Exercise 32.5 AC circuit with two resistors
32.6 RC and RLC series circuits
When an AC source is connected to multiple circuit elements, either in series or in parallel, applying the loop rule becomes more complicated than for DC circuits because we need to add several oscillating potential differences that may be out of phase with one another. For example, suppose we have a resistor and a capacitor in series with an AC source (Figure 32.52), known as an RC series circuit. The loop rule states that
ℰ =vR+vC. (32.28)
To compute the sum on the right side of this equation, we must add potential differences that vary sinusoidally at the same angular frequency v but are out of phase. The combined potential difference v of two potential differences v1 and v2 that oscillate at the same angular frequency is
v=V1 sin (vt+f1)+V2 sin (vt+f2), (32.29) where f1 and f2 are the initial phases of the two potential differences. Calcu- lating this sum algebraically gets very messy, but using phasors to calculate it simplifies things greatly.
Figure 32.53a shows the phasors that correspond to the two terms on the right in Eq. 32.29. Recall that the instantaneous value of the quantity represented by a rotating phasor equals the vertical component of the phasor (see Figure 32.11).
Therefore, v at any instant equals the sum of the vertical components of the pha- sors that represent v1 and v2. This sum is equal to the vertical component of the vector sum V1+V2 of the phasors, as shown in Figure 32.53b.
Note that the combined potential difference v oscillates at the same angular frequency as v1 and v2. Consequently, the three phasors V1, V2, and V1+V2 ro- tate as a unit at angular frequency v, as shown in Figure 32.54. The phase rela- tionship among the three phasors is constant, as is the phase relationship among the potential differences.
Figure 32.52 An RC series circuit, consisting of a resistor and a capacitor in series across the terminals of an AC source.
R
C ℰ
Figure 32.53 (a) Phasor diagram for a system of two oscillating potential differences v1 and v2. (b) Vector diagram indicating that the vertical component of the vector sum of the phasors equals the sum of the vertical components of the individual phasors.
(a) (b)
V1 V2
V1
V1 + V2
V2 v = v1 + v2
v = v1 + v2
v1
v2 vertical component of V1 + V2
vertical component of V2
vertical component of V1
v2 = V2 sin(vt + f2)
v1 = V1 sin(vt + f1) f2 f1
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The next example shows how to apply these principles to a specific situation.
To convince yourself that the phasor method is worthwhile, try adding the two original trigonometric functions algebraically after solving the problem using phasors!
Figure 32.54 Phasor diagram and graph showing time dependence of v1, v2, and v=v1+v2 from Figure 32.53.
All three phasors rotate as a unit at angular frequency v.
v
t V1
V1 + V2 V2
v = v1 + v2
v1 v2
Example 32.7 Adding phasors
Use phasors to determine the sum of the two oscillating poten- tial differences v1=(2.0 V) sin vt and v2=(3.0 V) cos vt.
❶ GettinG Started I begin by making a graph showing the time dependence of v1 and v2, and I draw the corresponding phasors V1 and V2 to the left of my graph (Figure 32.55). I add to my phasor diagram the phasor V1+V2, which is the phasor that represents the potential difference sum v1+v2 that I must determine. Using phasor V1+V2, I can sketch the time depen- dence of the sum v1+v2 by tracing out the projection of phasor V1+V2 onto the vertical axis of my V(vt) graph as this phasor rotates counterclockwise from the starting position I drew.
to the length of the phasor V1+V2, and from my sketch I see that the initial phase fi is given by the angle between V1+V2 and V1.
❸ execute plan The length of the phasor V1+V2 is given by the Pythagorean theorem applied to the right triangle containing fi in my phasor diagram:
A= 2V12+V22= 2(3.0 V)2+(2.0 V)2 = 213 V2=3.6 V.
The tangent of the angle between V1+V2 and V1 is then tan fi=V2
V1=3.0 V 2.0 V=1.5, so fi=tan-1(1.5)=56°.
Now that I have determined A and fi, I can write the sum of the two potential differences as
v1+v2=(3.6 V) sin (vt+56°). ✔
❹ evaluate reSult The amplitude of the sinusoidal function I obtained is 3.6 V, which is greater than the larger of the two pha- sors I added, as I expect. My answer shows that the sum of the two potential differences reaches its maximum when vt+fi=90°, or when vt=90° −fi=34°. This conclusion agrees with my phasor diagram: The phasor V1+V2 reaches the vertical position after it rotates through an angle of 90° −fi=90° −56° =34°.
(I could also verify my answer by adding the two original sine functions algebraically, but the trigonometry needed in that ap- proach is tedious.)
❷ deviSe plan To obtain an algebraic expression for v1+v2, I first write the oscillating potential differences in the form v1=V1 sin (vt+f1) and v2=V2 sin (vt+f2). Comparing these expressions with the given potential differences, I see that V1=2.0 V, f1=0, and V2=3.0 V. In order to determine f2, I use the trigonometric identity cos (vt)=sin (vt+p>2), and so my given information v2=(3.0 V) cos vt = (3.0 V) sin (vt+p>2) tells me that f2=p>2. The sum v1+v2 is a sinusoidally varying function that can be writ- ten as v1+v2=A sin (vt+fi). The amplitude A is equal
Figure 32.55
QUANTITATIVE TOOLS
32.13 Suppose you need to add two potential differences that are oscillating at different angular frequencies—say, 2 sin (vt) and 3 cos (2vt). Can you use the phasor method described above to determine the sum? Why or why not?
Let us now return to the RC series circuit of Figure 32.52 and construct a phasor diagram in order to determine the amplitude and phase of the current in terms of the amplitude of the source emf and the resistance and capacitance of the circuit elements. From the current, we can calculate the potential differences across the circuit elements.
Because the circuit contains only one loop, the time-dependent current i is the same throughout. Therefore, we begin by drawing a phasor that represents i (Figure 32.56a). We are free to choose the phase of this phasor because we have not yet specified the phase of any of the potential differences in the circuit. Also, the length we draw for phasor I is unimportant because it is the only current pha- sor for this circuit.
Next, we draw the phasors for vR and vC, the potential differences across the resistor and capacitor, respectively. We must get the relative phases right, and the lengths of the phasors must also be appropriately proportioned. Because the current is in phase with vR (Figure 32.45a), we draw the corresponding phasor as shown in Figure 32.56b; its length is VR=IR.
What about the phasor for vC? We found previously that the current in a capacitor leads the potential difference across the capacitor by 90° (Figure 32.48a), which means we must draw the phasor for vC 90° behind the phasor for i, as it is in Figure 32.56b. The length of this phasor is VC=IXC.
Finally, we need to draw the phasor for the emf supplied by the source. Phasor addition with the loop rule for this circuit (Eq. 32.28) tells us that the phasor ℰmax for the emf is the vector sum of the phasors VR and VC (Figure 32.56c). The amplitudes of the potential differences are related by
ℰmax2 =VR2+VC2. (32.30) If we substitute VR=IR (Eq. 32.6) and VC=IXC (Eq. 32.15), this becomes
ℰmax2 =(IR)2+(IXC)2=I2(R2+XC2)=I2 aR2+ 1
v2C2b. (32.31)
Solving for I gives I= ℰmax
2R2+1>v2C2. (32.32)
Figure 32.56 Steps involved in constructing a phasor diagram for the circuit in Figure 32.52. The diagram in part d indicates the phase of the current relative to the source emf.
I
VR
VC
I ℰmax
VR
VC I (a) Draw current phasor (b) Add phasors for vR and vC (c) Add phasor for emf
vR is in phase with current.
vC lags current by 90°.
ℰmax: vector sum of VR and VC Phase and length
are arbitrary.
f
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Remembering that ℰmax=Vload, we see that even though this load includes both resistive and reactive elements, I is still proportional to Vload! The constant of proportionality is called the impedance of the load and is denoted by Z:
I=ℰmax
Z . (32.33)
The impedance of the load is a property of the entire load. It is measured in ohms and depends on the frequency for any load that contains reactive elements.
Impedance plays the same role in AC circuits that resistance plays in DC cir- cuits. In fact, Eq. 32.33 can be thought of as the equivalent of Ohm’s law for AC circuits. Equation 32.32 shows that, for an RC series circuit, Z depends on both R and C:
ZRCK 2R2+1>v2C2 (RC series combination). (32.34) To express VR and VC in terms of ℰmax, R, C, and v, we use Eq. 32.32:
VR=IR= ℰmaxR
2R2+1>v2C2 (32.35) VC=IXC= ℰmax>vC
2R2+1>v2C2. (32.36) To calculate the phase constant f, the geometry shown in Figure 32.56c gives us, with Eqs. 32.6 and 32.15,
tan f= - VC
VR = - IXC
IR = - 1
vRC (32.37)
or f=tan-1 a- 1
vRCb (RC series circuit). (32.38) The negative value of f indicates that the current in an RC series circuit leads the emf, just as it does in an AC circuit with only a capacitor. As you can see in Figure 32.56c, however, the phase difference between the emf and the current in the RC series circuit is less than 90°.
Example 32.8 High-pass filter
A circuit that allows emfs in one angular-frequency range to pass through essentially unchanged but prevents emfs in other angular-frequency ranges from passing through is called a filter.
Such a circuit is useful in a variety of electronic devices, includ- ing audio electronics. An example of a filter, called a high-pass filter, is shown in Figure 32.57. Emfs that have angular frequen- cies above a certain angular frequency, called the cutoff angular frequency vc, pass through to the two output terminals marked vout, but the filter attenuates the amplitudes of emfs that have frequencies below the cutoff value. (a) Determine an expression that gives, in terms of R and C, the cutoff angular frequency vc at which VR=VC. (b) Determine the potential difference am- plitude vout across the output terminals for vWvc and for v Vvc.
❶ GettinG Started This circuit is the same as the one in Figure 32.52, which I used to determine expressions for VR (Eq. 32.35) and VC (Eq. 32.36) in terms of R and C, so I can use Figure 32.57 Example 32.8.
ℰ vout
C
R
(Continued)
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32.14 Interchange the resistor and the capacitor in Figure 32.57, and then show that the high-pass filter becomes a low-pass filter.
Filters can also be constructed by wiring an inductor and a resistor in series with an AC source. Such a circuit is called an RL series circuit and can be analyzed in exactly the manner we used to analyze an RC series circuit (see Example 32.9).
Finally, let’s analyze an RLC series circuit: a resistor, a capacitor, and an induc- tor all in series with an AC source (Figure 32.58). As with the RC series circuit, the instantaneous current i is the same in all three elements, and the sum of all the potential differences equals the emf of the source:
ℰ =vR+vL+vC. (32.39)
The phasor diagram for this circuit is constructed in Figure 32.59 for the case where VL7VC. As before, we begin with the phasors for i and vR, and then note those results. From Figure 32.57 I see that the potential differ-
ence vout is equal to the potential difference across the resistor, so Vout=VR.
❷ deviSe plan In order to determine the value of vc at which VR=VC, I equate the right sides of Eqs. 32.35 and 32.36. The resulting v factor in my expression then is the cutoff value vc. For part b, I know that Vout=VR. Therefore I can use Eq. 32.35 to determine Vout and then determine how Vout behaves in the limiting cases where vWvc and vV vc.
❸ execute plan (a) Equating the right sides of Eqs. 32.35 and 32.36, I get R=1>vC. Solving for v yields the desired cutoff angular frequency vc:
vc= 1 RC. ✔
(b) To obtain the values of Vout for vWvc and for v V vc, I first rewrite Eq. 32.35 in a form that contains vc:
Vout=VR= ℰmaxR 2R2+1>v2C2
= ℰmax
21+1>R2v2C2= ℰmax
21+vc2>v2. (1)
For vWvc, the second term in the square root vanishes and Eq. 1 reduces to Vout= ℰmax. ✔
For vVvc, the second term in the square root dominates, so I can ignore the first term. Equation 1 then becomes
Vout=VR= ℰmax
21+vc2>v2≈ ℰmax 2vc2>v2 =ℰmaxv
vc = ℰmaxvRC.
In the limit that the angular frequency v approaches zero, Vout approaches zero as well. ✔
❹ evaluate reSult The name high-pass filter makes sense be- cause this circuit allows emfs with an angular frequency higher than the cutoff angular frequency to pass through to the output but attenuates emfs of angular frequency lower than the cutoff angular frequency, preventing them from passing through to the output. It is the capacitor that does the actual passing or block- ing. It blocks low-angular-frequency emfs because for these emfs the capacitive reactance, XC=1>vC, is very high. For high- angular-frequency emfs, XC approaches zero, and so the capaci- tor passes the emf undiminished.
Figure 32.58 An RLC series circuit, consist- ing of a resistor, an inductor, and a capacitor in series across the terminals of an AC source.
R
ℰ L
C
Figure 32.59 Steps involved in constructing a phasor diagram for the RLC series circuit in Figure 32.58.
The diagram in part c indicates the phase of the current relative to the source emf.
(a) Begin with phasors for i and vR (in phase) (b) Add VC and VL (c) Add VL − VC and VR to obtain ℰmax ℰmax
VR VL − VC
I VR
VL − VC
I VR
VL
I VC vL leads i by 90°
vC lags i by 90°
If VL 7 VC, f is positive and current lags emf.
f
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that vC lags i by 90° and vL leads i by 90° (Figure 32.59a). As a result, the phasors VC and VL can be added directly (Figure 32.59b). Finally, the loop rule (Eq. 32.39) requires the phasor for the emf to equal the vector sum of the phasors for the potential differences, as shown in Figure 32.59c. Consequently, the amplitudes VR, VL, and VC must satisfy
ℰmax2 =VR2 +(VL−VC)2. (32.40) Rewriting Eq. 32.40 in terms of I, R (from Eq. 32.6), XL (from Eq. 32.27), and XC (from Eq. 32.15) gives
ℰmax2 =I2[R2+(XL−XC)2]=I2[R2+(vL−1>vC)2], (32.41)
and thus I= ℰmax
2R2+(vL−1>vC)2. (32.42) The impedance of the RLC series combination (in other words, the constant of proportionality between I and ℰmax) is therefore
ZRLCK 2R2+(vL−1>vC)2 (RLC series combination). (32.43)
Table 32.1 lists the impedances of various loads.
Figure 32.59c shows that the phase relationship between the current and the source emf depends on the relative magnitudes of VL and VC. The phase of the current relative to the emf is given by
tan f=VL−VC
VR =XL−XC R
=vL−1>vC
R (RLC series circuit). (32.44) If VL7VC, as it is in Figure 32.59, f is positive, meaning that the current lags the source emf. Here the inductor dominates the capacitor, and as a result the series combination of the inductor and capacitor behaves like an inductor. If VL6VC, f is negative, the inductor-capacitor combination is dominated by the capacitor, and the current leads the source emf, just as in an RC series circuit.
In general, when analyzing AC series circuits, follow the procedure shown in the Procedure box on page 886.
Table 32.1 Impedances of various types of loads (all elements in series)
Load Z
R R
L vL
C 1>vC
RC 2R2+(1>vC)2
RLC 2R2+(vL−1>vC)2
Note that impedances do not simply add the way resistances do. However, the impedance of any simpler load can be found from the impedance of the RLC combination; for example, ZRC=ZRLC without the term containing L.
QUANTITATIVE TOOLS
if there is no capacitor, ignore the term containing C;
and so on.
3. To determine the amplitude of the current, in the circuit, you can now use Eq. 32.42; to determine the phase of the current relative to the emf, use Eq. 32.44.
4. Determine the amplitude of the potential difference across any reactive element using V=XI, where X is the reactance of that element. For a resistor, use V=RI.
When analyzing AC series circuits, we generally know the properties of the various circuit elements (such as R, L, C, and ℰ) but not the potential differences across them. To determine these, follow this procedure:
1. To develop a feel for the problem and to help you eval- uate the answer, construct a phasor diagram for the circuit.
2. Determine the impedance of the load using Eq. 32.43. If there is no inductor, then ignore the term containing L;
Procedure: Analyzing AC series circuits
Example 32.9 RL series circuit
Consider the circuit shown in Figure 32.60. (a) Determine the cutoff angular frequency vc and the phase constant at which VR=VL. (b) Can this circuit be used as a low-pass or high-pass filter?
❸ execute plan (a) Ignoring the term containing C in Eq. 32.43 and substituting the result in Eq. 32.33, I get for the current amplitude
I= ℰmax 2R2+(vL)2.
I can now use Eq. 32.6 to calculate the amplitude of the potential difference across the resistor,
VR=IR= ℰmaxR
2R2+(vL)2, (1) and Eq. 32.27 to calculate the amplitude of the potential differ- ence across the inductor,
VL=IXL= ℰmaxvL
2R2+(vL)2, (2) where I have substituted vL for XL (Eq. 32.26). Equating the right sides of Eqs. 1 and 2 yields R=vL. Substituting vc for v in this equation and solving for vc give me for the cutoff angular fre- quency value at which VR=VL:
vc=R L. ✔
To determine the phase constant for the condition VR=VL, I substitute VR for VL in Eq. 32.44 and set VC equal to zero:
tan f=VL−VC
VR =VR
VR=1, so the phase constant is 45°. ✔
(b) Just as I did in Example 32.8, to obtain the limiting values of Vout, I first rewrite Eq. 1 in a form that contains vc:
Vout=VR= ℰmaxR
2R2+(vL)2= ℰmax
21+(vL)2>R2
= ℰmax
21+v2>vc2. (3)
❶ GettinG Started This example is similar to Example 32.8, with the capacitor of that example replaced by an inductor here.
As in Example 32.8, I see from the circuit diagram that the po- tential difference vout is equal to the potential difference across the resistor, so Vout=VR. I begin by drawing a phasor diagram for the circuit (Figure 32.61). I first draw phasors VR and I, which I know from Figure 32.45a are in phase. I then add VL, which leads I by 90° (Figure 32.51a). I make VL have the same length as VR because the problem asks about the circuit when VR=VL.
ℰ R vout
Figure 32.60 Example 32.9. L
Figure 32.61
❷ deviSe plan To determine the potential difference ampli- tudes VL and VR across the inductor and the resistor, I follow the procedure given in the Procedure box above. I then set these two amplitudes equal to each other in order to determine vc and the phase constant. To determine whether this circuit can be used as a low-pass or high-pass filter, I examine the behavior of Vout for vWvc and for vVvc.