Alternating-Current Circuits

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1 hapter 1 Alternating-urrent ircuits 1.1 A Sources Simple A circuits Purely esistive load Purely Inductive oad Purely apacitive oad The Series ircuit Impedance esonance Power in an A circuit Width of the Peak Transformer Parallel ircuit Summary Problem-Solving Tips Solved Problems Series ircuit Series ircuit esonance High-Pass Filter ircuit Filter onceptual Questions Additional Problems eactance of a apacitor and an Inductor Driven ircuit Near esonance ircuit Black Box Parallel ircuit ircuit Parallel ircuit Power Dissipation FM Antenna Driven ircuit

2 Alternating-urrent ircuits 1.1 A Sources In hapter 1 we learned that changing magnetic flux can induce an emf according to Faraday s law of induction. In particular, if a coil rotates in the presence of a magnetic field, the induced emf varies sinusoidally with time and leads to an alternating current (A), and provides a source of A power. The symbol for an A voltage source is An example of an A source is t () = sinωt (1.1.1) where the maximum value is called the amplitude. The voltage varies between and since a sine function varies between +1 and 1. A graph of voltage as a function of time is shown in Figure Figure Sinusoidal voltage source The sine function is periodic in time. This means that the value of the voltage at time t will be exactly the same at a later time t = t+ T where T is the period. The frequency, f, defined as f = 1/ T, has the unit of inverse seconds (s 1 ), or hertz (Hz). The angular frequency is defined to be ω = π f. When a voltage source is connected to an circuit, energy is provided to compensate the energy dissipation in the resistor, and the oscillation will no longer damp out. The oscillations of charge, current and potential difference are called driven or forced oscillations. After an initial transient time, an A current will flow in the circuit as a response to the driving voltage source. The current, written as 1-

3 It () = Isin( ωt φ) (1.1.) will oscillate with the same frequency as the voltage source, with an amplitude I and phase φ that depends on the driving frequency. 1. Simple A circuits Before examining the driven circuit, let s first consider the simple cases where only one circuit element (a resistor, an inductor or a capacitor) is connected to a sinusoidal voltage source Purely esistive load onsider a purely resistive circuit with a resistor connected to an A generator, as shown in Figure (As we shall see, a purely resistive circuit corresponds to infinite capacitance = and zero inductance =.) Applying Kirchhoff s loop rule yields Figure 1..1 A purely resistive circuit t () () t = t () I() t= (1..1) where ( t) = I( t) is the instantaneous voltage drop across the resistor. The instantaneous current in the resistor is given by I t () sin () t sinωt = = = I ωt (1..) where =, and I = is the maximum current. omparing Eq. (1..) with Eq. (1.1.), we find φ =, which means that I ( t) and ( t ) are in phase with each other, meaning that they reach their maximum or minimum values at the same time. The time dependence of the current and the voltage across the resistor is depicted in Figure 1..(a). 1-3

4 Figure 1.. (a) Time dependence of I ( t ) and ( t ) across the resistor. (b) Phasor diagram for the resistive circuit. The behavior of I ( t) and ( t ) can also be represented with a phasor diagram, as shown in Figure 1..(b). A phasor is a rotating vector having the following properties: (i) length: the length corresponds to the amplitude. (ii) angular speed: the vector rotates counterclockwise with an angular speed ω. (iii) projection: the projection of the vector along the vertical axis corresponds to the value of the alternating quantity at time t. We shall denote a phasor with an arrow above it. The phasor has a constant magnitude of. Its projection along the vertical direction is sin ω t, which is equal to ( t ), the voltage drop across the resistor at time t. A similar interpretation applies to I for the current passing through the resistor. From the phasor diagram, we readily see that both the current and the voltage are in phase with each other. The average value of current over one period can be obtained as: 1 1 I π t I t I t dt I t dt dt T T T T T T T () = () = sinω = sin This average vanishes because = (1..3) 1 T sinωt = sinωt dt T = (1..4) Similarly, one may find the following relations useful when averaging over one period: 1-4

5 1 T cosωt = cos t dt T ω = 1 T sinωtcosωt = sin tcos t dt T ω ω = 1 T 1 T π t 1 sin ωt = sin t dt sin dt T ω = T = T 1 T 1 T π t 1 cos ωt = cos ωt dt = cos dt T T = T (1..5) From the above, we see that the average of the square of the current is non-vanishing: 1 T 1 T 1 T π t 1 I() t = I () t dt = I sin ωt dt = I sin dt I = T T T T (1..6) It is convenient to define the root-mean-square (rms) current as I Irms = I () t = (1..7) In a similar manner, the rms voltage can be defined as rms = () t = (1..8) The rms voltage supplied to the domestic wall outlets in the United States is rms = 1 at a frequency f = 6 Hz. The power dissipated in the resistor is P t I t t I t () = () () = () (1..9) from which the average over one period is obtained as: 1 rms P() t = I() t = I= Irms= Irmsrms = (1..1) 1.. Purely Inductive oad onsider now a purely inductive circuit with an inductor connected to an A generator, as shown in Figure

6 Figure 1..3 A purely inductive circuit As we shall see below, a purely inductive circuit corresponds to infinite capacitance = and zero resistance =. Applying the modified Kirchhoff s rule for inductors, the circuit equation reads di t () () t = t () = (1..11) dt which implies where di t () = = sinωt (1..1) dt =. Integrating over the above equation, we find π I() t = di = sin ωt dt cosωt sin t = = ω ω ω (1..13) where we have used the trigonometric identity π cosωt sin = ωt (1..14) for rewriting the last expression. omparing Eq. (1..14) with Eq. (1.1.), we see that the amplitude of the current through the inductor is I = = (1..15) ω where = ω (1..16) is called the inductive reactance. It has SI units of ohms (Ω), just like resistance. However, unlike resistance, depends linearly on the angular frequency ω. Thus, the resistance to current flow increases with frequency. This is due to the fact that at higher 1-6

7 frequencies the current changes more rapidly than it does at lower frequencies. On the other hand, the inductive reactance vanishes as ω approaches zero. By comparing Eq. (1..14) to Eq. (1.1.), we also find the phase constant to be π φ = + (1..17) The current and voltage plots and the corresponding phasor diagram are shown in the Figure 1..4 below. Figure 1..4 (a) Time dependence of I ( t ) and ( t ) across the inductor. (b) Phasor diagram for the inductive circuit. As can be seen from the figures, the current I ( t ) is out of phase with ( ) t byφ = π / ; it reaches its maximum value after ( t ) does by one quarter of a cycle. Thus, we say that The current lags voltage by π / in a purely inductive circuit 1..3 Purely apacitive oad In the purely capacitive case, both resistance and inductance are zero. The circuit diagram is shown in Figure Figure 1..5 A purely capacitive circuit 1-7

8 Again, Kirchhoff s voltage rule implies Qt () t () () t = t () = (1..18) which yields Q() t = () t = () t = sinωt (1..19) where =. On the other hand, the current is dq π I() t ωcosωt ωsin =+ = = ωt+ dt (1..) where we have used the trigonometric identity π cosωt sin = ωt+ (1..1) The above equation indicates that the maximum value of the current is I = ω = (1..) where 1 = (1..3) ω is called the capacitance reactance. It also has SI units of ohms and represents the effective resistance for a purely capacitive circuit. Note that is inversely proportional to both and ω, and diverges as ω approaches zero. By comparing Eq. (1..1) to Eq. (1.1.), the phase constant is given by π φ = (1..4) The current and voltage plots and the corresponding phasor diagram are shown in the Figure 1..6 below. 1-8

9 Figure 1..6 (a) Time dependence of I ( t ) and ( t ) across the capacitor. (b) Phasor diagram for the capacitive circuit. Notice that at t =, the voltage across the capacitor is zero while the current in the circuit is at a maximum. In fact, I ( ) t reaches its maximum before ( t ) by one quarter of a cycle ( φ = π / ). Thus, we say that The current leads the voltage by π/ in a capacitive circuit 1.3 The Series ircuit onsider now the driven series circuit shown in Figure Figure Driven series ircuit Applying Kirchhoff s loop rule, we obtain di Q t () () t () t () t = t () I = (1.3.1) dt which leads to the following differential equation: 1-9

10 di Q + I+ = sin ωt (1.3.) dt Assuming that the capacitor is initially uncharged so that I = + dq / dt is proportional to the increase of charge in the capacitor, the above equation can be rewritten as One possible solution to Eq. (1.3.3) is dq dq Q sin + + = ωt (1.3.3) dt dt Qt () = Qcos( ωt φ) (1.3.4) where the amplitude and the phase are, respectively, / Q = = ω + ω ω + ω ω ( / ) ( 1/ ) ( 1/ ) = ω + ( ) (1.3.5) and 1 1 tanφ = ω = ω (1.3.6) The corresponding current is dq It () = + = I sin( ωt φ) (1.3.7) dt with an amplitude I = Qω = + ( ) (1.3.8) Notice that the current has the same amplitude and phase at all points in the series circuit. On the other hand, the instantaneous voltage across each of the three circuit elements, and has a different amplitude and phase relationship with the current, as can be seen from the phasor diagrams shown in Figure

11 Figure 1.3. Phasor diagrams for the relationships between current and voltage in (a) the resistor, (b) the inductor, and (c) the capacitor, of a series circuit. From Figure 1.3., the instantaneous voltages can be obtained as: () t = I sinωt= sinωt π () t = Isin ωt+ = cosωt π () t = Isin ωt = cosω t (1.3.9) where = I, = I, = I (1.3.1) are the amplitudes of the voltages across the circuit elements. The sum of all three voltages is equal to the instantaneous voltage supplied by the A source: t () = () t+ () t+ () t (1.3.11) Using the phasor representation, the above expression can also be written as = + + (1.3.1) as shown in Figure (a). Again we see that current phasor I leads the capacitive voltage phasor by π / but lags the inductive voltage phasor by π /. The three voltage phasors rotate counterclockwise as time passes, with their relative positions fixed. 1-11

12 Figure (a) Phasor diagram for the series circuit. (b) voltage relationship The relationship between different voltage amplitudes is depicted in Figure 1.3.3(b). From the Figure, we see that = = + + = + ( ) = ( I) + ( I I ) = I + ( ) (1.3.13) which leads to the same expression for I as that obtained in Eq. (1.3.7). It is crucial to note that the maximum amplitude of the A voltage source is not equal to the sum of the maximum voltage amplitudes across the three circuit elements: + + (1.3.14) This is due to the fact that the voltages are not in phase with one another, and they reach their maxima at different times Impedance We have already seen that the inductive reactance = ω and capacitance reactance = 1/ ω play the role of an effective resistance in the purely inductive and capacitive circuits, respectively. In the series circuit, the effective resistance is the impedance, defined as Z = + ( ) (1.3.15) The relationship between Z, and can be represented by the diagram shown in Figure 1.3.4: 1-1

13 Figure Diagrammatic representation of the relationship between Z, and The impedance also has SI units of ohms. In terms of Z, the current may be rewritten as It () sin( ωt φ). = (1.3.16) Z Notice that the impedance Z also depends on the angular frequency ω, as do and. Using Eq. (1.3.6) for the phase φ and Eq. (1.3.15) for Z, we may readily recover the limits for simple circuit (with only one element). A summary is provided in Table 1.1 below: Simple ircuit 1 = ω 1 = φ ω = tan Z = + ( ) purely resistive purely inductive π / purely capacitive π / Table 1.1 Simple-circuit limits of the series circuit 1.3. esonance Eq. (1.3.15) indicates that the amplitude of the current I = / Zreaches a maximum when Z is at a minimum. This occurs when =, or ω= 1/ ω, leading to 1 ω = (1.3.17) The phenomenon at which I reaches a maximum is called a resonance, and the frequency ω is called the resonant frequency. At resonance, the impedance becomes Z =, the amplitude of the current is 1-13

14 I = (1.3.18) and the phase is φ = (1.3.19) as can be seen from Eq. (1.3.5). The qualitative behavior is illustrated in Figure Figure The amplitude of the current as a function of ω in the driven circuit. 1.4 Power in an A circuit In the series circuit, the instantaneous power delivered by the A generator is given by Pt () = Itt () () = sin( ωt φ) sinωt= sin( ωt φ)sinωt Z Z = ( sin ωtcosφ sinωtcosωtsinφ) Z (1.4.1) where we have used the trigonometric identity sin( ωt φ) = sinωtcosφ cosωtsinφ (1.4.) The time average of the power is 1-14

15 1 1 Pt ( ) = sin ωtcos φ dt sinωtcosωtsin dt T T T Z T Z φ = cosφ sin ωt sinφ sinωtcosωt Z Z 1 = cosφ Z (1.4.3) where Eqs. (1..5) and (1..7) have been used. In terms of the rms quantities, the average power can be rewritten as 1 rms Pt ( ) = cosφ = cosφ = Irmsrms cosφ (1.4.4) Z Z The quantity cosφ is called the power factor. From Figure 1.3.4, one can readily show that Thus, we may rewrite Pt () as cosφ = (1.4.5) Z rms Pt () = Irmsrms = Irms = Irms Z Z (1.4.6) In Figure 1.4.1, we plot the average power as a function of the driving angular frequency ω. Figure Average power as a function of frequency in a driven series circuit. We see that Pt () attains the maximum when cosφ = 1, or Z =, which is the resonance condition. At resonance, we have 1-15

16 rms P = I max rmsrms = (1.4.7) Width of the Peak The peak has a line width. One way to characterize the width is to define ω = ω ω, + where ω ± are the values of the driving angular frequency such that the power is equal to half its maximum power at resonance. This is called full width at half maximum, as illustrated in Figure The width ω increases with resistance. To find Figure 1.4. Width of the peak ω, it is instructive to first rewrite the average power Pt () as Pt () 1 1 ω = = + ( ω 1/ ω) ω + ( ω ω ) (1.4.8) with Pt () = /. The condition for finding ω ± is max 1 1 ω Pt () = Pt () = = max ω± 4 ω + ( ω ω ) ω± (1.4.9) which gives ω ( ω ω ) = (1.4.1) Taking square roots yields two solutions, which we analyze separately. case 1: Taking the positive root leads to 1-16

17 ω + ω+ ω =+ (1.4.11) Solving the quadratic equation, the solution with positive root is ω+ = + + ω 4 (1.4.1) ase : Taking the negative root of Eq. (1.4.1) gives ω ω ω = (1.4.13) The solution to this quadratic equation with positive root is ω = + + ω 4 (1.4.14) The width at half maximum is then ω = ω+ ω = (1.4.15) Once the width ω is known, the quality factor Q (not to be confused with charge) can be obtained as ω Q = = ω ω (1.4.16) omparing the above equation with Eq. ( ), we see that both expressions agree with each other in the limit where the resistance is small, and ω = ω ( / ) ω. 1.5 Transformer A transformer is a device used to increase or decrease the A voltage in a circuit. A typical device consists of two coils of wire, a primary and a secondary, wound around an iron core, as illustrated in Figure The primary coil, with turns, is connected to alternating voltage source () t 1. The secondary coil has N turns and is connected to a load resistance. The way transformers operate is based on the principle that an N

18 alternating current in the primary coil will induce an alternating emf on the secondary coil due to their mutual inductance. Figure A transformer In the primary circuit, neglecting the small resistance in the coil, Faraday s law of induction implies dφ = (1.5.1) B 1 N1 dt where Φ B is the magnetic flux through one turn of the primary coil. The iron core, which extends from the primary to the secondary coils, serves to increase the magnetic field produced by the current in the primary coil and ensure that nearly all the magnetic flux through the primary coil also passes through each turn of the secondary coil. Thus, the voltage (or induced emf) across the secondary coil is dφ = (1.5.) B N dt In the case of an ideal transformer, power loss due to Joule heating can be ignored, so that the power supplied by the primary coil is completely transferred to the secondary coil: I = I (1.5.3) 1 1 In addition, no magnetic flux leaks out from the iron core, and the flux Φ B through each turn is the same in both the primary and the secondary coils. ombining the two expressions, we are lead to the transformer equation: N = (1.5.4) N 1 1 By combining the two equations above, the transformation of currents in the two coils may be obtained as: 1-18

19 I N = I = I 1 1 N1 (1.5.5) Thus, we see that the ratio of the output voltage to the input voltage is determined by the turn ratio N / N 1. If N > N1, then > 1, which means that the output voltage in the second coil is greater than the input voltage in the primary coil. A transformer with N <, then, and > N1 is called a step-up transformer. On the other hand, if N N1 < N is called a step- the output voltage is smaller than the input. A transformer with N down transformer. 1 < Parallel ircuit onsider the parallel circuit illustrated in Figure The A voltage source is t () = sinωt. Figure Parallel circuit. Unlike the series circuit, the instantaneous voltages across all three circuit elements,, and are the same, and each voltage is in phase with the current through the resistor. However, the currents through each element will be different. In analyzing this circuit, we make use of the results discussed in Sections The current in the resistor is where I / =. The voltage across the inductor is t () I () t = = sinωt = I sinωt (1.6.1) which gives di () = () = sinω = (1.6.) t t t dt π π I t t dt t t I ωt (1.6.3) t ( ) = sin ω ' ' = cosω = sin ω = sin ω 1-19

20 where I / = and = ω is the inductive reactance. Similarly, the voltage across the capacitor is () t = sin ωt = Q()/ t, which implies dq π π I () t = = ω cosωt = sin ωt+ = I sin ωt+ dt (1.6.4) where I / = and = 1/ ω is the capacitive reactance. Using Kirchhoff s junction rule, the total current in the circuit is simply the sum of all three currents. It () = I() t+ I() t+ I () t π π = Isinωt+ Isin ωt + Isin ωt+ (1.6.5) The currents can be represented with the phasor diagram shown in Figure Figure 1.6. Phasor diagram for the parallel circuit From the phasor diagram, we see that I = I + I + I (1.6.6) and the maximum amplitude of the total current, I, can be obtained as I = I = I + I + I = I + ( I I ) = + ω = + ω (1.6.7) 1-

21 Note however, since I ( t ), I ( t) and I ( t) are not in phase with one another, I is not equal to the sum of the maximum amplitudes of the three currents: I I + I + I (1.6.8) With I / = Z, the (inverse) impedance of the circuit is given by Z ω = + ω = + (1.6.9) The relationship between Z,, and is shown in Figure Figure elationship between Z,, and in a parallel circuit. From the figure or the phasor diagram shown in Figure 1.6., we see that the phase can be obtained as I I φ = = = = ω I ω (1.6.1) tan The resonance condition for the parallel circuit is given by φ =, which implies The resonant frequency is 1 1 = (1.6.11) 1 ω = (1.6.1) which is the same as for the series circuit. From Eq. (1.6.9), we readily see that 1/Z is minimum (or Z is maximum) at resonance. The current in the inductor exactly 1-1

22 cancels out the current in the capacitor, so that the total current in the circuit reaches a minimum, and is equal to the current in the resistor: I = (1.6.13) As in the series circuit, power is dissipated only through the resistor. The average power is Z P() t = I() t () t = I() t = sin ωt = = (1.6.14) Z Thus, the power factor in this case is Pt () Z 1 power factor = = = = cosφ /Z 1+ ω ω (1.6.15) 1.7 Summary In an A circuit with a sinusoidal voltage source t () = sinωt, the current is given by It () = I sin( ωt φ), where I is the amplitude and φ is the phase constant. For simple circuit with only one element (a resistor, a capacitor or an inductor) connected to the voltage source, the results are as follows: ircuit Elements esistance /eactance = ω 1 = ω urrent Amplitude I I I Phase angle φ = = = π / current lags voltage by 9 π / current leads voltage by 9 where is the inductive reactance and is the capacitive reactance. For circuits which have more than one circuit element connected in series, the results are 1-

23 ircuit Elements Impedance Z urrent Amplitude Phase angle φ ( ) I I I = = = ( ) π < φ < π < φ < φ > if > φ < if < where Z is the impedance Z of the circuit. For a series circuit, we have ( ) Z = + The phase angle between the voltage and the current in an A circuit is 1 φ = tan In the parallel circuit, the impedance is given by = + ω = + ω Z and the phase is φ ω ω 1 1 = tan = tan The rms (root mean square) voltage and current in an A circuit are given by rms I =, Irms = The average power of an A circuit is Pt () = I cosφ rms rms where cosφ is known as the power factor. The resonant frequency ω is 1-3

24 ω = 1 At resonance, the current in the series circuit reaches the maximum, but the current in the parallel circuit is at a minimum. The transformer equation is N N = 1 1 where is the voltage source in the primary coil with turns, and is the 1 N1 N 1 output voltage in the secondary coil with turns. A transformer with N > N is called a step-up transformer, and a transformer with transformer. N < N 1 is called a step-down 1.8 Problem-Solving Tips In this chapter, we have seen how phasors provide a powerful tool for analyzing the A circuits. Below are some important tips: 1. Keep in mind the phase relationships for simple circuits (1) For a resistor, the voltage and the phase are always in phase. () For an inductor, the current lags the voltage by 9. (3) For a capacitor, the current leads to voltage by 9.. When circuit elements are connected in series, the instantaneous current is the same for all elements, and the instantaneous voltages across the elements are out of phase. On the other hand, when circuit elements are connected in parallel, the instantaneous voltage is the same for all elements, and the instantaneous currents across the elements are out of phase. 3. For series connection, draw a phasor diagram for the voltages. The amplitudes of the voltage drop across all the circuit elements involved should be represented with phasors. In Figure the phasor diagram for a series circuit is shown for both the inductive case > and the capacitive case <. 1-4

25 Figure Phasor diagram for the series circuit for (a) > and (b) <. From Figure 1.8.1(a), we see that > in the inductive case and leads I by a phaseφ. On the other hand, in the capacitive case shown in Figure 1.8.1(b), > and I leads by a phaseφ. 4. When =, or φ =, the circuit is at resonance. The corresponding resonant frequency isω = 1/, and the power delivered to the resistor is a maximum. 5. For parallel connection, draw a phasor diagram for the currents. The amplitudes of the currents across all the circuit elements involved should be represented with phasors. In Figure 1.8. the phasor diagram for a parallel circuit is shown for both the inductive case > and the capacitive case <. Figure 1.8. Phasor diagram for the parallel circuit for (a) > and (b) <. From Figure 1.8.(a), we see that I > I in the inductive case and leads I by a phaseφ. On the other hand, in the capacitive case shown in Figure 1.8.(b), I > I and I leads by a phaseφ. 1-5

26 1.9 Solved Problems Series ircuit A series circuit with = 16 mh, = 1 µ F, and = 4.Ω is connected to a sinusoidal voltage t ( ) ( ) = 4. sinωt, with ω = rad/s. (a) What is the impedance of the circuit? (b) et the current at any instant in the circuit be I ( t ) I ( ωt φ ) (c) What is the phaseφ? Solution: (a) The impedance of a series circuit is given by ( ) Z = sin. Find I. = + (1.9.1) where and = ω (1.9.) 1 = (1.9.3) ω are the inductive reactance and the capacitive reactance, respectively. Since the general expression of the voltage source is t () = sin( ωt), where is the maximum output voltage and ω is the angular frequency, we have = 4 and ω = rad/s. Thus, the impedance Z becomes Z = (4. Ω ) + ( rad/s)(.16 H) = 43.9Ω 1 6 ( rad/s)(1 1 F) (1.9.4) (b) With = 4., the amplitude of the current is given by I Z A 43.9Ω = = = (1.9.5) (c) The phase between the current and the voltage is determined by 1-6

27 1 ω 1 1 tan tan ω φ = = 1 ( rad/s)(.16 H) 6 ( rad/s 1 )( 1 1 F ) = tan = Ω (1.9.6) 1.9. Series ircuit Suppose an A generator with ( t) ( 15) sin ( 1t) circuit with = 4. Ω, = 8. mh, and = is connected to a series = 5. µ F, as shown in Figure Figure series circuit (a) alculate, and, the maximum of the voltage drops across each circuit element. (b) alculate the maximum potential difference across the inductor and the capacitor between points b and d shown in Figure Solutions: (a) The inductive reactance, capacitive reactance and the impedance of the circuit are given by 1 1 = = = Ω (1.9.7) ω ( 1 rad/s)( F) 3 ( )( ) = ω= 1 rad/s 8. 1 H = 8. Ω (1.9.8) and 1-7

28 ( ) ( ) ( ) Z = + = 4. Ω + 8. Ω Ω = 196 Ω (1.9.9) respectively. Therefore, the corresponding maximum current amplitude is I Z A 196 Ω = = = (1.9.1) The maximum voltage across the resistance would be just the product of maximum current and the resistance: ( )( ) = I=.765 A 4. Ω = 3.6 (1.9.11) Similarly, the maximum voltage across the inductor is ( )( ) = I =.765 A 8. Ω = 6.1 (1.9.1) and the maximum voltage across the capacitor is ( )( ) = I =.765 A Ω = 153 (1.9.13) Note that the maximum input voltage is related to, and by = + ( ) (1.9.14) (b) From b to d, the maximum voltage would be the difference between and : = + = = = 147 bd (1.9.15) esonance A sinusoidal voltage () t ( ) = sinωt is applied to a series circuit with = 1. mh, = 1 nf and =. Ω. Find the following quantities: (a) the resonant frequency, (b) the amplitude of the current at resonance, (c) the quality factor Q of the circuit, and 1-8

29 (d) the amplitude of the voltage across the inductor at the resonant frequency. Solution: (a) The resonant frequency for the circuit is given by f ω π π π 1. 1 H 1 1 F = = = = 3 9 ( )( ) 533Hz (1.9.16) (b) At resonance, the current is I 1.A. Ω = = = (1.9.17) (c) The quality factor Q of the circuit is given by Q ω 1 3 ( )( ) π 533 s 1. 1 H 15.8 = = = (. Ω) (1.9.18) (d) At resonance, the amplitude of the voltage across the inductor is 1 3 ( ) π ( )( ) = I = Iω= 1. A 533 s 1. 1 H = (1.9.19) High-Pass Filter An high-pass filter (circuit that filters out low-frequency A currents) can be represented by the circuit in Figure 1.9., where is the internal resistance of the inductor. Figure 1.9. filter (a) Find / 1, the ratio of the maximum output voltage to the maximum input voltage

30 (b) Suppose r = 15. Ω, = 1Ω and = 5 mh / 1 = 1/.. Find the frequency at which Solution: (a) The impedance for the input circuit is 1 = ( + ) + where Z r Z = + for the output circuit. The maximum current is given by = ω and I 1 = = Z 1 ( + r) + (1.9.) Similarly, the maximum output voltage is related to the output impedance by = I Z = I + (1.9.1) This implies (b) For / 1 = 1/, we have + = 1 ( + r) ( + r) 4 ( + r) = = (1.9.) (1.9.3) Since = ω= π f, the frequency which yields this ratio is f ( ) ( 1. Ω+ 15. Ω) 4( 1. Ω) 1 = = = 5.51Hz (1.9.4) π π.5 H ircuit onsider the circuit shown in Figure The sinusoidal voltage source is t () = sinωt. If both switches and S are closed initially, find the following S1 quantities, ignoring the transient effect and assuming that,, and ω are known: 1-3

31 Figure (a) the current I( t) as a function of time, (b) the average power delivered to the circuit, (c) the current as a function of time a long time after only S 1 is opened. (d) the capacitance if both S1 and S are opened for a long time, with the current and voltage in phase. (e) the impedance of the circuit when both and S are opened. S1 (f) the maximum energy stored in the capacitor during oscillations. (g) the maximum energy stored in the inductor during oscillations. (h) the phase difference between the current and the voltage if the frequency of t ( ) doubled. is (i) the frequency at which the inductive reactance reactance. is equal to half the capacitive Solutions: (a) When both switches S 1 and S are closed, the current goes through only the generator and the resistor, so the total impedance of the circuit is and the current is (b) The average power is given by I = ωt (1.9.5) () t sin Pt () = I () tt () = sin ωt = (1.9.6) 1-31

32 (c) If only S 1 is opened, after a long time the current will pass through the generator, the resistor and the inductor. For this circuit, the impedance becomes Z 1 1 = = + + ω (1.9.7) and the phase angle φ is φ tan ω 1 = (1.9.8) Thus, the current as a function of time is It I t t + ω 1 ( ) = sin( ω φ) = sin ω tan ω (1.9.9) Note that in the limit of vanishing resistance =, φ = π /, and we recover the expected result for a purely inductive circuit. (d) If both switches are opened, then this would be a driven circuit, with the phase angle φ given by 1 ω tanφ = = ω (1.9.3) If the current and the voltage are in phase, then φ =, implying tanφ =. et the corresponding angular frequency be ω ; we then obtain and the capacitance is 1 ω = ω (1.9.31) 1 = ω (1.9.3) (e) From (d), we see that when both switches are opened, the circuit is at resonance with =. Thus, the impedance of the circuit becomes Z ( ) = + = (1.9.33) (f) The electric energy stored in the capacitor is 1-3

33 1 1 ( ) It attains maximum when the current is at its maximum I : UE = = I (1.9.34) 1 1 1,max = = = ω U I (1.9.35) where we have used ω = 1/. (g) The maximum energy stored in the inductor is given by U 1 = = (1.9.36),max I (h) If the frequency of the voltage source is doubled, i.e., ω = ω = 1/, then the phase becomes φ ( / ) ( / ) ω 1/ ω 3 tan tan tan = = = (1.9.37) (i) If the inductive reactance is one-half the capacitive reactance, then 1 = ω = 1 1 (1.9.38) ω 1 ω ω = = (1.9.39) Filter The circuit shown in Figure represents an filter. Figure

34 et the inductance be = 4 mh, and the input voltage ( ) ω = rad/s. in =. sinωt, where (a) What is the value of such that the output voltage lags behind the input voltage by3.? (b) Find the ratio of the amplitude of the output and the input voltages. What type of filter is this circuit, high-pass or low-pass? (c) If the positions of the resistor and the inductor are switched, would the circuit be a high-pass or a low-pass filter? Solutions: (a) The phase relationship between and is given by I ω tanφ = = = (1.9.4) I Thus, we have ( )( ) ω rad/s.4h = = = 139 Ω (1.9.41) tanφ tan 3. (b) The ratio is given by = = = cosφ = cos 3. =.866. out in in + (1.9.4) The circuit is a low-pass filter, since the ratio / decreases with increasing ω. (c) In this case, the circuit diagram is out in Figure high-pass filter The ratio of the output voltage to the input voltage would be 1-34

35 = = = = 1 + in in + + ω ω out ω 1/ The circuit is a high-pass filter, since the ratioout / in approaches one in the large-ω limit. 1.1 onceptual Questions 1. onsider a purely capacitive circuit (a capacitor connected to an A source). (a) How does the capacitive reactance change if the driving frequency is doubled? halved? (b) Are there any times when the capacitor is supplying power to the A source?. If the applied voltage leads the current in a series circuit, is the frequency above or below resonance? 3. onsider the phasor diagram shown in Figure for an circuit. (a) Is the driving frequency above or below the resonant frequency? (b) Draw the phasor associated with the amplitude of the applied voltage. (c) Give an estimate of the phase φ between the applied A voltage and the current. 4. How does the power factor in an circuit change with resistance, inductance and capacitance? 5. an a battery be used as the primary voltage source in a transformer? 1-35

36 6. If the power factor in an circuit is cosφ = 1/, can you tell whether the current leading or lagging the voltage? Explain Additional Problems eactance of a apacitor and an Inductor (a) A =.5 µ F capacitor is connected, as shown in Figure (a), to an A generator with = 3. What is the amplitude I of the resulting alternating current if the angular frequency ω is (i) 1 rad/s, and (ii) 1 rad/s? Figure (a) A purely capacitive circuit, and (b) a purely inductive circuit. (b) A 45-mH inductor is connected, as shown in Figure 1.1.1(b), to an A generator with = 3. The inductor has a reactance = 13 Ω. What must be (i) the applied angular frequency ω and (ii) the applied frequency f for this to be true? (iii) What is the amplitude I of the resulting alternating current? (c) At what frequency f would our.5-µf capacitor and our 45-mH inductor have the same reactance? What would this reactance be? How would this frequency compare to the natural resonant frequency of free oscillations if the components were connected as an oscillator with zero resistance? Driven ircuit Near esonance The circuit shown in Figure contains an inductor, a capacitor, and a resistor in series with an A generator which provides a source of sinusoidally varying emf t () = sinωt. 1-36

37 Figure This emf drives current It () = I sin( ωt φ) through the circuit at angular frequencyω. (a) At what angular frequency ω will the circuit resonate with maximum response, as measured by the amplitude I of the current in the circuit? What is the value of the maximum current amplitude I max? (b) What is the value of the phase angle φ between frequency? t ( ) and I( t) at this resonant (c) Suppose the frequency ω is increased from the resonance value until the amplitude I of the current decreases from I max to I / max. Now what is the phase difference φ between the emf and the current? Does the current lead or lag the emf? ircuit 3 A series circuit with = 4. 1 Ωand =.4 µ F is connected to an A voltage source t ( ) = (1 ) sinωt, with ω = rad/s. (a) What is the rms current in the circuit? (b) What is the phase between the voltage and the current? (c) Find the power dissipated in the circuit. (d) Find the voltage drop both across the resistor and the capacitor Black Box An A voltage source is connected to a black box which contains a circuit, as shown in Figure

38 Figure A black box connected to an A voltage source. The elements in the circuit and their arrangement, however, are unknown. Measurements outside the black box provide the following information: t ( ) = (8 )sin ωt (a) Does the current lead or lag the voltage? It ( ) = (1.6 A)sin( ωt+ 45 ) (b) Is the circuit in the black box largely capacitive or inductive? (c) Is the circuit in the black box at resonance? (d) What is the power factor? (e) Does the box contain a resistor? A capacitor? An inductor? (f) ompute the average power delivered to the black box by the A source Parallel ircuit onsider the parallel circuit shown in Figure Figure Parallel circuit The A voltage source is t () = sinωt. (a) Find the current across the resistor. 1-38

39 (b) Find the current across the inductor. (c) What is the magnitude of the total current? (d) Find the impedance of the circuit. (e) What is the phase angle between the current and the voltage? ircuit Suppose at t = the capacitor in the circuit is fully charged to Q. At a later time t= T /6, where T is the period of the oscillation, find the ratio of each of the following quantities to its maximum value: (a) charge on the capacitor, (b) energy stored in the capacitor, (c) current in the inductor, and (d) energy in the inductor Parallel ircuit onsider the parallel circuit shown in Figure Figure Parallel circuit The A voltage source is t () = sinωt. (a) Find the current across the resistor. (b) Find the current across the capacitor. (c) What is the magnitude of the total current? 1-39

40 (d) Find the impedance of the circuit. (e) What is the phase angle between the current and the voltage? Power Dissipation A series circuit with = 1. Ω, = 4 mh and =. µ F is connected to an A voltage source which has a maximum amplitude = 1. (a) What is the resonant frequency ω? (b) Find the rms current at resonance. (c) et the driving frequency be ω = 4 rad/s. ompute,, Z and φ FM Antenna 6 An FM antenna circuit (shown in Figure ) has an inductance = 1 H, a 1 capacitance = 1 F and a resistance = 1Ω. A radio signal induces a 5 sinusoidally alternating emf in the antenna with an amplitude of 1. Figure (a) For what angular frequency ω (radians/sec) of the incoming waves will the circuit be in tune -- that is, for what ω will the current in the circuit be a maximum. (b) What is the quality factor Q of the resonance? (c) Assuming that the incoming wave is in tune, what will be the amplitude of the current in the circuit at this in tune frequency. (d) What is the amplitude of the potential difference across the capacitor at this in tune frequency? 1-4

41 Driven ircuit Suppose you want a series circuit to tune to your favorite FM radio station that broadcasts at a frequency of 89.7 MHz. You would like to avoid the obnoxious station that broadcasts at 89.5 MHz. In order to achieve this, for a given input voltage signal from your antenna, you want the width of your resonance to be narrow enough at 89.7 MHz such that the current flowing in your circuit will be 1 times less at 89.5 MHz than at 89.7 MHz. You cannot avoid having a resistance of =. 1Ω, and practical considerations also dictate that you use the minimum possible. (a) In terms of your circuit parameters,, and, what is the amplitude of your current in your circuit as a function of the angular frequency of the input signal? (b) What is the angular frequency of the input signal at the desired resonance? (c) What values of and must you use? (d) What is the quality factor for this resonance? (e) Show that at resonance, the ratio of the amplitude of the voltage across the inductor with the driving signal amplitude is the quality of the resonance. (f) Show that at resonance the ratio of the amplitude of the voltage across the capacitor with the driving signal amplitude is the quality of the resonance. (g) What is the time averaged power at resonance that the signal delivers to the circuit? (h) What is the phase shift for the input signal at 89.5 MHz? (i) What is the time averaged power for the input signal at 89.5 MHz? (j) Is the circuit capacitive or inductive at 89.5 MHz? 1-41

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