Remember that a full cycle is 360? - a “complete trip” around a circle.)

    VrmsFor a capacitive load the ratio is not a constant independent of the frequency of the generator. It can be Irms

    shown that in fact

    V11rms !! so that .XCIfCfC22;;rms

    111VUnits check: . See Figure 23.2 on page 727 of your text. !!!!ohmsCC11(((HzFAssVV

    For a purely capacitive load the average power delivered to the circuit by the generator is zero. The reason for this is that the instantaneous voltage and current in the circuit are exactly 90? out of phase. Over one cycle the generator delivers as much power to the capacitor as it gets back from the capacitor. (Remember that over a

    generator cycle the capacitor will charge then discharge.)

Example

    Two stripped wires from the end of a lamp cord are soldered to the terminals of a 200 F capacitor. The lamp cord, which has a standard electric plug on the other end, is then plugged into a 120 V, 60 Hz AC outlet.

    (Do not try this at home.)

a. Find the reactance of the capacitor.

    11 !!!XXX; ; 13.3 ohmsCCC42fC;?260210;;;;

    b. Find the rms current drawn from the wall outlet.

    V120 Vrms VIXIII!!!!; ; ; 9.02 ArmsrmsCrmsrmsrmsX13.3 (C

     p. 67

    Inductors and Inductive Reactance

    Now let us connect just an inductor of inductance L across an AC generator. In this case the generator is said to

    have a purely inductive load. The phase constant is and we write 2

    ?: vVtiItVIX!?!??!sin, sin and ;;rmsrmsL,,2??

where X is called the inductive reactance. Inductive reactance, like resistance, is measured in ohms. L

Since the angle for the instantaneous current is smaller than the angle for the instantaneous voltage by /2

    radians or 90?, the current is said to lag the voltage by 90? or lag the voltage by a quarter cycle. (Alternatively

    one can say that the voltage leads the current by 90? or the voltage leads the current by a quarter cycle.)

    VrmsFor an inductive load the ratio is not a constant independent of the frequency of the generator. It can be Irms

    shown that in fact

    Vrms !;!;2 so that 2.fLXfLLIrms

    1VVUnits check: . See Figure 23.6 on page 729 of your text. HzH(!(!!ohmsAsAs

    For a purely inductive load the average power delivered to the circuit by the generator is zero. The reason for

    this is that the instantaneous voltage and current in the circuit are exactly 90? out of phase. Over one cycle the generator delivers as much power to the inductor as it gets back from the inductor. (Remember that over a

    generator cycle the induced emf in the inductor will reverse direction.)

Example

Two stripped wires from the end of a lamp cord are soldered to the terminals of a 200 mH inductor. The lamp

    cord, which has a standard electric plug on the other end, is then plugged into a 120 V, 60 Hz AC outlet.

    (Do not try this at home.)

a. Find the reactance of the inductor.

     XfLXX!;!;!2; 2600.200; 75.4 ohms;;;;LLL

b. Find the rms current drawn from the wall outlet.

    V120 Vrms VIXIII!!!!; ; ; 1.59 ArmsrmsLrmsrmsrmsX75.4 (L

     p. 68

    RCL Series Circuits

An RCL series circuit consists of a resistor, a capacitor, an inductor and an AC generator connected in series.

    See the figure.

    The mathematical analysis of this circuit requires the solution of a differential equation. However, there is a way to solve the circuit using a geometrical device that is analogous to a vector. This device is called a phasor (or rotor). A phasor is a vector

    whose tail sits at the origin of an xy-coordinate system. The phasor rotates counterclockwise about the origin with angular

    frequency ? (the angular frequency of the AC generator). The Vtsin?;;phasor represents either voltage or current, and its y-component is the instantaneous value of the quantity it represents.

    We will assume that at any instant the current through each circuit element is given by

     I

    . iIt!??,sin;;

    { ??,t

    The current phasor has length I and makes an angle of ?t - with

    respect to the x-axis. At any instant its y-component equals the current in the circuit.

    Now consider the voltage phasor of the resistor. The instantaneous voltage across the resistor is just

     iRIRtvVt!??,!??,sin or sin;;;;RRVR I The length of the resistor’s voltage phasor is the voltage amplitude V. R

    At any instant the angle it makes with the x-axis is ?t - . The y-{ ??,tcomponent of this phasor is then

    , Vtsin??,;;R

    which is the instantaneous voltage across the resistor. Note that the current and voltage across the resistor are in phase. Hence the voltage

    phasor for the resistor lies on top the current phasor.

     VR I Now consider the voltage phasor for the capacitor. Here it is

    critically important to remember the phase relationship between the current and voltage for a capacitor. Does the current lead or lag the ??,t voltage in a capacitor? By how many degrees? The current leads the

    voltage by 90?. Since the phasors rotate counterclockwise, the voltage phasor for the capacitor must lie 90? clockwise from the current

    VC phasor.

     p. 69

    VL Now consider the voltage phasor for the inductor. It is critically VR important to remember the phase relationship between the current and I voltage for an inductor. Does the current lead or lag the voltage in an

    inductor? By how many degrees? The current lags the voltage by 90?.

     ??,tSince the phasors rotate counterclockwise, the voltage phasor for the

    inductor must lie 90? counterclockwise from the current phasor.

Note that the voltage phasors for the inductor and the capacitor lie VC along the same line. (We have arbitrarily assumed that V is larger L

    than V.) Using the rules of vector addition we may combine them to c

    obtain the next diagram.

By Kirchhoff’s loop rule the voltage drops across the capacitor, VR resistor and inductor must, at any instant, equal the voltage rise I

    across the generator. This will be satisfied if the vector sum of the VVLCV V and the V phasors matches the voltage phasor of the LCR ??,tgenerator. See the last diagram below.

    From the last diagram we obtain some very important relationships. In particular, note that

    22222VVVVVVVV!??!?? or ;;;;LCRLCRV

    since , and we can writeVIXVIXVIR!!!LLCCR

     2222VR VIXIXIRVIXXR!??!?? or ;;;;;;LCLCI VV2LC2?t or where VIZZXXR!!??;;LC

     ??,t

    Z is called the impedance of the circuit and is measured in ohms. Note that we have dropped the “rms” subscripts for the voltage

    and the current in the V = IX formulas above because the formulas are also valid if we replace each rms value with its corresponding amplitude (the square root of 2 cancels from both sides of each equation).

We can now find a formula for the phase of the current. From the right triangle with sides V, V and V - V in RLC

    the diagram above we have

    VVIXIXXX???LCLCLC tan so that tan,!!,!VIRRR

     p. 70

    Average Power

On average, only the resistance in the RCL series circuit consumes power. The average rate of power

    consumption is given by

    2 PIRrmsZ - XXLC The triangle at the right is useful to remember since one can quickly obtain the formulas that were derived above from it: R

    XX22LC ZXXR!??,! and tan;;LCR

    R2also note that from which we obtain !,!,!,!,cos so that cos and cos or cosRZPIZPIIZ;;rmsrmsrmsZ

     PIV!,cosrmsrms

    cos is called the power factor of the RCL circuit.

    XXLCUsing the formula we make the following observations and definitions: tan,!R

    If and the circuit is said to have an inductive load. XX,,,, 0LC

    If and the circuit is said to have a capacitive load. XX,,,, 0LC

    If and the circuit is said to have a resistive load. XX!,!, 0LC

Example

A series RCL circuit has a 75.0 ( resistor, a 20.0 F capacitor and a 55.0 mH inductor connected across an 800

    volt rms AC generator operating at 128 Hz.

    a. Is the load on the circuit inductive, capacitive or resistive? What is the phase angle ?

    2XfLX!;!;?!(2; 21285.51044.2 ;;;;LL

    11XX!!!(; 61.2 CC52fC21282.010;?;;;;

    Since the load is . The phase angle isXXcapacitive CL

    XX??44.261.2?:?:LC,!,!arctan; arctan,,,,R75.0????

    ,!?,!?0.223 rad or 12.8.?

     p. 71

    b. What is the rms current in the circuit?

    To answer this question we must determine the circuit’s impedance Z then use I = V/Z: rmsrms

    2222ZXXRZ!??!??; 44.261.275.0;;;;;;LC

     V800 VrmsZIII!(!!!76.9 . ; ; 10.4 ArmsrmsrmsZ76.9 (

c. Write the formula for the current in the circuit as a function of time.

    iItIcurrentamplitude!??,sin where is the .;;

    IIII!!!2: 10.41.414; 14.7 A;;rms

    ?!;?!;?!2: 2128; 804 rad/sf;;

    itNotetheuseofradians!??14.7sin8040.223 .;;

    (t in seconds and i in amperes.)

    d. Find the rms voltage across each circuit element.

    VIRVV!!(!; 10.4 A75.0 ; 780 V;;;;RrmsrmsRrmsRrms

     VIXVV!!(!; 10.4 A61.2 ; 636 V;;;;CrmsrmsCRrmsCrms

    VIXVV!!(!; 10.4 A44.2 ; 460 V;;;;LrmsrmsLRrmsLrms

    Question: Shouldn’t these voltages add to 800 V?

    Answer: No. One must take into account the phase of the voltage across each element. See part e.

e. Find the instantaneous voltage across each circuit element at t = 0 seconds.

    viRvv!!!; 14.775.0sin0.223; 244 V;;;;;;RRR

     viXvv!!?!?; 14.761.2sin0.2231.57; 877 V;;;;;;?;C/2CCC

    viXvv!!?!; 14.744.2sin0.2231.57; 633 V;;;;;;?;L/2LLL

    Question: Why do these voltages add to zero?

    Answer: Their sum is in agreement with Kirchhoff’s loop rule; the voltage across the generator is

    vVtvtt!?!!!sin or 8002sin8040 at 0 s. ;;;;

f. Find the average power delivered to the circuit by the generator.

    PIVP!,!?cos; 10.4 A800 Vcos0.223;;;;;;;;rmsrms

    P8.11 kW

     p. 72

    The Limiting Behavior of Capacitors and Inductors

Unlike a resistor, which has a constant resistance R independent of the ac frequency, capacitors and

    inductors have reactances that do depend on it.

    The inductive reactance is given by

     XfL!;2L

    If f is large, so is X, and the inductor acts almost like an open circuit. If f is small, so is X, and the inductor LLacts almost like a short circuit.

    Vr

    R

    V= VLVgeffin= VoutL

    Frequency = f

This circuit can be regarded as a high-pass filter. At very-high frequencies the inductor has a high reactance

    and acts almost like an open circuit. Thus, the current is low, the voltage drop in the resistor is low, and V out= V. At very-low frequencies the inductor has a low reactance and acts like a short circuit. The output in

    voltage is virtually zero. Hence, the circuit passes high-frequency AC voltages but stops low-frequency AC

    voltages.

    The capacitive reactance is given by

    1 XC2fC

    If f is large, X is small, and the capacitor acts almost like a short circuit. If f is small, X is large, and the CCcapacitor acts almost like an open circuit.

    Vr

    R

    V= VCVgeffin= Vcout

    Frequency = f

     p. 73

    This circuit can be regarded as a low-pass filter. At very low frequencies the capacitor has a high reactance

    and is almost like an open circuit. Thus, the current is low, the voltage drop in the resistor is low, and V = out

    V. At very high frequencies the capacitor has a low reactance and acts like a short circuit. The output in

    voltage is virtually zero. Hence this circuit passes low-frequency AC voltages but stops high-frequency AC

    voltages.

Example

Suppose that an RC circuit (as shown in the last diagram above) is used in a crossover network in a 2-way

    stereo speaker. (A 2-way stereo speaker has a small speaker a “tweeter” – for high frequencies and a large speaker a “woofer” – for low frequencies. A crossover network in the speaker system directs low frequencies to the woofer and high frequencies to the tweeter). In the last diagram above V is the voltage in

    supplied by the speaker output jacks of a stereo receiver; V is the voltage to be delivered to the woofer. If out

    R is 30 ohms, find the capacitance C so that the amplitude of frequency 8,000 Hz is reduced to half its value

    at output.

    VVXVXininCoutCVIXIV!!!!; ; ; outCout2222ZVXRXR??inCC

    11X22C!?!?; Square both sides.XRXCC2222XRC

    11312222222 XRXRXXR?!!)!; ;;CCcC4443

    21122XXX!!(!(!30300 ; 17.3 ; ;;CCC32fC

    11!;!!:8,000 Hz. ; 1.15 FCCCf; 2800017.32fX;;;;C

Remark. The frequency whose amplitude is reduced to half by the crossover network is called the crossover

    frequency. In the above example 8,000 Hz is the crossover frequency.

Example

    Estimate the impedance of the circuit shown at the left for a generator frequency of

    a. 1,000,000 Hz

    b. 0.001 Hz

a. For a high frequency the inductors act like open circuits and the

    capacitor acts like a short circuit, effectively producing the circuit shown in the diagram on the next page.

     p. 74

    The impedance is now just the net resistance of the circuit.

    Since the resistors are in series,

     RRRRRZ!?!?!(!(; 1k2k; 3 k, 3 k12

    b. For a low frequency the inductors act as short circuits and the

    capacitor acts as an open circuit, effectively producing the

    circuit shown in the diagram below.

    The impedance is now just the net resistance of the circuit.

    Since the resistors are in parallel,

    11111113!?!?!!(; ; ; 0.67 kR RRRRR1k2k2k12

    Z!(0.67 k

    As the frequency of the AC generator is changed from very

    low values to very high values the impedance of the circuit

    will increase from the lower limit of 0.67 k( to the upper

    limit of 3 k(.

    22Note: The formula for impedance we found earlier, , does not apply to the given ZXXR!??;;LC

    circuit in this example because the circuit elements are not connected in series! The formulas for the

    reactances, however, always apply.

Electrical Resonance

    For an RCL series circuit the current amplitude is given by

    VV I!!22ZXXR??;;LC

where V is the voltage amplitude. If V , R, C, and L are fixed and the frequency of the AC generator is variable,

    we can change the reactances of the inductor and capacitor by changing the frequency of the generator. As the

    frequency of the generator changes, so does the impedance Z of the circuit and the current amplitude I. If we

    look at the above formula we see that Z can be minimized (made as small as possible) by making the reactances

    equal to one another. The current amplitude I will then be maximized (made as large as possible). If XX and LC

    these conditions are met, electrical resonance is said to occur in the circuit. The RCL series circuit is said to be

    at resonance. For resonance,

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