# Chapter 12 Driven RLC Circuits

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1 hapter Driven ircuits. A Sources A ircuits with a Source and One ircuit Element Purely esistive oad 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 Parallel ircuit Power Dissipation FM Antenna Driven ircuit

2 Driven ircuits. A Sources In hapter we learned that changing magnetic flux could 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 V (t) = V sin(ωt), (..) where the maximum value V is called the amplitude. The voltage varies between V and V since a sine function varies between + and. A graph of voltage as a function of time is shown in Figure... The phase of the voltage source is φ V = ωt, (the phase constant is zero in Eq. (..)). 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 = / T, has the unit of inverse seconds ( s - ), or hertz ( Hz ). The angular frequency is defined to be ω = π f. When a voltage source is connected to a 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 in the circuit is also sinusoidal, -

3 I( t) = I sin( ωt φ), (..) and will oscillate with the same angular frequency ω as the voltage source, has amplitude I, phase φ I = ωt φ, and phase constant φ that depends on the driving angular frequency. Note that the phase constant is equal to the phase difference between the voltage source and the current Δφ φ V φ I = ωt (ωt φ) = φ. (..3). A ircuits with a Source and One ircuit Element Before examining the driven circuit, let s first consider cases where only one circuit element (a resistor, an inductor or a capacitor) is connected to a sinusoidal voltage source... Purely esistive oad onsider a purely resistive circuit with a resistor connected to an A generator with A source voltage given by V (t) = V sin(ωt), as shown in Figure... (As we shall see, a purely resistive circuit corresponds to infinite capacitance = and zero inductance =.) Figure.. A purely resistive circuit We would like to find the current through the resistor, I (t) = I sin(ωt φ ). (..) Applying Kirchhoff s loop rule yields V (t) I (t) =, (..) where V (t) = I (t) is the instantaneous voltage drop across the resistor. The instantaneous current in the resistor is given by -3

4 I (t) = V (t) = V sin(ωt) = I sin(ωt). (..3) omparing Eq. (..3) with Eq. (..), we find that the amplitude is I = V = V (..4) where V = V, and =. (..5) The quantity is called the resistive reactance, to be consistent with nomenclature that will introduce shortly for capacitive and inductive elements, but it is just the resistance. The key point to recognize is that the amplitude of the current is independent of the driving angular frequency. Because φ =, I ( t ) and V ( t ) are in phase with each other, i.e. they reach their maximum or minimum values at the same time, the phase constant is zero, φ =. (..6) The time dependence of the current and the voltage across the resistor is depicted in Figure..(a). (a) (b) Figure.. (a) Time dependence of I ( t ) and V ( t ) across the resistor. (b) Phasor diagram for the resistive circuit. The behavior of I ( t ) and V ( t ) can also be represented with a phasor diagram, as shown in Figure..(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. -4

5 (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 V has a constant magnitude of V. Its projection along the vertical direction is V sin(ωt), which is equal to V ( 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: I (t) = T T I (t)dt = T T I sin(ωt) dt = I T T sin πt T dt =. (..7) This average vanishes because sin(ωt) = T T sin(ωt) dt =. (..8) Similarly, one may find the following relations useful when averaging over one period, cos(ωt) = T sin(ωt)cos(ωt) = T sin (ωt) = T cos (ωt) = T T cos(ωt) dt =, T sin(ωt)cos(ωt) dt =, T T sin (ωt) dt = T πt sin T T dt =, cos (ωt) dt = T πt cos T T dt =. (..9) From the above, we see that the average of the square of the current is non-vanishing: π t I t I t dt I t dt I dt I T T T T T T T ( ) = ( ) = sin ω = sin =. (..) It is convenient to define the root-mean-square (rms) current as I Irms = I ( t) = (..) -5

6 In a similar manner, the rms voltage can be defined as V Vrms = V ( t) =. (..) The rms voltage supplied to the domestic wall outlets in the United States is V = V at a frequency f = 6 Hz. rms The power dissipated in the resistor is P t = I t V t = I t. (..3) ( ) ( ) ( ) ( ) The average power over one period is then Vrms P ( t) = I ( t) = I = Irms = IrmsVrms =. (..4).. Purely Inductive oad onsider now a purely inductive circuit with an inductor connected to an A generator with A source voltage given by V (t) = V sin(ωt), as shown in Figure..3. As we shall see below, a purely inductive circuit corresponds to infinite capacitance = and zero resistance =. Figure..3 A purely inductive circuit We would like to find the current in the circuit, I (t) = I sin(ωt φ ). (..5) Applying the modified Kirchhoff s rule for inductors, the circuit equation yields di V ( t) V ( t) = V ( t) =. (..6) dt -6

7 where we define V (t) = di / dt. earranging yields di dt = V (t) = V sin(ωt), (..7) where V = V. Integrating Eq. (..7), we find the current is I (t) = di = V sin(ωt) dt = I sin(ωt π / ). = V ω cos(ωt) = V ω sin(ωt π / ), (..8) where we have used the trigonometric identity cos(ωt) = sin(ωt π / ). omparing Eq. (..8) with Eq. (..5), we find that the amplitude is I = V ω = V, (..9) where the inductive reactance,, is given by = ω. (..) The inductive reactance has SI units of ohms ( Ω ), just like resistance. However, unlike resistance, depends linearly on the angular frequency ω. Thus, the inductance reactance to current flow increases with frequency. This is due to the fact that at higher frequencies the current changes more rapidly than it does at lower frequencies. On the other hand, the inductive reactance vanishes as ω approaches zero. The phase constant, φ, can also be determined by comparing Eq. (..8) to Eq. (..5), and is given by φ = + π. (..) The current and voltage plots and the corresponding phasor diagram are shown in Figure

8 (a) (b) Figure..4 (a) Time dependence of I ( t ) and V ( 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 V ( t ) by φ = π / ; it reaches its maximum value one quarter of a cycle later than V ( t ). The current lags voltage by π / in a purely inductive circuit The word lag means that the plot of I ( t ) is shifted to the right of the plot of V ( t ) in Figure..4 (a), whereas in the phasor diagram the phasor I (t) is behind the phasor for V (t) as they rotate counterclockwise in Figure..4(b)...3 Purely apacitive oad onsider now a purely capacitive circuit with a capacitor connected to an A generator with A source voltage given by V (t) = V sin(ωt). In the purely capacitive case, both resistance and inductance are zero. The circuit diagram is shown in Figure..5. Figure..5 A purely capacitive circuit We would like to find the current in the circuit, -8

9 Again, Kirchhoff s loop rule yields The charge on the capacitor is therefore where V = V. The current is I (t) = I sin(ωt φ ). (..) Q( t) V ( t) V ( t) = V ( t) =. (..3) Q(t) = V (t) = V (t) = V sin(ωt), (..4) I (t) = + dq dt = ωv cos(ωt) = ωv sin(ωt + π / ), (..5) = I sin(ωt + π / ), where we have used the trigonometric identity cosωt = sin(ωt + π / ). The maximum value of the current can be determined by comparing Eq. (..5) to Eq. (..), I V = ωv =, (..6) where the capacitance reactance,, is =. (..7) ω The capacitive reactance 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. The phase constant can be determined by comparing Eq. (..5) to Eq. (..), and is φ = π. (..8) The current and voltage plots and the corresponding phasor diagram are shown in the Figure..6 below. -9

10 (a) (b) Figure..6 (a) Time dependence of I ( t ) and V ( 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 one quarter of a cycle earlier than V ( ) t. The current leads the voltage by π/ in a capacitive circuit The word lead means that the plot of I (t) is shifted to the left of the plot of V (t) in Figure..6 (a), whereas in the phasor diagram the phasor I (t) is ahead the phasor for V (t) as they rotate counterclockwise in Figure..6(b)..3 The Series ircuit onsider now the driven series circuit with V (t) = V sin(ωt + φ) shown in Figure.3.. Figure.3. Driven series ircuit -

11 We would like to find the current in the circuit, I(t) = I sin(ωt). (.3.) Notice that we have added a phase constant φ to our previous expressions for V (t) and I(t) when we were analyzing single element driven circuits. Applying Kirchhoff s modified loop rule, we obtain V (t) V (t) V (t) V (t) =. (.3.) We can rewrite Eq. (.3.) using V (t) = I, V (t) = di / dt, and V (t) = Q / as di dt + I + Q = V sin(ωt + φ). (.3.3) Differentiate Eq. (.3.3), using I = + dq / dt, and divide through by, yields what is called a second order damped linear driven differential equation, d I dt + di dt + I = ωv cos(ωt + φ). (.3.4) We shall find the amplitude, I, of the current, and phase constant φ which is the phase shift between the voltage source and the current by examining the phasors associates with the three circuit elements, and. The instantaneous voltages across each of the three circuit elements,, and has a different amplitude and phase compared to the current, as can be seen from the phasor diagrams shown in Figure.3.. (a) (b) (c) Figure.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. -

12 Using the phasor representation, Eq. (.3.) can be written as V = V + V + V (.3.5) as shown in Figure.3.3(a). Again we see that current phasor I leads the capacitive voltage phasor V by π / but lags the inductive voltage phasor V by π /. The three voltage phasors rotate counterclockwise as time increases, with their relative positions fixed. Figure.3.3 (a) Phasor diagram for the series circuit. (b) voltage relationship The relationship between different voltage amplitudes is depicted in Figure.3.3(b). From Figure.3.3, we see that the amplitude satisfies V = V = V + V + V = V + (V V ) = (I ) + (I I ) (.3.6) = I + ( ). Therefore the amplitude of the current is I = V + ( ). (.3.7) Using Eqs. (..5), (..), and (..7) for the reactances, Eq. (.3.7) becomes I = V + (ω ω ), series circuit. (.3.8) From Figure.3.3(b), we can determine that the phase constant satisfies -

13 tanφ = = ω ω. (.3.9) Therefore the phase constant is φ = tan ω ω, series circuit. (.3.) It is crucial to note that the maximum amplitude of the A voltage source V is not equal to the sum of the maximum voltage amplitudes across the three circuit elements: V V + V + V (.3.) This is due to the fact that the voltages are not in phase with one another, and they reach their maxima at different times..3. Impedance We have already seen that the inductive reactance = ω, and the capacitive reactance = / 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 = + ( ) (.3.) The relationship between Z,,, and can be represented by the diagram shown in Figure.3.4: Figure.3.4 Vector representation of the relationship between Z,,, and The impedance also has SI units of ohms. In terms of Z, the current (Eqs. (.3.) and (.3.7)) may be rewritten as. -3

14 I(t) = V sin(ωt) (.3.3) Z Notice that the impedance Z also depends on the angular frequency ω, as do. and Using Eq. (.3.9) for the phase constant φ and Eq. (.3.) for Z, we may readily recover the limits for simple circuit (with only one element). A summary is provided in Table. below: Simple ircuit = ω = φ = tan Z = + ( ) ω purely resistive purely inductive purely capacitive.3. esonance π / π / Table. Simple-circuit limits of the series circuit In a driven series circuit, the amplitude of the current (Eq. (.3.8)) has a maximum value, a resonance, which occurs at the resonant angular frequency ω. Because the amplitude I of the current is inversely proportionate to Z (Eq. (.3.3), the maximum of I occurs when Z is minimum. This occurs at an angular frequency ω such that =, Therefore, the resonant angular frequency is ω = ω. (.3.4) ω =. (.3.5) At resonance, the impedance becomes Z =, and the amplitude of the current is and the phase constant is zero, (Eq. (.3.)), I V =, (.3.6) -4

15 φ =. (.3.7) A qualitative plot of the amplitude of the current as a function of driving angular frequency for two driven circuits, with different values of resistance, > is illustrated in Figure.3.5. The amplitude is larger for smaller a smaller value of resistance. Figure.3.5 The amplitude of the current as a function of ω in the driven circuit, for two different values of the resistance..4 Power in an A circuit In the series circuit, the instantaneous power delivered by the A generator is given by P(t) = I(t)V (t) = V Z sin(ωt) V sin(ωt + φ) = V sin(ωt)sin(ωt + φ) Z = V Z (sin (ωt)cosφ + sin(ωt)cos(ωt)sinφ) where we have used the trigonometric identity The time average of the power is (.4.) sin(ωt + φ) = sin(ωt)cosφ + cos(ωt)sinφ. (.4.) P(ω) = T T V Z sin (ωt)cosφ dt + T T V sin(ωt)cos(ωt)sinφ dt = Z V cosφ.(.4.3) Z -5

16 where we have used the integral results in Eq. (..9). In terms of the rms quantities, V rms = V / and I rms = V rms / Z, the time-averaged power can be written as P(ω) = V Z cosφ = V rms Z cosφ = I V cosφ (.4.4) rms rms The quantity cosφ is called the power factor. From Figure.3.4, one can readily show that cosφ =. (.4.5) Z Thus, we may rewrite P(ω) as where P(ω) = I rms (ω), (.4.6) I rms (ω) = V + (ω ω ), (.4.7) In Figure.4., we plot the time-averaged power as a function of the driving angular frequency ω for two driven circuits, with different values of resistance, >. Figure.4. Average power as a function of frequency in a driven series circuit. We see that P(ω) attains the maximum value when cosφ =, or Z =, which is the resonance condition. At resonance, we have P(ω ) = I rms V rms = V rms. (.4.8) -6

17 .4. Width of the Peak The peak has a line width. One way to characterize the width is to define Δω = 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.4.. The width Δ ω increases with resistance. To find Figure.4. Width of the peak Δ ω, it is instructive to first rewrite the average power P(ω) as P(ω) = V + (ω / ω) = V ω ω + (ω ω ), (.4.9) with P(ω ) = V /. The condition for finding ω ± is P((ω )) = P(ω ) V ± 4 = V ω ± ω ± + (ω ± ω ). (.4.) Eq. (.4.) reduces to ω ( ω ω ) =. (.4.) Taking square roots yields two solutions, which we analyze separately. ase : Taking the positive root leads to ω + ω ω = + (.4.) + -7

18 We can solve this quadratic equation, and the solution with positive root is ω + = + + ω 4. (.4.3) ase : Taking the negative root of Eq. (.4.) yields ω ω ω =. (.4.4) The solution to this quadratic equation with positive root is ω = + + ω 4. (.4.5) The width at half maximum is then the difference Once the width Δω = ω+ ω =. (.4.6) Δ ω is known, the quality factor Q qual is defined as Q qual = ω Δω = ω. (.4.7) omparing the above equation with Eq. (..7), we see that both expressions agree with each other in the limit where the resistance is small, and ω = ω ( / ) ω..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.5.. The primary coil, with N turns, is connected to alternating voltage source V ( t ). The secondary coil has N turns and is connected to a load with resistance. The way transformers operate is based on the principle that an alternating current in the primary coil will induce an alternating emf on the secondary coil due to their mutual inductance. -8

19 Figure.5. A transformer In the primary circuit, neglecting the small resistance in the coil, Faraday s law of induction implies dφ B V = N, (.5.) 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 ensures 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 =. (.5.) B V 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 V = I V. (.5.3) 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, V V N =. (.5.4) N By combining the two equations above, the transformation of currents in the two coils may be obtained as I = V V I = N N I. (.5.5) -9

20 Thus, we see that the ratio of the output voltage to the input voltage is determined by the turn ratio N / N. If N > N, then V > V, which means that the output voltage in the second coil is greater than the input voltage in the primary coil. A transformer with N > N is called a step-up transformer. On the other hand, if N < N, then V < V, and the output voltage is smaller than the input. A transformer with N < N is called a stepdown transformer..6 Parallel ircuit onsider the parallel circuit illustrated in Figure.6.. The A voltage source is V (t) = V sin(ωt). Figure.6. 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..4. The current in the resistor is I (t) = V (t) = V sin(ωt) = I sin(ωt). (.6.) where I / = V. The voltage across the inductor is V (t) = V (t) = V sin(ωt) = di dt. (.6.) Integrating Eq. (.6.) yields t V I (t) = sin(ωt ')dt ' = V ω cos(ωt) = V sin(ωt π / ) = I sin(ωt π / ),(.6.3) -

21 where I / = V and = ω is the inductive reactance. Similarly, the voltage across the capacitor is V (t) = V sin(ωt) = Q(t) /, which implies I (t) = dq dt = ωv cos(ωt) = V sin(ωt + π / ) = I sin(ωt + π / ), (.6.4) where I / = V and = / ω is the capacitive reactance. Using Kirchhoff s junction rule, the total current in the circuit is simply the sum of all three currents. I(t) = I (t) + I (t) + I (t) = I sin(ωt) + I sin(ωt π / ) + I sin(ωt + π / ). (.6.5) The currents can be represented with the phasor diagram shown in Figure.6.. Figure.6. Phasor diagram for the parallel circuit From the phasor diagram, the phasors satisfy the vector addition condition, I = I + I + I. (.6.6) The amplitude of the current, I, can be obtained as I = I = I + I + I = = V + ω ω I + (I I ) = V +. (.6.7) -

22 Therefore the amplitude of the current is I = V +, parallel circuit. (.6.8) From the phasor diagram shown in Figure.6., we see that the tangent of the phase constant can be obtained as V V I I = = = = ω. (.6.9) I ω tanφ V Therefore the phase constant is φ = tan ω ω, parallel circuit. (.6.) Because I ( t ), I ( t ) and I ( t ) are not in phase with one another, I is not equal to the sum of the amplitudes of the three currents, I I + I + I. (.6.) With I / = V Z, the (inverse) impedance of the circuit is given by Z = + ω ω = + (.6.) The relationship between Z,, and is shown in Figure.6.3. Figure.6.3 elationship between Z,, and in a parallel circuit. -

23 The resonance condition for the parallel circuit is given by φ =, which implies =. (.6.3) We can solve Eq. (.6.3) for the resonant angular frequency and find that ω =. (.6.4) as in the series circuit. From Eq. (.6.), we readily see that / Z is minimum (or Z is maximum) at resonance. The current in the inductor exactly cancels out the current in the capacitor, so that the current in the circuit reaches a minimum, and is equal to the current in the resistor, V I = (.6.5) As in the series circuit, power is dissipated only through the resistor. The timeaveraged power is P(t) = I (t)v (t) = I (t) = V sin (ωt) = V = V Z Thus, the power factor in this case is Z. (.6.6) P( t) Z power factor = = = = cosφ. (.6.7) V / Z + ω ω -3

24 .7 Summary In an A circuit with a sinusoidal voltage source V (t) = V sin(ωt), the current is given by I( t) = I sin( ωt φ), where I is the amplitude and φ is the phase constant (phase difference between the voltage source and the current). 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 = ω urrent Amplitude I I I V Phase constant φ = V = V = 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 ircuit Elements Impedance Z urrent Amplitude Phase constant φ ( ) I I I = = = V + V + V + ( ) 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 constant (phase difference between the voltage and the current) in an A circuit is φ = tan. -4

25 In the parallel circuit, the impedance is given by = + ω = + Z ω, and the phase constant is φ ω. = tan = tan ω The rms (root mean square) voltage and current in an A circuit are given by V rms V I =, Irms =. The average power of an A circuit is P( t) = I V cosφ, rms rms where cosφ is known as the power factor. The resonant angular frequency ω is ω =. 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 V V N =, N where V is the voltage source in the primary coil with N turns, and V is the output voltage in the secondary coil with N turns. A transformer with N > N is called a step-up transformer, and a transformer with N < N is called a step-down transformer. -5

26 .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:. Keep in mind the phase relationships for simple circuits () 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 the 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.8. the phasor diagram for a series circuit is shown for both the inductive case > and the capacitive case <. (a) (b) Figure.8. Phasor diagram for the series circuit for (a) > and (b) <. From Figure.8.(a), we see that V > V in the inductive case and V leads I by a phase constant φ. For the capacitive case shown in Figure.8.(b), V > V and I leads V by a phase constant φ. -6

27 4. When V = V, or φ =, the circuit is at resonance. The corresponding resonant angular frequency is ω = / maximum., and the power delivered to the resistor is a 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.8. the phasor diagram for a parallel circuit is shown for both the inductive case > and the capacitive case <. (a) (b) Figure.8. Phasor diagram for the parallel circuit for (a) > and (b) <. From Figure.8.(a), we see that I > I in the inductive case and V leads I by a phase constant φ. For the capacitive case shown in Figure.8.(b), I > I and I leads V by a phase constant φ..9 Solved Problems.9. Series ircuit A series circuit with = 6 mh, = µ F, and = 4.Ω is connected to a sinusoidal voltage V (t) = (4. V)sin(ωt), 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 sin(ωt φ). Find I. (c) What is the phase constant φ? -7

28 Solution: (a) The impedance of a series circuit is given by where ( ) Z = +, (.9.) = ω and = / ω are the inductive and capacitive reactances, respectively. The voltage source is V ( t) = V sin( ωt), where V is the maximum output voltage and ω is the angular frequency, with V = 4 V and ω = rad/s. Thus, the impedance Z is Z = (4. Ω) + ( rad/s)(.6 H) ( rad/s)( 6 F) = 43.9Ω. (.9.) (b) With V = 4.V, the amplitude of the current is given by I V Z 4. V.9A. (.9.3) 43.9Ω = = = (c) The phase constant (the phase difference between the voltage source and the current) is given by φ = tan ω = tan ω ( rad/s)(.6 H) ( rad/s)( 6 F) = tan 4. Ω.9. Series ircuit = 4.. (.9.4) Suppose an A generator with V (t) = (5V)sin(t) is connected to a series circuit with = 4. Ω, = 8. mh, and = 5. µ F, as shown in Figure

29 (a) alculate V, V and V circuit element. Figure.9. series circuit, the maximum values of the voltage drop across each (b) alculate the maximum potential difference across the inductor and the capacitor between points b and d shown in Figure.9.. Solutions: (a) The inductive reactance, capacitive reactance and the impedance of the circuit are given by = ω = = Ω, (.9.5) ( rad/s)(5. 6 F) The inductance is = ω = ( rad/s)(8. 3 H) = 8. Ω. (.9.6) Z = + ( ) = (4. Ω) + (8. Ω Ω) = 96 Ω. (.9.7) The corresponding maximum current amplitude is I V Z 5 V.765A. (.9.8) 96 Ω = = = The maximum voltage across the resistance is the product of maximum current and the resistance, V = I = (.765 A)(4. Ω) = 3.6 V. (.9.9) Similarly, the maximum voltage across the inductor is -9

30 The maximum voltage across the capacitor is V = I = (.765 A)(8. Ω) = 6. V. (.9.) V = I = (.765 A)( Ω) = 53 V. (.9.) Note that the maximum input voltage V is related to V, V and V by V = V + ( V V ). (.9.) (b) From b to d, the maximum voltage would be the difference between V and V, V = V + V = V V = 6. V 53 V = 47 V. (.9.3) bd.9.3 esonance A sinusoidal voltage V (t) = ( V)sin(ωt) is applied to a series circuit with =. mh, = nf, and =. Ω. Find the following quantities: (a) the resonant frequency, (b) the amplitude of the current at resonance, (c) the quality factor Q qual of the circuit, and (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 (. 3 H)( 9 F) = 533Hz. (.9.4) (b) At resonance, the current is I V V.A. (.9.5). Ω = = = -3

31 (c) The quality factor Q qual of the circuit is given by Q qual = ω = π(533 s )(. 3 H) = 5.8. (.9.6) (. Ω) (d) At resonance, the amplitude of the voltage across the inductor is V = I = I ω = (. A)π(533 s )(. 3 H) = V. (.9.7).9.4 High-Pass Filter A high-pass filter (circuit that filters out low-frequency A currents) can be represented by the circuit in Figure.9., where is the internal resistance of the inductor. Figure.9. filter (a) Find V / V, the ratio of the maximum output voltage V to the maximum input voltage V. (b) Suppose r = 5. Ω, = Ω, and = 5 mh. Find the frequency at which V / V = /. Solution: (a) The impedance for the input circuit is impedance for the output circuit is Z ( r) = + + where = ω. The Z = +. The maximum current is given by I V V = = Z ( + r) +. (.9.8) Similarly, the maximum output voltage is related to the output impedance by -3

32 This implies V = I Z = I +. (.9.9) V + = V ( + r) +. (.9.) (b) For V / V = /, we have + ( + r) 4 ( + r) = =. (.9.) Because = ω = π f, the frequency that yields this ratio is f = π = π(.5 H) (. Ω + 5. Ω) 4(. Ω) 3 = 5.5Hz. (.9.).9.5 ircuit Figure.9.3 onsider the circuit shown in Figure.9.3. The sinusoidal voltage source is V (t) = V sin(ωt). If both switches S and S are closed initially, find the following quantities, ignoring the transient effect and assuming that,, V, and ω are known. (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 is opened. -3

33 (d) The capacitance if both S and S are opened for a long time, with the current and voltage in phase. (e) The impedance of the circuit when both S and S are opened. (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 V ( t ) is doubled. (i) The frequency at which the inductive reactance reactance Solutions:. is equal to half the capacitive (a) When both switches S and S are closed, the current only goes through 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 (t) = V sin(ωt). (.9.3) P(t) = I (t)v (t) = V sin (ωt) = V. (.9.4) (c) If only S 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 = = + + ω, (.9.5) and the phase constant φ is ω φ = tan. (.9.6) Thus, the current as a function of time is I(t) = I sin(ωt φ) = V + ω sin[ωt tan (ω / )]. (.9.7) -33

34 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 constant φ given by φ = tan ω = tan ω. (.9.8) If the current and the voltage are in phase, then φ =, implying tanφ =. et the corresponding angular frequency be ω ; we then obtain Therefore the capacitance is ω = ω. (.9.9) = ω. (.9.3) (e) From (d), when both switches are opened, the circuit is at resonance with =. Thus, the impedance of the circuit becomes Z = + ( ) =. (.9.3) (f) The electric energy stored in the capacitor is It attains maximum when the current is at its maximum I, ( ) U E = V = I. (.9.3) V V,max = = = ω U I, (.9.33) where we have used ω = /. (g) The maximum energy stored in the inductor is given by U V = =. (.9.34),max I -34

35 (h) If the frequency of the voltage source is doubled, i.e., ω = ω = /, then the phase constant becomes ω / ω φ = tan = ( / ) ( / ) 3 tan = tan. (.9.35) (i) If the inductive reactance is one-half the capacitive reactance, = ω = ω. (.9.36) This occurs when eh angular frequency is ω ω = =. (.9.37).9.6 Filter The circuit shown in Figure.9.4 represents a filter. Figure.9.4 et the inductance be = 4 mh, and the input voltage V in = (.V)sin(ωt), where ω = rad/s. (a) What is the value of such that the output voltage lags behind the input voltage by 3.? (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 were switched, would the circuit be a high-pass or a low-pass filter? -35

36 Solutions: (a) Because the output voltage V out is measured across the resistor, it is in phase with the current. Therefore the phase difference φ between V out and V in is equal to the phase constant and satisfies Thus, we have (b) The ratio is given by tanφ = V V = I I = ω. (.9.38) = ω tanφ = (rad/s)(.4h) = 39Ω. (.9.39) tan 3. V V = = = cosφ = cos 3. =.866. out Vin Vin + (.9.4) The circuit is a low-pass filter, since the ratio Vout / Vin decreases with increasing ω. (c) In this case, the circuit diagram is Figure.9.5 high-pass filter The ratio of the output voltage to the input voltage would be = = = = + V in Vin + + ω ω Vout V ω /. The circuit is a high-pass filter, since the ratio Vout / Vin approaches one in the limit as ω >> /. -36

37 . onceptual Questions. 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 a series circuit. Figure.. (a) Is the driving frequency above or below the resonant frequency? (b) Draw the phasor V associated with the amplitude of the applied voltage. (c) Give an estimate of the phase constant φ between the applied A voltage and the current. 4. How does the power factor in a circuit change with resistance, inductance, and capacitance? 5. an a battery be used as the primary voltage source in a transformer? 6. If the power factor in a circuit is cosφ = /, can you tell whether the current leading or lagging the voltage? Explain. -37

38 . 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 V (t) = V sin(ωt) with V = 3 V. What is the amplitude I of the resulting alternating current if the angular frequency ω is (i) rad/s, and (ii) rad/s? (a) (b) Figure.. (a) A purely capacitive circuit, and (b) a purely inductive circuit. (b) A 45-mH inductor is connected, as shown in Figure..(b), to an A generator V (t) = V sin(ωt) with V = 3 V. The inductor has a reactance = 3 Ω. (i) What is the applied angular frequency ω and, (ii) the applied frequency f, in order that = 3 Ω? (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 V (t) = V sin(ωt). This emf drives current I( t) = I sin( ωt φ) through the circuit at angular frequency ω. -38

39 Figure.. (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 constant φ (the phase difference between V ( t ) and I( t ) ) at this resonant angular frequency? (c) Suppose the angular frequency ω is increased from the resonance value until the amplitude I of the current decreases from I max to I / max. What is new value of the phase difference φ between the emf and the current? Does the current lead or lag the emf?..3 ircuit 3 A series circuit with = 4. Ω and =.4 µ F is connected to an A voltage source V (t) = ( V)sin(ωt), with ω = rad/s. (a) What is the rms current in the circuit? (b) What is the phase difference 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. -39

40 ..4 Black Box An A voltage source is connected to a black box which contains a circuit, as shown in Figure..3. Figure..3 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: V (t) = (8 V)sin(ωt), and I(t) = (.6 A)sin(ωt + 45 ). (a) Does the current lead or lag the voltage? (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...5 Parallel ircuit onsider the parallel circuit shown in Figure..4. Figure..4 Parallel circuit -4

41 The A voltage source is V (t) = V sin(ωt). (a) Find the current across the resistor. (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 difference between the current and the voltage?..6 Parallel ircuit onsider the parallel circuit shown in Figure..5. Figure..5 Parallel circuit The A voltage source is V (t) = V sin(ωt). (a) Find the current across the resistor. (b) Find the current across the capacitor. (c) What is the magnitude of the total current? (d) Find the impedance of the circuit. (e) What is the phase difference between the current and the voltage?..7 Power Dissipation A series circuit with =. Ω, = 4 mh, and =. µ F is connected to an A voltage source V (t) = V sin(ωt) that has amplitude V = V. -4

42 (a) What is the resonant angular frequency ω? (b) Find the rms current at resonance. (c) et the driving angular frequency be ω = 4 rad/s. ompute,, Z, and φ...8 FM Antenna 6 An FM antenna circuit (shown in Figure..6) has an inductance = H, a capacitance = F, and a resistance = Ω. A radio signal induces a sinusoidally alternating emf in the antenna with amplitude 5 V. Figure..6 (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 qual 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 angular frequency. (d) What is the amplitude of the potential difference across the capacitor at this in tune angular frequency?..9 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 station that broadcasts at 89.5MHz. 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 times less at 89.5MHz than at -4

43 89.7 MHz. You cannot avoid having a resistance of =. Ω considerations also dictate that you use the minimum possible., and practical (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 constant for the input signal at 89.5MHz? (i) What is the time-averaged power for the input signal at 89.5MHz? (j) Is the circuit capacitive or inductive at 89.5MHz? -43

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