Chapter 28 Alternating Current Circuits

Transcription

1 History teaches us that the searching spirit of man required thousands of years for the discovery of the fundamental principles of the sciences, on which the superstructure was then raised in a comparatively short time. But these very fundamental propositions are nevertheless so clear and simple, that the discovery of them reminds us, in more than one respect, of Columbus s egg Julius Robert Mayer 28.1 Introduction Up to now the only electric circuits considered were DC circuits. It is only natural that this historical sequence should be followed because the original power source available for a circuit was the DC battery. However, with the knowledge gained through electromagnetic induction, it soon became apparent that by mechanically rotating a loop of wire in an external magnetic field, a sinusoidally varying electromagnetic force could be obtained in the rotating loop of wire. This varying emf gives rise to a varying alternating current. This new AC generator was described in section Recall that the alternating emf was given by or where E = ωab sin ωt (27.30) E = Emax sin ωt (27.32 Emax = ωab (27.31) Here ω is the angular speed of the rotating coil, A is the cross-sectional area of the coil, and B is the magnetic field that the coil is rotated through. The resulting alternating current in the coil was found to be i = Emax sin ωt = imax sin ωt (27.35) R The angular speed ω of the current loop is related to the frequency f of the alternating current by ω = 2πf (28.1) 28.2 The Effective Current and Voltage in an AC Circuit In contrast to a direct current, in which charge flows in one direction, in an alternating current the charge flows first in one direction and then in the opposite direction. The alternating current i varies sinusoidally with time, as shown in figure 28.1(a).

2 Figure 28.1 An alternating current The current i starts from zero and increases in one direction until it reaches its maximum value imax. It then decreases to zero. It starts to increase again, but now the charge is flowing in the opposite direction. This is shown as the negative portion of the sine wave. The current increases negatively to imax and then decreases to zero, where the cycle begins again. We can express the alternating current i as i = imax sin(2πft) (28.2) where f is the frequency of the alternating current and t is the time. With a DC current, the current is specified as having a certain value, let us say 5 A. It always has that value because it is a constant current. We would also like to describe an AC current by a single number, such as a current of 5 A. However, with an AC current, we cannot say that the current has only one value; it has many values because the current is constantly changing with time. It would certainly be desirable to be able to refer to a single value for the alternating current. We therefore need to find a different way to describe the alternating current in order to consider it as having a single value, like a DC current. We cannot simply take the average of an AC current, because the average current is zero; that is, half the time the current is positive, half the time the current is negative. We need to look at some other way to describe the effects of the alternating current. Consider the heating effect of the alternating current. The power P dissipated in a resistor R in a DC circuit carrying a current I was found in chapter 24 to be P = I 2 R (24.18) Using the same procedure, the power dissipated in an AC circuit is P = i 2 R = [imax sin(2πft)] 2 R 28-2

3 = i 2 max sin 2 (2πft) R (28.3) Equation 28.3 shows that the power dissipated across the resistor is a function of the time t. That is, the power dissipated is not a constant as in the DC circuit but varies with time. Let us, instead, consider the average value of the power generated, that is, P avg = i 2 max[sin 2 (2πft)]avgR Although the average value of sin(2πft) is zero, the average value of its square is not, as we can see from figure 28.1(b). We find the average value of sin 2 (2πft) by using the trigonometric identity sin 2 θ = 1 (1 cos 2θ) 2 Applying that identity to the problem Its average value is sin 2 (2πft) = 1 [1 cos(4πft)] 2 [sin 2 (2πft)]avg = 1 {1 [cos(4πft)]avg} 2 However, the average value of the cosine is zero. Hence, [sin 2 (2πft)]avg = 1 2 We can also see this in figure 28.1(b) by noting that the straight line labeled i 2 = 0.5 cuts the sin 2 θ curve in half. Thus the shaded area above 0.5 is seen to be equal to the shaded area below 0.5. Using this value of 1/2 for the average value of the sin 2 (2πft), the average power expended in an AC circuit is found from equation 28.3 as Pavg = 1 i 2 maxr (28.4) 2 This average power expended in an AC circuit, with a current varying from imax to +i max, is not equivalent to the power expended in a DC circuit carrying the constant current imax. To find the equivalence between the power expended in an AC circuit and the power expended in a DC circuit, we introduce an effective current. An effective current ieff is defined as a constant current that generates heat in a resistor R at the same rate as an alternating current. Therefore, the average power generated in an AC circuit is equal to the power generated in a DC circuit carrying an effective current ieff, that is, (P)DC = (Pavg)AC 28-3

4 Hence, i 2 effr = 1 i 2 maxr 2 i = 1 i = i (28.5) eff 2 max max Equation 28.5 gives the effective value of an alternating current. The effective value is equivalent to a constant current of this value. It is 70.7% of the maximum or peak value of the AC current. Even though the current is constantly changing with time, the current can be referred to in terms of a single value, its effective or equivalent value. The effective value of an AC current is equivalent to a constant DC current of this value. If an alternating current is applied to a resistor, the voltage drop across the resistance R is V = ir Using the definition of the alternating current in equation 28.2, i = imax sin(2πft), we can write this as V = imaxr sin(2πft) But imaxr = Vmax. Hence, the alternating voltage across the resistor is given by Using the same analogy as for current, we obtain V = Vmax sin(2πft) (28.6) V = 1 V = V (28.7) eff 2 max max Equation 28.7 gives the effective value of the voltage in an AC circuit. The effective voltage is a constant value of the voltage that produces the same effect as the alternating voltage. Thus, the constantly changing alternating voltage is described in terms of the single valued, constant effective voltage. Whenever we discuss current or voltage in an AC circuit, we mean its effective value. AC ammeters and voltmeters measure the effective value of current and voltage, respectively. Since this effective voltage or current is the square root of the mean or average value of the voltage or current squared, it is often called the root-meansquared voltage or current, or simply the rms voltage or current. Hence, an alternative notation is occasionally used with I eff replaced by Irms, and Veff replaced by Vrms. 28-4

5 Example 28.1 The effective value of an AC current. An AC current in a circuit varies from 3.50 A to A. Find the effective value of this current. Solution The effective value of the current, found from equation 28.5, is i eff = 0.707imax = (0.707)(3.50 A) = 2.47 A Thus, even though the current is varying with time, if an AC ammeter were placed in the circuit it would read the single value of 2.47 A. To go to this Interactive Example click on this sentence. Example 28.2 Finding the maximum value when the effective value is known. What is the maximum voltage in a 60.0-Hz, 120-V line? Solution The effective voltage is 120 V and the maximum voltage, found from equation 28.7, is V eff = Vmax Vmax = Veff = 120 V = 170 V Hence, even though an AC voltmeter would read the single value of 120 V, the actual voltage would be varying between 170 V and +170 V. To go to this Interactive Example click on this sentence. In the analysis of an AC circuit, we will depart from our usual custom of studying the simplest cases first and then building up to the most general case. 28-5

6 Instead, we will start with the general case, an RLC series circuit (i.e., a circuit with a resistance R, an inductor L, and a capacitor C, connected in series) and take as special cases the simpler examples of an RC circuit and an RL circuit. By now, the student should be advanced enough to treat the general case first, and to be able to see the beauty in the general case, in that it contains all the other examples as special cases. The LC circuit will be left as a problem for the student to do An RLC Series Circuit Consider a circuit containing a resistor R, an inductor L, and a capacitor C, connected in series. This combination, shown in figure 28.2, is called an RLC series circuit. Because R, L, and C are in series, we would expect the total applied Figure 28.2 An RLC series circuit. voltage to be equal to the sum of the voltages across R, L, and C. But if we placed a voltmeter across R, L, and C, we would find that V VR + VL + VC (28.8) The effective voltages do not add up algebraically because the sinusoidal voltages are not in phase with one another. Phase Relations The different phase relations of the current and voltages in an AC circuit can be seen with the help of figure To understand these phase relations let us look at 28-6

7 the instantaneous voltage drop across each circuit component. Figure 28.3 Phase relations in a series RLC circuit. Phase Relationship for a Resistor The instantaneous voltage drop across the resistor is given by VR = ir (28.9) But the current in an AC circuit was given by equation as Hence i = i max sin ωt VR = imax R sin ωt (28.10) Equation gives the instantaneous voltage drop across the resistor as a function of time. Notice that VR varies with the sin ωt, just as the current does, and hence the voltage drop across the resistor VR is everywhere in phase with the current i in the circuit. Figure 28.3(a) shows the AC current i in the circuit, while part b of the diagram shows the voltage drop VR across the resistor. Notice that the voltage VR is 28-7

8 everywhere in phase with the current i. That is, the location of the maximum and minimum values of the voltage VR coincide with the maximum and minimum values of the current i. This was to be expected since the current i was multiplied by the constant value R. Phase Relationship for an Inductor Recall that the voltage drop across an inductor was given in chapter 27 as VL = Ldi (28.11) dt Using equation for the current i, this becomes VL = Ld (imax sin ωt) dt VL = L imax d (sin ωt) = L imax cos ωt (ω) dt VL = (ωl)imax cos ωt (28.12) Equation gives the voltage drop across the inductor as a function of time, and is plotted in figure 28.3(c). Since the cosine function is 90 0 out of phase with the sine function, the voltage across the inductor will be 90 0 out of phase with the current in the circuit. Notice in the figure that the voltage across the inductance VL leads the voltage across the resistance VR and the current i by In other words, VL leads VR or i, because the peak of the VL wave occurs before the peak of the VR wave. To emphasize this difference in phase between the cosine and sine function the cosine term 1 is sometimes written as cos ωt = sin(ωt + π/2) Hence equation can also be written in the form VL = (ωl)imax sin(ωt + π/2) (28.13) In this form it is very obvious that VL is 90 0 or π/2 rad out of phase with the current i in the circuit Phase Relationship for a Capacitor The potential drop across the capacitor is 1 1 From the trigonometric formula sin(ωt + θ) = sin ωt cos θ + cos ωt sin θ we get sin(ωt + π/2) = sin ωt cos π/2 + cos ωt sin π/2 sin(ωt + π/2) = sin ωt (0) + cos ωt (1) sin(ωt + π/2) = cos ωt 28-8

9 The current i in the circuit is given by hence VC = q (28.14) C i = dq/dt dq = i dt and the charge q is found by integrating, that is, q 0 max imax sinω ( ω) ( cosω ) q = dq = idt = i sinωtdt i ω max q = t dt = t + C Where C is a constant of integration that would depend upon some initial charge on the capacitor if there was one. Since we assumed that there is no initial charge on the capacitor, we set this constant equal to zero. We then obtain Combining equations and we get ω q = i max cos t (28.15) V C = 1 C i max cos t (28.16) Equation gives the voltage drop VC across the capacitor as a function of time and is plotted in figure 28.3(d). Again, since the cosine function is 90 0 out of phase with the sine function, the voltage across the capacitor is also 90 0 out of phase with the current in the circuit. Hence, we see that the VC curve is one-quarter of a cycle behind the current curve. That is, the voltage drop across the capacitor VC lags behind the voltage across the resistance VR and the current i in the AC circuit by In other words, VC lags VR or i, because the peak of the VC wave occurs after the peak of the VR wave. Also notice that because of the minus sign in equation the voltage VC across the capacitor is out of phase with the voltage across the inductor VL. We can again emphasize this difference in phase between the cosine and sine function by writing the cosine term 2 as 2 From the trigonometric formula we get sin(ωt θ) = sin ωt cos θ cos ωt sin θ sin(ωt π/2) = sin ωt cos π/2 cos ωt sin π/2 sin(ωt π/2) = sin ωt (0) cos ωt (1) sin(ωt π/2) = cos ωt 28-9

10 cos ωt = sin(ωt π/2) Hence equation can also be written in the form V C = 1 C i max sin( t /2) (28.17) From equation we see that VL is + π/2 rad or before i, while equation shows that VC is π/2 rad or 90 0 after i. In these forms it is also very obvious that VC and VL are or π rad out of phase with each other. An old mnemonic device to remember whether the emf across an inductor or capacitor leads or lags the current in an AC circuit, is the phrase ELI the ICE man. Meaning, in an inductor L, the emf E leads the current I in the circuit, whereas in a capacitor C, the emf E lags behind the current I in the circuit. Ohm s Law for an RLC Series Circuit Equation 28.8 stated that the applied effective voltage in the circuit was not equal to the algebraic sum of the effective voltages, that is, V VR + VL + VC (28.8) If we were to place a voltmeter across the circuit, we would not read the algebraic sum of these values, because, as just seen, the voltage across the inductor VL and the voltage across the capacitor VC are out of phase with the voltage across the resistor VR. The instantaneous values of VR, VL, and VC do add up by Kirchhoff s rule. It is the phase relations that causes the effective values not to add up algebraically. Since VL and VC are out of phase with each other, as seen in figure 28.3, and VL is 90 0 out of phase with VR, we can treat each of these voltages as though they were vectors, as shown in figure 28.4(a). Taking the x-axis as the reference direction, we place VR along the x-axis. VL then lies along the positive y- axis, which, if we think of a counterclockwise rotation of the vector, is before the x-axis, and VC lies along the negative y-axis, which is 90 0 after the x-axis. The sum of these voltages becomes the vector sum shown in figure 28.4(b) and its magnitude is given by the Pythagorean theorem as Figure 28.4 Vector diagram of the voltages in an AC circuit

11 V = V R 2 + (V L V C ) 2 (28.18) The voltage across RLC is, of course, also equal to the impressed voltage from the AC source. The angle φ between the applied voltage V and the voltage across the resistor VR is called the phase angle, and is found from the figure to be = tan 1 V L V C V R (28.19) Recall that the voltage across the resistor is in phase with the current in the circuit. Hence, the phase angle is a measure of how much the applied voltage V leads or lags the AC current in the circuit. If φ is above the x-axis, φ is positive, VL is greater than VC, and the voltage leads the current by φ degrees. A circuit that has a positive value of φ is called an inductive circuit. If φ lies below the x-axis, φ is negative, VC is greater than VL, and the voltage lags the current by φ degrees. A circuit that has a negative value of φ is called a capacitive circuit. The potential drop across the resistor is given by Ohm s law as VR = ir (28.20) and we saw the voltage drop across the resistor was also given by V R = imax R sin ωt (28.10) In the same way, the voltage drop across the inductor was shown to be which can also be written as which looks like the form VL = (ωl)imax cos ωt (28.12) VL = (ωl)i VR = R i Hence the quantity ωl can be viewed as some type of AC resistance. Let us define it as the inductive reactance and designate it as XL = ωl. But since ω = 2πf this becomes XL = ωl = 2πfL (28.21) Equation gives the inductive reactance of the inductor in the AC circuit. The inductive reactance XL is the inductive analogue to resistance, that is, it also has the effect of impeding current in a circuit. The inductive reactance, which is measured in ohms, is a type of AC resistance but one that varies with the inductance L of the circuit and the frequency f of the alternating current in the circuit. The inductance L of the circuit is measured in henries (H)

12 The effective voltage drop across the inductor can now be given by VL = ixl (28.22) Note that in practice, most inductors also have some resistance. We will assume here that the resistance is small enough to be ignored. In the laboratory, it is a good idea to measure the resistance of the coil you are using to see if the assumption is valid. Using the same analogy, we found the voltage drop across the capacitor as which can also be written as which again looks like the form V C = 1 C i max cos t V C = 1 C i VR = R i (28.16) Hence the quantity (1/ωC) can also be viewed as another type of AC resistance. Let us define it as the capacitive reactance and designate it as XC = 1/ωC. But since ω = 2πf this becomes XC = 1 = 1 (28.23) ωc 2πfC Equation gives the capacitive reactance XC of the capacitor in the AC circuit. The capacitive reactance XC is the capacitive analogue to resistance, that is, it also has the effect of impeding current in a circuit. The capacitive reactance, which is also measured in ohms, is a type of AC resistance but one that varies with the capacitance C of the circuit and the frequency f of the alternating current in the circuit. The capacitance C of the circuit is measured in Farads (F). The potential drop across the capacitor, using the same analogy, is now given by VC = ixc (28.24) Substituting equations 28.20, 28.22, and back into equation 28.18, yields V = (ir) 2 + (ix L ix C ) 2 or V i = R2 + (X L X C ) 2 Just as V/i was defined as the resistance R of a DC circuit, we will now define the ratio as the impedance Z of an AC circuit, that is, Z = R 2 + (X L X C ) 2 (28.25) 28-12

13 With this definition, Ohm s law for an AC circuit takes the simple form i = V Z (28.26) Therefore, in order to solve for the current in an AC circuit, we reduce it to the same simple form as a DC circuit. Instead of simply knowing the resistance, as in a DC circuit, we must now know the impedance of an AC circuit for its solution. The impedance is a function of the frequency of the AC circuit, through its inductive and capacitive reactances. It is often helpful to draw an impedance diagram, which can be deduced from the voltage diagram, and is shown in figure Figure 28.5 Impedance diagram for an AC circuit. A complete analysis of the series RLC circuit is given in example Interactive tutorial 56 at the end of the chapter will not only give a complete analysis of the series RLC circuit but it will also plot all the voltages and current to show the phase relationships in the circuit. Example 28.3 An RLC series circuit. The RLC series circuit shown in figure 28.6 has a resistance R = 400 Ω, an inductor L = 5.00 H, a capacitor C = 3.00 µf, and they are connected to a 110-V, 60.0-Hz line. Find (a) the inductive reactance XL, (b) the capacitive reactance XC, (c) the impedance Z of the circuit, (d) the current i in the circuit, (e) the voltage drop VR across R, (f) the voltage drop VL across L, (g) the voltage drop VC across C, (h) the total voltage V across RLC, and (i) the phase angle φ. Solution a. The inductive reactance, found from equation 28.21, is 28-13

14 Figure 28.6 An example of an AC circuit. XL = 2πfL = V/(A/s) s (5.00 H) H V = 1890 A V/A = 1890 Ω b. The capacitive reactance, found from equation 28.23, is XC = 1 = 1 ( F ) 2πfC 2π(60.0 1/s)( F) (C/V) V C/s = 884 C/s A V/A = 884 Ω c. The impedance of the circuit, found from equation 28.25, is Z = R 2 + (X L X C ) 2 = (400 ) 2 + ( ) 2 = 1080 Ω d. The effective current i in the circuit, found from equation 28.26, is i = V Z 28-14

15 = 110 V 1080 Ω = A e. The voltage drop across R, found from equation 28.20, is VR = ir = (0.102 A)(400 Ω) = 40.8 V f. The voltage drop across L, found from equation 28.22, is VL = ixl = (0.102 A)(1890 Ω) = 193 V g. The voltage drop across C, found from equation 28.24, is VC = ixc = (0.102 A)(884 Ω) = 90.2 V h. The total voltage across R, L, and C in series, found from equation 28.18, is V = V 2 R + (V L V C ) 2 = (40.8 V) 2 + (193 V 90.2 V) 2 = 110 V which is, of course, equal to the applied voltage. Notice that the voltages are added vectorially and not algebraically. i. The phase angle, found from equation 28.19, is = tan 1 V L V C V R = tan V 90.2 V 40.8 V = This means that the applied voltage leads the current in the circuit by , and the phase relation is shown in figure Since φ is a positive angle the circuit is called an inductive circuit

16 Figure 28.7 The phase angle between the applied voltage and the current in the AC circuit. To go to this Interactive Example click on this sentence. Example 28.4 An RC series circuit. A 110-V, 60.0-Hz, AC line is connected across a resistance of 1000 Ω and a capacitor of 1.00 µf, as shown in figure Find (a) the capacitive reactance, (b) the impedance, (c) the current in the circuit, (d) the voltage drop VR across the resistor, (e) the voltage drop VC across the capacitor, (f) the total voltage across R and C, and (g) the phase angle φ between the voltage and current. Figure 28.8 An RC series circuit. Solution a. The capacitive reactance XC, found from equation 28.23, is XC = 1 = 1 2πfC 2π(60.0 1/s)( F) = 2650 Ω b. The impedance Z is found from equation Because there is no induction in this RC circuit, XL = 0. Therefore, the impedance becomes Z = R 2 + X C

17 Z = (1000 ) 2 + (2650 ) 2 = 2830 Ω c. The effective current i in the circuit comes from Ohm s law, equation 28.26, and is i = V = 110 V Z 2830 Ω = A d. The voltage drop VR across the resistor, found from equation 28.20, is VR = ir = ( A)(1000 Ω) = 38.9 V e. The voltage drop VC across the capacitor, found from equation 28.24, is VC = ixc = ( A)(2650 Ω) = 103 V f. The total voltage drop across R and C in series is found from equation Since there is no inductance in this circuit, VL = 0. Therefore, V = V 2 2 R + V C = (38.9 V) 2 + (103 V) 2 = 110 V Note that the voltage across R and C in series is the same as the applied voltage, which it should be. Because of the phase difference of the voltages, they add as vectors rather than as algebraic quantities. g. The phase angle φ between the voltage and the current in the circuit is found from equation with VL = 0. Therefore, = tan 1 V C V R = tan V 38.9 V = This phase angle is represented in figures 28.10(a) and 28.10(b). The voltage in the circuit lags the current in the circuit by Since φ is a negative quantity, the circuit is called a capacitive circuit

18 Figure 28.9 Phase relation for an RC circuit. To go to this Interactive Example click on this sentence. Example 28.5 An RL series circuit. A 110-V, 60-Hz, AC line is connected across a resistance of 1000 Ω and an inductor of 5.00 H, as shown in figure Find (a) the inductive reactance, (b) the impedance, (c) the current in the circuit, (d) the voltage drop VR across the resistor, (e) the voltage drop VL across the inductor, (f) the total voltage drop V across R and L, and (g) the phase angle between the voltage V and the current I. Solution a. The inductive reactance XL, found from equation 28.21, is XL = 2πfL = 2π(60 1/s)(5.00 H) = 1890 Ω Figure An RL series circuit. b. The impedance Z is found from equation 28.25, with XC = 0. Therefore, 28-18

19 Z = R 2 + X L 2 = (1000 ) 2 + (1890 ) 2 = 2140 Ω c. The current i in the circuit, found from Ohm s law, equation 28.26, is i = V = 110 V Z 2140 Ω = A d. The voltage drop VR across the resistor, found from equation 28.20, is VR = ir = ( A)(1000 Ω) = 51.4 V e. The voltage drop VL across the inductor, found from equation 28.22, is VL = ixl = ( A)(1890 Ω) = 97.1 V f. The total voltage drop V across R and L is found from equation with VC = 0. Therefore, V = V R 2 + V L 2 = (51.4 V) 2 + (97.1 V) 2 = 110 V Note that the voltage drop across R and L is the same as the applied voltage V, as is expected. g. The phase angle φ between the current and voltage in the circuit is found from equation with VC = 0. Therefore, = tan 1 V L V R = tan V 51.4 V = Because this is a positive angle it means that the voltage leads the current in the circuit by and the circuit is an inductive circuit. This is shown in figures 28.12(a) and 28.12(b)

20 Figure Phase relations in an RL circuit. To go to this Interactive Example click on this sentence Resonance in an RLC Series Circuit Not only does the instantaneous value of the AC current change with time, but the effective value of the AC current even changes with the frequency of the AC source. When is this current a maximum in the RLC series circuit? The current is given by Ohm s law as i = V = V (28.27) Z R 2 + (X L X C ) 2 Notice from the form of the equation that the current i is a maximum whenever the impedance Z is a minimum. This condition is called resonance. As we can see from equation 28.27, Z is a minimum whenever XL = XC. In that case the impedance Z is equal to the resistance R. To determine the frequency at which resonance occurs, we set the inductive reactance equal to the capacitive reactance and get XL = XC 2πfL = 1 2πfC Solving for the resonant frequency we get f 0 = 1 2 LC (28.28) where f 0 is the frequency at which resonance occurs. Notice that it depends on the value of the inductance L and capacitance C. At this frequency, the maximum current that can flow in the circuit is obtained. If we plot the current in the circuit as a function of the frequency of the impressed voltage source, we obtain figure 28-20

21 At the frequency f0, the maximum current is obtained. At resonance, VL and VC are equal, but are out of phase. Resonance is used in the tuning circuit of a radio and in variable oscillators such as are found in physics laboratories. Figure Resonant frequency of an RLC series circuit. Example 28.6 Resonant frequency. Find the resonant frequency of the RLC series circuit in example Solution The resonant frequency, found from equation 28.28, is f 0 = 1 2 LC 1 H F 2π 5.00 H F V/ A/s C/V ( ) ( ) ( ) = 41.1 A s C C/s A 1 = 41.1 s 2 = 41.1 cycles/s = 41.1 Hz To go to this Interactive Example click on this sentence Power in an AC Circuit A current in an inductor causes a magnetic field to exist within that coil as discussed in chapter 26. In chapter 27 we saw that the work done in moving a charge through a coil shows up as stored energy in the magnetic field of the coil. If the coil is a perfect inductor, that is, its resistance is zero or at least negligible, all the energy applied to the coil is stored in the magnetic field of the coil and no 28-21

22 energy is lost in the coil. When the applied emf goes to zero, the energy in the magnetic field of the coil is returned to the circuit by a current. Hence, there is no energy lost in a perfect inductor. Whatever energy the inductor absorbs in one half of a cycle, it returns to the circuit in the next half-cycle. The capacitor acts similarly. As charge is stored on the plates of the capacitor, the energy used to place that charge on the capacitor is stored in the electric field between the plates of the capacitor. The energy absorbed by the electric field in one half-cycle is returned to the circuit in the next half-cycle. Therefore, there is no power consumed by the inductor or the capacitor in an AC circuit. Only across the resistance R of the circuit, is power dissipated by Joule heating. The power lost across the resistor is given by But as we can see in figure 28.4, P = ivr VR = V cos φ where φ is the phase angle between the voltage and the current i in the circuit. Therefore, the power dissipated in an AC circuit is P = iv cos φ (28.29) The quantity cos φ is called the power factor of the circuit, and is given by PF = cos φ (28.30) We can obtain further insight into the meaning of the power factor from the geometry in figure 28.5 as PF = cos φ = R (28.31) Z If we multiply both numerator and denominator of equation by i 2, we get PF = cos φ = i 2 R = i 2 R i 2 Z ivapplied The quantity i 2 R is the power dissipated by the circuit, whereas ivapplied is the power supplied to the circuit by the AC generator. Therefore, PF = cos φ = Power consumed (28.32) Power supplied Hence, the power factor is the ratio of how much power is actually used in the circuit to the total power that must be supplied to the circuit. The power factor is 28-22

23 quite often expressed as a percentage. For example, if the phase angle φ = , the power factor is PF = cos = = 86.6% This means that 86.6% of the power supplied to the circuit is actually consumed in the circuit. The rest is returned to the generator. A power factor close to 1 (or 100%) is usually desired for AC circuits. If a circuit has a power factor of 50%, and 500 W of power are actually used in the circuit, the AC generator would have to be able to deliver 1000 W. Thus, a larger and more expensive generator would be necessary to supply energy to a circuit with a low power factor. If the power factor was 1 (or 100%) a generator rated at 500 W would be adequate for the circuit. Example 28.7 Power factor in an AC circuit. In the RLC series circuit of example 28.3, find (a) the power factor, (b) the power consumed, and (c) the total power that must be supplied to the circuit. a. In example 28.3 the phase angle φ was found to be The power factor of the circuit, found from equation 28.30, is b. The power dissipated in the circuit is c. The power applied to the circuit is Solution PF = cos φ = cos = = 36.9% P = i 2 R = (0.102 A) 2 (400 Ω) = 4.16 W As a check, note that P app = ivapp = (0.102 A)(110 V) = 11.2 W Power consumed = 4.16 W = = 37.1% Power supplied 11.2 W which agrees with the original value, within round-off errors of the calculations. To go to this Interactive Example click on this sentence

24 28.6 An RLC Parallel Circuit An AC circuit with a resistor, inductor, and capacitor in parallel, called an RLC parallel circuit, is shown in figure Since R, L, and C are in parallel Figure A parallel RLC circuit. the voltage across each of them is the same. This means that the voltages are not only equal, but they are also in phase with one another. Thus, V = VR = VL = VC Because all these voltages are in phase, it is reasonable to expect that some of the currents will not be in phase. The impressed voltage V, or the voltage across the resistor VR, will be used as the reference to measure all phase differences. The current IR through R is in phase with V as shown in figures 28.14(a) and 28.14(b). The current through the inductor is out of phase with the current through the resistor. We can see this phase difference by observing the phase relations in the series circuit of figure 28.3, where the voltage VL leads VR and the current i by Therefore, if VL leads i by 90 0, then we can equally well say that i lags VL by Hence, the current in the inductor part of the series circuit lags the voltage across the inductor by For the parallel circuit, the current IL through the inductor must also lag the voltage across the inductor by But since the voltage across the inductor is in phase with the voltage across the resistance, which, in turn, is in phase with the current IR in the resistor, it follows that IL, the current through the inductor, must lag the current IR in the resistor by This is shown in figure 28.14(c). The current through the capacitor is also out of phase with the current through the resistor. For the series circuit, the voltage VC lags VR and i by 90 0, as observed in figure Therefore, the current in the series circuit leads the voltage across the capacitor by Thus, in the parallel circuit of figure 28.13, the current through the capacitor must also lead the voltage across the capacitor. Since the 28-24

25 Figure Phase relations in a parallel RLC circuit. voltage across the capacitor is in phase with the voltage across the resistor, which is in phase with the current IR through the resistor, then the current through the capacitor IC must lead the current through the resistor IR by This is shown in figure 28.14(d). Figure can be summarized by saying that IL lags IR by 90 0 IC leads IR by 90 0 IC is out of phase with IL Because the currents in the parallel RLC circuit are out of phase with each other, they can be drawn as vectors in figure 28.15(a). The total resultant current IT, found from the geometry of figure 28.15(b), is I T = I R 2 + (I C I L ) 2 (28.33) 28-25

26 Figure Currents in a parallel circuit are added vectorially. The individual currents, found by Ohm s law, are IR = V (28.34) R IL = V (28.35) XL IC = V (28.36) The phase angle φ that the total current leads or lags the current in the resistor, and hence the applied voltage, found from figure 28.15(b), is XC tan φ = IC IL IR = tan 1 I C I L I R (28.37) If we substitute equations 28.34, 28.35, and into equation 28.37, we also obtain = tan 1 V V XC X L V/R (28.38) = tan 1 R XC R X L The total impedance of the circuit, found from Ohm s law, is Z = V I T (28.39) 28-26

27 Example 28.8 An RLC parallel circuit. A resistor of 400 Ω, an inductor of 5.00 H, and a capacitor of 3.00 µf are connected in parallel to a 110-V, 60.0-Hz line. Find (a) the inductive reactance, (b) the capacitive reactance, (c) the current through the resistor IR, (d) the current through the inductor IL, (e) the current through the capacitor IC, (f) the total current in the circuit IT, (g) the phase angle φ, and (h) the total impedance of the circuit. Solution a. The inductive reactance is b. The capacitive reactance is XL = 2πfL = 2π(60.0 1/s)(5.00 H) = 1890 Ω XC = 1 2πfC = 1 2π(60.0 1/s)( F) = 884 Ω c. The current through the resistor IR, found from equation 28.34, is IR = V R = 110 V 400 Ω = A d. The current IL through the inductor, found from equation 28.35, is IL = V XL = 110 V 1890 Ω = A e. The current IC through the capacitor, found from equation 28.36, is IC = V XC = 110 V 884 Ω 28-27

28 = A f. The total current IT in the circuit, found from equation 28.33, is I T = I R 2 + (I C I L ) 2 = (0.275 A) 2 + (0.124 A A) 2 = A g. The phase angle φ, found from equation 28.37, is = tan 1 = tan 1 I C I L I R A A A = The total current in the circuit leads the applied voltage by h. The total impedance of the circuit, found from equation 28.39, is Z = V IT = 110 V A = 389 Ω To go to this Interactive Example click on this sentence. Example 28.9 An RC parallel circuit. A 110-V, 60.0-Hz, AC line is connected in parallel to a resistor of 1000 Ω and a capacitor of 1.00 µf, as shown in figure Find (a) the capacitive reactance, (b) the current IR through the resistor, (c) the current IC through the capacitor, (d) the total current in the circuit, (e) the phase angle φ, and (f) the total impedance of the circuit. a. The capacitive reactance is found to be Solution XC = 1 2πfC 28-28

29 Figure A parallel RC circuit. = 1 2π(60.0 1/s)( F) = 2650 Ω b. The current in the resistor, found from equation 28.34, is IR = V R = 110 V 1000 Ω = A c. The current IC through the capacitor, found from equation 28.36, is IC = V XC = 110 V 2650 Ω = A d. The total current in the circuit is found from equation with IL = 0, that is, I T = I 2 2 R + I C = (0.110 A) 2 + ( A) 2 = A e. The phase angle φ is found from equation with IL = 0, that is, = tan 1 I C IR = tan =

30 f. The total impedance of the circuit, found from equation 28.39, is Z = V IT = 110 V A = 932 Ω To go to this Interactive Example click on this sentence The Transformer The transformer is a device, based on the principle of mutual induction studied in chapter 25, that changes electrical energy at a particular voltage and current in one coil into a different voltage and current in a second coil. The simplest transformer is the coaxial solenoids shown in figure 27.6 and reproduced here as figure 28.17(a). The schematic diagram for the transformer is shown in figure 28.17(b). The first solenoid coil is called the primary coil, whereas coil 2 is called the secondary. If a varying emf is applied across coil 1 causing a changing current in coil 1 and hence a changing magnetic flux, an emf is induced in coil 2 by this changing magnetic flux Φ C, and is given by Figure A transformer. E2 = N2 dφc dt If we were to change magnetic flux in coil 2 by the same amount, this would induce an emf in coil 1 given by 28-30

31 E1 = N1 dφc dt Since the changing magnetic flux is the same in each coil Hence, dφc = E2 = E1 dt N2 N1 E2 = E1 (28.40) N2 N1 Therefore, the emf induced in the secondary of the transformer is given by E 2 = N2 E1 (28.41) If N2, the number of turns in the secondary, is greater than N1, the number of turns in the primary, then the ratio of N2/N1 is greater than 1. Therefore, the induced emf in the secondary E2 is greater than the impressed emf in the primary E1. When N2 is greater than N1, the transformer is called a step-up transformer because it stepsup or increases the voltage from the primary to the secondary. When N2 is less than N1, the ratio N2/N1 is less than 1, and the induced emf in the secondary is less than the applied emf across the primary. In this case, the transformer is called a step-down transformer because it has reduced the voltage from the primary to the secondary. The law of conservation of energy states that the power put into coil 1 is equal to the power taken out of coil 2. Because the product of an emf E with the current i represents power, we must have N1 E 2i2 = E 1i1 (28.42) If an alternating emf and current are impressed across coil 1 by an AC generator, then an alternating emf and current are found in coil 2, where the relation between the emfs and currents is given by equation (Of course, the emfs and currents are given by their effective values.) Solving for the ratio of the emf of coil 2 to the emf of coil 1 gives E2 = i1 (28.43) E1 i2 Equation shows that the ratio of the emfs in a transformer is reciprocal to the ratio of the currents. Note, that if equation is written in the form 28-31

32 E2 = N2 E1 N1 then we can combine it with equation as E2 = i1 = N2 (28.44) E1 i2 N1 Equation shows that the current in the secondary winding of a transformer is given by i2 = N1 i1 (28.45) N2 Notice, that if N2 is greater than N1 (a step-up transformer), the current in the secondary is less than the current in the primary. Comparing equations and shows that the currents are reciprocal to the emfs. This means that in a stepup transformer, E2 is greater than E1, but the current i2 is less than i1. Hence, we do not get something for nothing from the transformer. The increase in voltage comes with a decrease in current, such that the power into the circuit is equal to the power out. A picture of a typical transformer is shown in figure 28.17(c). Example The voltage in a step-up transformer. A 120-V line from an AC generator is connected to the primary of a transformer of 50 turns. If the secondary of the transformer has 100 turns, what voltage will be found across the secondary of the transformer? Solution The voltage across the secondary, found from equation 28.41, is E2 = N2 E1 = 100 (120 V) = 240 V N1 50 To go to this Interactive Example click on this sentence. Example The current in a step-up transformer. If the current in the primary of example is 3.00 A, what is the current in the secondary? 28-32

33 Solution The current in the secondary, found from equation 28.45, is i2 = N1 i1 = 50 (3.00 A) = 1.50 A N2 100 To go to this Interactive Example click on this sentence. Example A step-down transformer. You wish to make a transformer that can take the 120 V from your wall outlet and drop it down to 24.0 V to operate a toy electric train. If there are 100 turns of wire in the primary how many turns do you need for the secondary? Solution The number of turns in the secondary can be found from equation as N2 = E2 N1 E1 N2 = E2 N1 E V 100 turns 120 V = 20.0 turns = ( ) To go to this Interactive Example click on this sentence. Having analyzed an AC and a DC circuit, it is obvious that the AC circuit is more difficult to deal with than a DC circuit. What then is the great advantage of AC power over DC power that it is used so extensively? Two of the main reasons for the desirability of AC power are the ease in generating AC power with the AC generator described in chapter 25 and, second, the ease of transmitting this AC power from the power station to the home by means of transformers. Power is lost in transmission lines by Joule heating in the form P = I 2 R. The smaller the value of the current in the line the smaller is the energy loss. As an example, suppose a particular power station generated 100,000 W by having a current of 20.0 A at

34 V. If the transmission line has a resistance of 20.0 Ω, the power lost to heat during transmission is P = I 2 R = (20.0 A) 2 (20.0 Ω) = 8000 W If the power output from the power station is connected to a step-up transformer that has a turns ratio of 1:10 from the primary to the secondary, then the voltage at the secondary becomes E2 = N2 E1 The current in the secondary is N V 1 = 50,000 V = ( ) i2 = N1 i1 N = 2.00 A = ( 20.0 A) The power lost in the transmission line due to Joule heating is now P = i 2 R = (2.00 A) 2 (20.0 Ω) = 80.0 W Therefore, instead of transmitting the 100,000 W of power at 20.0 A and 5000 V, with a loss of 8000 W, the same 100,000 W can be transmitted at 2.00 A and 50,000 V with a power loss of only 80.0 W, a saving of 7920 W. When the transmission line reaches the local substation another transformer steps-down this voltage to lower values, which are then transmitted to your local electric pole. Another transformer steps-down this voltage further to give you the desired current and voltage for your home. Before we leave this chapter it is perhaps appropriate to compare the effects of electromagnetic induction when used in a DC circuit and in an AC circuit. In chapter 27 we saw that a changing current in one coil induced an emf in the second coil. In most cases in chapter 27, we considered a DC current rising from zero to some maximum value, remaining constant for a time, and then decreasing to zero. The induced emf existed only for the short time that the current was rising or falling. The transformer reaches its maximum utilization in an AC circuit because the instantaneous value of an AC current is continually changing with time. Hence, there is a continuously alternating emf and current induced in the second coil. The transformer is thus an excellent practical example of electromagnetic induction when used in an AC circuit

35 Have You Ever Wondered? An Essay on the Application of Physics Metal Detectors at Airports Have you ever wondered how airport security can determine if a terrorist passenger is trying to sneak a gun or a bomb aboard an aircraft? To determine if a gun, or any piece of iron, is being carried by a passenger, each passenger is required to walk through a metal detector, figure 1(a). The metal detector is essentially a coil of wire that the passenger walks through. The coil is connected to an AC source, as shown in figure 1(b). An AC voltmeter is connected across the coil to record the voltage VLA across the coil. The subscript A has been placed on VL to indicate that this is the voltage across the coil when air is in the space within the coil. The voltage drop across the coil, found by equation 28.22, is (28H.1) VLA = ixla where XLA is the inductive reactance given by equation as XLA = 2πfLA (28H.2) Here LA is the inductance of the coil with air in the space within the coil. For simplicity in our discussion, let us assume that the coil of the metal detector can be Figure 1 Every passenger must walk through the metal detector. considered to be a solenoid with air in the space inside the coil. The inductance for such a coil is found from equation as L A = µ0aln 2 (28H.3) 28-35

36 where µ0 is the permeability of free space or air, A is the area of the coil, l is the length of the coil, and n is the number of turns of wire per unit length. If an iron rod is now placed within the coil, the permeability µ0 in equation 28H.3 must be replaced with the permeability µ of the iron rod. Recall that when dealing with electric fields ε 0 was the permittivity of free space, and when dealing with different materials the permittivity of the medium was given by ε = κε 0. In the same way, the permeability of a medium is given by µ = Km µ0 (28H.4) where Km is called the relative permeability of the medium. For soft iron Km 5000, and the inductance of the coil with iron in it is now given by LI = µaln 2 = (5000µ0)Aln 2 = 5000LA (28H.5) The inductive reactance of the coil with iron in the space within the coil now becomes XLI, XLI = 2πfLI = 2πf(5000LA) = (5000)(2πfLA) = (5000)XLA The voltage drop across the iron core solenoid is now VLI = ixli = i(5000xla) = 5000(iXLA) = 5000VLA (28H.6) Thus by placing an iron rod into the solenoid the voltage drop across the solenoid has increased by a factor of 5000 over the voltage across the solenoid when air was in the space. Hence moving iron into the coil is relatively easy to detect. This is the principle of the metal detector. In practice, the iron to be detected does not fill up the entire space within the coil and hence the voltage drop across the coil is not as great as in this simplified model. However, the principle is still the same and the introduction of iron into the coil produces an increase in voltage across the coil. In our simplified description we have used a voltmeter to detect the presence of metal. This increased voltage could also be used to trigger an alarm or a bell to indicate the presence of the metal. Hence, if a terrorist passenger tries to carry a concealed gun through the detector coil, the increase in the permeability of the coil due to the presence of the gun causes an alarm to sound, indicating the presence of concealed metal on the passenger

37 The Language of Physics Effective current That constant current that generates heat in a resistor at the same rate as an alternating current. The effective value of an alternating current is equal to 70.7% of the maximum or peak value of the AC current. An AC ammeter measures the effective current in the circuit. The effective current is sometimes called the rms current (p ). Effective voltage The constant value of the voltage that produces the same effect as an alternating voltage. The effective value of an alternating voltage is equal to 70.7% of the maximum or peak value of the AC voltage. An AC voltmeter measures the effective voltage. The effective voltage is sometimes called the rms voltage (p ). RLC series circuit A circuit containing a resistor, an inductor, and a capacitor in series. The voltage across the resistor is in phase with the current in the circuit. The voltage across the inductor in the circuit leads the current in the circuit by 90 0, whereas the voltage across the capacitor lags the current in the circuit by 90 0 (p ). Phase angle A measure of how much the applied voltage leads or lags the current in an AC circuit (p ). Inductive reactance The inductive analogue to resistance in an AC circuit; that is, an inductive reactance tends to impede the flow of charge in an AC circuit. The inductive reactance is proportional to the frequency of the AC source; a high-frequency AC source causes a high reactance, whereas a low-frequency source causes a low reactance (p ). Capacitive reactance The capacitive analogue to resistance in an AC circuit; that is, a capacitive reactance tends to impede the flow of charge in an AC circuit. The capacitive reactance is inversely proportional to the frequency of the AC source; a highfrequency AC source causes a low reactance, whereas a low-frequency source causes a high reactance (p ). Impedance A measure of the opposition to the flow of charges in an AC circuit. It is composed of the resistance, inductive reactance, and the capacitive reactance of the circuit. It is the AC analogue to resistance in a DC circuit (p )

38 Resonance That condition in an AC circuit where the maximum current is obtained. It occurs when the inductive reactance is equal to the capacitive reactance (p ). Resonant frequency The frequency of an AC circuit that causes resonance in the circuit. It depends on the inductance and capacitance of the circuit (p ). Power factor The ratio of the power consumed in an AC circuit to the power applied to the circuit. It is also equal to the cosine of the phase angle (p ). RLC parallel circuit An AC circuit in which the resistor, inductor, and capacitor are placed in parallel with each other. In such a circuit, the voltages are all in phase but the currents are out of phase with each other. The current in the capacitor leads the current in the resistor by 90 0, whereas the current in the inductor lags the current in the resistor by 90 0 (p ). Transformer A device, based on the principle of mutual induction, that changes electrical energy at a particular voltage and current in the primary coil into a different voltage and current in the secondary coil (p ). Summary of Important Equations AC voltage E = E max sin ωt (27.32) V = Vmax sin(2πft) (28.6) AC current i = imax sin ωt (27.35) i = i max sin(2πft) (28.2) Relation of angular speed and frequency of sine wave ω = 2πf (28.1) Effective current in an AC circuit ieff = 0.707imax (28.5) Effective voltage in an AC circuit Veff = 0.707Vmax (28.7) Inductive reactance XL = 2πfL (28.21) Capacitive reactance XC = 1 (28.23) 2πfC 28-38

39 Impedance of an AC series circuit Z = R 2 + (X L X C ) 2 (28.25) Ohm s law for an AC circuit i = V (28.26) Z Voltage drop across a resistor VR = ir (28.20) Voltage drop across an inductor VL = ixl (28.22) Voltage drop across a capacitor VC = ixc (28.24) Total voltage in an AC series circuit V = V R 2 + (V L V C ) 2 (28.18) Phase angle between voltage and φ = tan 1 VL VC (28.19) current in an AC series circuit VR Resonant frequency of an AC series circuit f 0 = 1 2 LC (28.28) Power dissipated in an AC circuit P = iv cos φ (28.29) Power factor PF = cos φ = R (28.31) Z Power factor PF = Power consumed (28.32) Power supplied Total current in a parallel RLC circuit I T = I R 2 + (I C I L ) 2 (28.33) Current in resistor for parallel RLC circuit IR = V (28.34) R Current in inductor for parallel RLC circuit IL = V (28.35) XL Current in capacitor for parallel RLC circuit IC = V (28.36) Phase angle for parallel RLC circuit φ = tan 1 IC IL (28.37) IR Total impedance for parallel RLC circuit Z = V (28.39) IT Transformer E2 = i1 = N2 (28.44) E1 i2 N1 XC 28-39

40 Questions for Chapter 28 *1. When Americans visit Europe, they are told to get an adapter for their personal appliances such as electric razors, hair driers, and the like. Why is this necessary? What is the difference between European electricity and American electricity? 2. What does doubling the frequency do to the inductive reactance? The capacitive reactance? 3. Can you use the d Arsonval galvanometer to measure AC currents or voltages? *4. How can you measure an AC current or voltage? *5. What does increasing the area of the plates of a capacitor do to the resonant frequency in an RLC series circuit? How is this used in a variable capacitor of a tuned circuit? 6. What is the phase angle of an RLC series circuit when the inductive reactance equals the capacitive reactance? What is the power factor in this case? *7. How is an LC circuit like a mass attached to a vibrating spring? Compare the equation for resonance for a vibrating spring and the LC circuit. Can you make an analogue between them? *8. How can an RLC circuit be used to filter signals of a special frequency? *9. In many transformers, the wires are wrapped around an iron core rather than a hollow air core. Why do you think this is done? Problems for Chapter The Effective Current and Voltage in an AC Circuit 1. The effective voltage in an AC circuit is measured as 50.0 V. What is the maximum value of the voltage? 2. The peak-to-peak voltage of an AC voltage is measured on an oscilloscope to be 100 V. What is the effective voltage? 3. The effective current in an AC circuit is measured to be 5.00 A. What is the maximum value of the current? 4. An AC current varies from 5.65 A to A. Find the effective value of the current. What DC current would give the same effect? 28.3 An RLC Series Circuit 5. Find the inductive reactance of a 5.00-H coil when a 400-Hz AC voltage is impressed across it. 6. At what frequency will a 50.0-mH inductor have a reactance of 800 Ω? 7. A 110-V, 60.0-Hz, AC line is connected to a 6.55-mH coil. Find the current through the coil. 8. A coil has an impedance of 800 Ω and an inductive reactance of 600 Ω. Find the resistance of the coil

41 9. Find the capacitive reactance of a 10.0-µF capacitor at 60.0 Hz. 10. A 55.5-V, 400-Hz, AC line is connected to a 6-pF capacitor. Find the current in the circuit. 11. At what frequency will the inductive reactance of a 2.00-H inductor be (a) 20 Ω, (b) 200 Ω, and (c) 2000 Ω? Is it easier for an inductor to pass low-frequency or high-frequency signals? 12. At what frequency will the capacitive reactance of a 5.00-µF capacitor be (a) 10 Ω, (b) 100 Ω, and (c) 1000 Ω? Is it easier for a capacitor to pass low-frequency or high-frequency signals? 13. A 2.00-mH inductor is connected to a 110-V, 60.0-Hz line. Find (a) the inductive reactance and (b) the current through the inductor. 14. A 2.00-µF capacitor is connected to a 50.0-V, 40.0-Hz, AC line. Find the current flowing in the capacitor circuit. 15. Find the impedance of a series circuit of R = 1000 Ω, L = 5.00 mh, and C = 10.0 µf, if the AC source is at a frequency of 60.0 Hz. 16. A resistor R = 500 Ω, an inductor L = 20.0 mh, and a capacitor C = 6.00 µf are connected in series. Find the impedance if the source is 110 V at 400 Hz. *17. An RLC series circuit has R = 1800 Ω, L = 4.89 mh, C = 4.78 µf, and f = 60.0 Hz. Find (a) the phase angle between the applied voltage and the current in the circuit, (b) the phase angle between the voltage across the inductor and the applied voltage, and (c) the phase angle between the voltage across the capacitance and the applied voltage. 18. If the impedance of an RLC series circuit of R = 800 Ω, L = 50.0 mh, and C = 3.00 µf is 1178 Ω, find the phase angle between the current in the circuit and the applied voltage. 19. In an RLC series circuit, R = 200 Ω, C = 10.0 µf, and the frequency f = 70.0 Hz. What inductance would result in the potential across the RLC combination leading the current by ? 20. A 240-V, 50.0-Hz, AC line is connected in series with a resistor of 958 Ω and a capacitor of 9.50 µf. Find (a) the impedance of the circuit and (b) the phase angle of the circuit. 21. A 120-V, 50.0-Hz, AC line is connected to an RL series circuit of R = 280 Ω and L = 5.75 mh. Find the impedance of the circuit and the phase angle between the applied voltage and the current in the circuit. 22. In an RLC series circuit, VL = 12.0 V, VC = 24.0 V, and R = 160 Ω. If the potential across the RLC combination lags the current by , find the current in the circuit. 23. A 400-Ω resistor is connected in series with an unknown inductor to a 110-V, 60.0-Hz, AC line. If the current in the circuit is A, what is the value of the inductance of the inductor? 28.4 Resonance in an RLC Series Circuit 24. A 2.00-mH inductor is connected in series with a 10.0-µF capacitor to an AC line of variable frequency. At what frequency will resonance occur? 28-41

42 25. What value of inductance is necessary to tune an RLC circuit to a frequency of MHz if the capacitor has a value of 5.00 pf? 26. At what frequency is the inductive reactance equal to the capacitive reactance if XL = 850 Ω and C = 6.00 µf? 27. In an RLC series circuit, E = 120 V, C = 15.0 pf, and the resonant frequency is 50.0 Hz. (a) Find the inductance in the circuit. (b) If the current is 10.0 ma at resonance, find the resistance in the circuit. (c) If the resonant frequency is changed to 60.0 Hz by varying the capacitance, find the new value of the capacitance. 28. If R = 800 Ω, L = 50.0 mh, and C = 3.00 µf in an RLC series circuit of E = 120 V, plot the current I as a function of the frequency, for f = 100 to 800 Hz in intervals of 100 Hz. Find the resonant frequency from the graph and compare with the resonant frequency obtained from the equation. 29. A local FM radio station transmits at a frequency of 94.3 MHz. What value of capacitance is needed in a LC series circuit of L = 1.00 H in order to have resonance? 28.5 Power in an AC Circuit 30. A 5.00-A current flows in a series RLC circuit and leads the 120 V applied voltage by Find the power factor of the circuit and the power. 31. The power factor in an RLC series circuit is If the resistance of the circuit is 60.0 Ω, find the difference between the reactances, XL XC An RLC Parallel Circuit *32. A 220-V, 40.0-Hz, AC line is connected across a parallel RLC circuit of R = 900 Ω, L = 7.00 H, and C = 8.00 µf. Find (a) the inductive reactance, (b) the capacitive reactance, (c) the current through the resistor, (d) the current through the inductor, (e) the current through the capacitor, (f) the total current in the circuit, (g) the phase angle φ, and (h) the total impedance. *33. A 110-V, 60.0-Hz, AC line is connected across a resistance of 1000 Ω, which is in parallel with an inductor of 5.00 H. Find (a) the inductive reactance, (b) the current IR through the resistor, (c) the current IL through the inductor, (d) the total current in the circuit, (e) the phase angle φ, and (f) the total impedance of the circuit. *34. A 110-V, 60.0-Hz, AC line is connected across a 1.00-µF capacitor in parallel with a 5.00-H inductor. Find (a) the capacitive reactance, (b) the inductive reactance, (c) the current IC through the capacitor, (d) the current IL through the inductor, (e) the total current IT in the circuit, (f) the phase angle φ, and (g) the total impedance of the circuit. *35. A 110-V, 60.0-Hz, AC line is connected across a resistance of 500 Ω in parallel with a capacitor of 4.00 µf. Find (a) the capacitive reactance, (b) the current through R, (c) the current through C, (d) the total current in the circuit, (e) the phase angle φ, and (f) the total impedance of the circuit

43 28.7 The Transformer 36. A transformer has 100 turns in the primary coil and 250 turns in secondary. If the primary of the transformer is connected to an alternating voltage generator of 400 V, what voltage will appear across the secondary? 37. A transformer has 100 turns in the primary coil and 250 turns in the secondary. If there is an alternating current in the primary of the transformer of 3.00 A, what current will appear in the secondary? 38. If there are 200 turns of wire in the primary of a transformer connected to a 120-V outlet, how many turns do you need in the secondary in order to produce 6.00 V there? 39. A transformer has a current in its primary winding of 2.50 A and a current in its secondary winding of A. (a) If there are 20 turns in the primary coil, find the number of turns in the secondary coil. (b) If the resistance of the secondary coil is 100 Ω, find the applied emf in the primary coil and the induced emf in the secondary coil. (Assume 100% efficiency.) 40. Find the turns ratio for the transformer in the diagram such that a spark is given off at the points AB. The distance from point A to point B is 5.00 mm. Diagram for problem A power station generates 200 kw at 100 A and 2000 V. If the transmission line has a resistance of 10.0 Ω, how much power will be lost in transmission? If a step-up transformer of 1:10 is used, what will the power loss be then? Additional Problems 42. RC and RL circuits are sometimes used as timing circuits; show that the units of RC and L/R have the unit of time. *43. A 220-V, 40.0-Hz, AC line is connected across a series RLC circuit of R = 900 Ω, L = 7.00 H, and C = 8.00 µf. Find (a) the inductive reactance; (b) the capacitive reactance; (c) the impedance of the circuit; (d) the current in the circuit; (e) the voltage drop across R, L, and C; (f) the total voltage across the circuit; (g) the phase angle φ; and (h) the resonant frequency. *44. A 60.0-V, 100-Hz, AC line is connected across a series RLC circuit of R = 1000 Ω, L = 10.0 H, and C = 2.00 µf. Find (a) the inductive reactance, (b) the capacitive reactance, (c) the impedance of the circuit, (d) the current in the circuit, (e) the voltage drop across R, (f) the voltage drop across L, (g) the voltage drop across C, (h) the total voltage drop across the circuit, (i) the phase angle, (j) the power factor, and (k) the resonant frequency

44 *45. A 110-V, 60.0-Hz, AC line is connected across a resistor of 1800 Ω, in series with a capacitor of 3.00 µf. Find (a) the capacitive reactance, (b) the impedance, (c) the current in the circuit, (d) the voltage drop across R, (e) the voltage drop across C, (f) the total voltage measured across R and C in series, (g) the phase angle φ, and (h) the power factor. *46. A 120-V, 60.0-Hz, AC line is connected across a series circuit of a resistance of 500 Ω and an inductor of 10.0 H. Find (a) the inductive reactance, (b) the impedance, (c) the current in the circuit, (d) the voltage drop across R, (e) the voltage drop across L, (f) the total voltage drop across the circuit, (g) the phase angle φ, and (h) the power factor. *47. Three inductors are connected in series to an AC source as shown in the diagram. The inductors are shielded so that a changing flux in one inductor does not cause a mutual inductance in another inductor. Show that the equivalent inductance of the inductors in series is given by L = L1 + L2 + L3 Diagram for problem 47. Diagram for problem 48. *48. Three shielded inductors are connected in parallel as shown in the diagram. The inductors are shielded so that a changing flux in one inductor does not cause a mutual inductance in another inductor. Show that the equivalent inductance of the inductors in parallel is given by 1 = L L1 L2 L3 *49. In the circuit shown R = 1300 Ω, C1 = 5.00 µf, C2 = 8.00 µf, L = 6.78 mh, and the applied voltage is 110 V at 60.0 Hz. Find the current through each circuit element

45 Diagram for problem 49. Diagram for problem 50. *50. In the circuit shown R1 = 978 Ω, R2 = 560 Ω, L1 = 6.78 mh, L2 = 3.25 mh, C = 5.00 µf, and the applied voltage is 110 V at 60.0 Hz. Find the current through each circuit element. *51. In the circuit shown R 1 = 535 Ω, R2 = 350 Ω, L1 = 3.25 mh, L2 = 2.56 mh, C 1 = 5.00 µf, C2 = 7.50 µf, and the applied voltage is 110 V at 60.0 Hz. Find the current through each circuit element. Diagram for problem 51. Diagram for problem The diagram is an example of a low-pass filter. Show that very low-input frequencies pass through the circuit while high frequencies do not. That is, show that high-output voltages are obtained across AB for low-input frequencies, while very low-output voltages are obtained for high-input frequencies. 53. The diagram is an example of a high-pass filter. Show that very highinput frequencies pass through the circuit while low frequencies do not. That is, show that high-output voltages are obtained across AB for high-input frequencies, while very low-output voltages are obtained for low-input frequencies

46 Diagram for problem 53. Diagram for problem 54. *54. Find the turns ratio between the primary and secondary coils in the diagram, to produce a spark across AB, a distance of 5.00 mm, and find the value of C to give a spark rate of 5.00 Hz, 10.0 Hz, and 30.0 Hz. The value of the inductance is 5.00 H, R = 500 Ω, and the peak voltage is 1200 V. Interactive Tutorials 55. Effective current in an AC circuit. An alternating current of frequency f = 60 Hz varies from 2.54 A to 2.54 A. Find the effective value of this current. Plot the current and show the effective current on the plot. 56. RLC series circuit. An RLC series circuit has a resistance R = 800 Ω, inductance L = 2.00 H, capacitance C = F, and they are connected to an applied voltage V = 110 V, operating at a frequency f = 60 Hz. Find (a) the inductive reactance XL, (b) the capacitive reactance XC, (c) the impedance Z of the circuit, (d) the current i in the circuit, (e) the voltage drop VR across the resistor, (f) the voltage drop VL across the inductor, (g) the voltage drop VC across the capacitor, (h) the total voltage drop across RLC, (i) the phase angle φ, and (j) the resonant frequency f0. (k) Plot the curves of the voltage across R, L, and C individually and the voltage across the combination. Show the phase relations for all these voltages. 57. Resonance. An RLC series circuit has a resistance R = 800 Ω, inductance L = 2.00 H, capacitance C = F, and they are connected to an applied voltage V = 110 V, operating at a frequency f = 60 Hz. Plot the current flow i as a function of the frequency f of the AC source. From this graph pick out the resonant frequency of the circuit and compare it to the calculated value for the resonant frequency. 58. RLC parallel circuit. An RLC parallel circuit has a resistance R = 800 Ω, inductance L = 2.00 H, capacitance C = F, and they are connected to an applied voltage V = 110 V, operating at a frequency f = 60 Hz. Find (a) the inductive reactance XL, (b) the capacitive reactance XC, (c) the current IR through the resistor, (d) the current IL through the inductor, (e) the current IC through the capacitor, 28-46