Chapter 33 - Electromagnetic Oscillations and Alternating Current

Transcription

1 hapter 33 - Electromagnetic Oscillations and Alternating urrent Problem Set # - due: h 33-3, 0, 7, 8, 3, 34, 39, 4, 56, 59, 6, 77, 80, 84, 85, 87 ecture Outline The ircuit The ircuit 3 Basic A ircuits 4 Frequency Filtering ircuits 5 The Tuning ircuit 6 Power in A ircuits 7 Transformers We will use the loop and junction theorems to analyze the behavior of circuits that exhibit oscillatory behavior We will start by allowing the circuits to oscillate at their natural frequency, then we will examine the effect of forcing them to oscillate at other frequencies The ircuit ε a b c In the circuit at the left, the capacitor is charged when the switch is connected from a to b Then the switch is connected from b to c The capacitor begins to lose its charge as it attempts to get current through the inductor The inductor begins to build up a magnetic field This field peaks as the charge on the capacitor goes to zero Then the energy in the inductor begins to create current to charge the capacitor back up All of this is explained by the loop theorem which requires that the potential difference across the capacitor always equal the potential difference across the inductor, V c = V Q di = Q = d Q d Q = Q This differential equation is the SHM equation, d x = k m x = ω x x = Acos( +δ) By analogy, Q = Q m cos( +δ) where ω = The ircuit Example : Show that the above expression for the charge on the capacitor satisfies the loop theorem Q = Q m cos( +δ) dq = Q m ωsin( +δ) d Q = Q m ω cos( +δ) = ω Q Substituting into the loop theorem, ω Q / = Q / ω = 33-

2 Example : For an circuit find (a)the peak current in terms of the peak charge on the capacitor and (b)the fraction of the period between the time the capacitor's charge peaks and the current peaks (a)from the definition of current, I dq = Q m ω sin( +δ) = Q m ω (b)first let's define Q=Q m when t=0 This establishes the value of δ, Q = Q m cos( +δ) Q m = Q m cos(δ) cos(δ) = δ = 0 Note that I = when sin( +δ) = +δ = π = π t = π ecall, ω = π T t = The ircuit π π T t = 4 T ω Q m Q ε a b c The capacitor is charged by connecting a to b Then the switch is moved from b to c The oscillations of the circuit are damped out by the heat created in the resistor As usual, the loop theorem gives the details, V + V + V = 0 di + I + Q = 0 d Q + dq + Q = 0 If we assume that is fairly small then we would expect damped oscillations of the form, Q = Q m e γ t cosω d t where Q m is the initial charge on the capacitor, α is the damping constant, and ω d is the frequency of oscillation A graph of t such a curve is shown at the left To check to see if this guess solves the differential equation from the loop theorem, we need the derivatives, dq = Q m e γt ω d sinω d t +γ cosω d t and d Q = Q m e γt {( ω d γ )cos ω d t γω d sin ω d t} { } Substituting into the loop theorem, Q m e γt { ( ω d γ )cosω d t γω d sinω d t} + { ω d sinω d t + γ cosω d t} cosω t d = 0 Sorting things out, ( ω d γ ) + γ cosω t + ω d { γω d d }sinω d t = 0 If this is to be zero for all values of t, then the coefficients of the sine and cosine must be zero independently The sine coefficient gives, ω d γω d = 0 γ = 33-

3 The cosine coefficient requires, ( ω d γ ) + γ = 0 ω d γ Solving for ω d, ω d = γ = ( ) + γ Physics 4B ecture Notes = 0 ω d γ + γ = 0 ω d = Q = Q m e t cosω d t where ω d = The ircuit Example 3: Find the maximum allowed resistance for oscillations to occur Oscillations will occur as long as ω d is real, so the critical resistance is when, = c c = 4 Example 4: Find the fraction of energy lost per oscillation for an circuit with small At some initial time t the charge is, Q o = Q m e t cosω d t We want to choose this time so that the charge on the capacitor is peaking so, t n π where n is an integer The initial charge is, Q o = Q m e n π ω d The energy stored in the capacitor at this time is, U o = After one oscillation, t ( n +) π, Q = Q ω m e d The fraction of energy lost in one oscillation is, U = U o U = U o U o Q m e n π Q m Q m e e n π Q o ( n+) π ( n+) π For small use the Taylor expansion, e x = x + x = Q m e, and U = = e π n π Q m e ( n +) π U π = π U o ω d Now we define the "Q-factor" or "Quality" of the circuit as U U o π Q Q = ω d Note: high Q means low loss and low Q means high loss 33-3

4 3 Basic A icuits Instead of allowing circuits to oscillate at the frequency they choose, we can force them to oscillate at any frequency we choose by using an A power supply The response of the circuit depends on whether it contains resistors, capacitors, or inductors esistors V,max= i,max i V ε= sin V,max = i,max = V i t The circuit at the left shows a sinusoidally varying power supply connected to a resistor The resulting current in the circuit is given by Ohm's ule, V = i i = V By the loop theorem, the voltage on the resistor must always ε equal the voltage from the power supply so, m i = sin = sin The maximum current through the resistor is, i,max = Note that both the current and the voltage vary with time as the sin We say they are "in phase" This is shown in the graph at the left The sketch at the left is called a "phasor diagram" Vectors representing the maximum voltage and maximum current rotate counter-clockwise and make at angle with the positive x-axis The vertical component of each vector represents the instantaneous value of voltage or current The ratio of the maximum voltage to the maximum current is defined to be the "impedance" Χ The Definition of Impedance For a resistor, Χ Χ = εm = Χ = The Impedance of a esistor 33-4

5 apacitors ε= sin Physics 4B ecture Notes The circuit at the left shows a sinusoidally varying power supply connected to a capacitor The resulting current in the circuit can be found from the definition of capacitance, V,max= i,max i,max = ω i V V,max= Χ = ω V t i Q = V and the definition of current i = dq By the loop theorem, the voltage on the capacitor must always equal the voltage from the power supply so, i = d V ( ) = d ( sin) = ω cos The maximum current through the capacitor is, i,max = ω Note that the current and the voltage are out of phase The current peaks at an earlier time than the voltage This is shown in the graph at the left The phasor diagram for this circuit is shown at the left The current leads the voltage by 90 The ratio of the maximum voltage to the maximum current is defined to be the impedance, Χ Χ = ω = ω The Impedance of a apacitor Inductors V,max = i,max ε= sin i V t The circuit at the left shows a sinusoidally varying power supply connected to an inductor The resulting current in the circuit can be found from the definition of self-inductance V = di By the loop theorem, the voltage on the inductor must always equal the voltage from the power supply so, sin = di ε m i = cos The ω maximum current through the inductor is, i,max = Note that the current and the voltage ω are again out of phase The current peaks at a later time than the voltage This is shown in the graph at the left 33-5

6 The phasor diagram for this circuit is also shown at the left The current lags V V,max = the voltage by 90 The ratio of the maximum voltage to the maximum current is defined to be the impedance, Χ Χ = εm ω = ω i,max = ω Χ = ω The Impedance of an Inductor i Now that we see that the impedance of inductors and capacitors depends on the frequency of the power source we can begin to design circuits that have a frequency dependence such as crossovers and tuners The concept of phasor diagrams cam be extended to more complex circuits 4 Frequency Filtering ircuits Χ = ω Χ = ω Χ = Inductors stop high frequency and pass low frequency apacitors stop low frequency and pass high frequency esistors don't care about frequency Example 5: A stereo signal contains frequencies from 00Hz to 00kHz A tweeter has a resistance of 800Ω and functions well between 500kHz and 00kHz (a)design a circuit which will use an inductor to filter out the low frequencies (b)find the inductance so that it has the same peak current as the tweeter at 500kHz (c)find the ratio of the peak currents at 00kHz, (d)at 00kHz (e)include the woofer in the circuit (a)the power supply will send a variety of frequencies toward the speaker Putting an inductor across the speaker allows the low frequencies to bypass the tweeter (b)the voltage across the inductor must always equal the voltage across the tweeter according to the loop theorem so, = i,max Χ = i,max Χ At 500kHz the peak current is the same in both the resistor and the inductor so, i,max = i,max which means that Χ = Χ ω = = ω = πf = 800 π(5000) = 055mH (c)it's still true that = i,max Χ = i,max Χ i,max i,max = Χ = πf Χ (d)again, = i,max Χ = i,max Χ i,max i,max = π(0)(055) 800 = Χ = πf Χ stereo = 00 = π()(055) 800 = 000 tweeter =800Ω 33-6

7 (e)the woofer should go in series with the inductor since the low frequencies prefer this leg of the circuit Usually a capacitor is put in the leg with the tweeter to adjust the frequency that produces equal currents This type of circuit is called a "crossover" stereo woofer tweeter =800Ω =800Ω 5 The Tuning ircuit i (t) In the series circuit, the inductor stops the high frequencies and the capacitor stops the low frequencies Only frequencies in between can make it through the circuit et's find out i(t) i (t) the relationship between the peak current and the frequency The junction theorem requires the current at any time to be equal in all the circuit elements, i(t) = i (t) = i (t) = i (t) The loop theorem i (t) demands that the sum of the voltage drops at any time add up to the supply voltage, ε(t) = V (t) + V (t) + V (t) V,max εm φ V,max - V,max V,max i max = Z Since the voltages are out of phase we need to add up the phasors According to the phasor diagram, = V,max + ( V,max V,max ) Using the definition of impedance, = ( i max ) + ( i max Χ i max Χ ) Factoring out the peak current, V,max = i max + ( Χ Χ ) Using the definition of impedance, Χ = + ( Χ I Χ ) m Usually, the symbol Z is used for the total impedance of a circuit Z = + ( Χ Χ ) Series ircuit: Impedance The phase angle between the voltage and current can be found from the definition of tangent, tan φ = V,max V,max V,max = i max Χ i max Χ i max = Χ Χ tan φ = Χ Χ Series ircuit: Phase Angle 33-7

8 Example 6: An series circuit contains a 00Ω resistor, a 80mH inductor and a 600µF capacitor connected to a 0V-rms 600Hz power supply Find the rms current and voltage for each circuit element Since all the circuit elements are in series they have the same current To find this current we need the total impedance, Z = ( ) The impedance for each element is, =00Ω, + Χ Χ Χ = ω = πf = π(600)(080) = 679Ω and Χ = ω = πf = π(600)(600x0 6 ) = 44Ω The total impedance is, Z = (00) ( ) = 30Ω The current can be found by the definition of impedance, Χ = = ε rms I rms I rms = ε rms Z = 0 30 = 355A The rms voltage on each one can be found from their impedances: V,rms = i rms = 70V, V,rms = i rms Χ = 4V and V,rms = i rms Χ = 57V Why don't these add up to the supply voltage? These voltages don't happen at the same time How can they be greater than the supply voltage? The other voltages are negative at these times i m,max i max The current that passes through the circuit depends on the frequency This is called the "frequency response" ω o antenna ω = i max + ( Χ Χ ) i max = ( ) + ω ω The peak current is plotted against the frequency at the left Note that the capacitor cuts off the low frequencies and the inductor cuts off the high frequencies The frequency for maximum current is called the "resonance frequency" This occurs when ω ω = 0 ω o = ω o = Series ircuit: esonance Frequency The fact that only a small range of frequencies make it through this circuit make it ideal for tuning as in a radio A typical tuning circuit is shown at the left The tuning is usually accomplished by varying the capacitance 33-8

9 Example 7: Find the full wih of the resonance curve when the peak current is half its maximum value i max = i m,max Using the definition of impedance, the total impedance of the circuit and the fact that the term in parentheses is zero at resonance, ( ) = + ω ω ( ) + ω o ω o ε ( ω) = m + ω anceling the driving voltage and squaring both sides, ( ) = 4 ( ω ) = 3 ω + ω ω ω ω = 3 Assume the frequencies we want, ω, are symmetrically located about the resonance frequency, ω = ω o ± ω = ω ω o ± ω o ω = ± ω m ω o ω o ω ω o ω o Substituting back into the previous equation, ± ω ω ω o m ω o = 3 ω o ω o Since ω o = ω o ( ω o ) ± ω ω o = ω ω o, we get, 0 ± ω o ( ω o ) = 3 ω o ω o + ( ω o ) = 3 ω = ± 3 The wih of the curve only depends on and not on adios use capacitors to tune so that the stations can be equally spaced i m,max i m,max i max ω ω ω ω o ω ω = 3 Series ircuit: Full Wih Half Maximum Example 8: Design an series circuit that is an FM tuner The frequency of a typical FM station is 00MHz The spacing between stations is 000MHz Assume the circuit requires a 500mA peak current and the antenna induces a 00mV emf The resistance can be found from the relationship between the peak current and peak voltage, ε ε i max = m + ω ω = m = 00mV i m,max 500mA = 00Ω ( ) i m,max = The inductance is fixed by the spacing between stations, ω = 3 = 3 ω = 3(00) π(000x0 6 ) = 76µH The capacitance is used to tune for the desired frequency, ω o = = 4π f o = 4π (00x0 8 ) (76x0 6 ) = 0098pF 33-9

10 6 Power in A ircuits V,max εm φ V,max V,max V,max - V,max i max = Z Energy supplied to capacitors is stored in the electric field between the plates This energy will be returned in full when the capacitor discharges Energy supplied to inductors is stored in the magnetic field of the coils This energy is also returned in full when the field collapses The only place where energy is "lost" to the circuit is in the resistor where heat is produced and not returned The power consumed by a resistor is P = iv as before The only difference in an A circuit is that the current and voltage oscillate with time, P(t) = ( i max sin) ( V,max sin) = i max V,max sin So the peak power consumed is, P = i max V,max In terms of the peak voltage supplied to the circuit, P = i max cosφ The cosφ is called the "power factor" Using Ohm's ule, P = i max V,max = i max i max suspect The average power consumed is, P av = i = i rms ( ) = i max as you might P av = i rms Average Power onsumed Example 9: For the circuit of example, find the average power delivered (a)at 60Hz and (b)at the resonance frequency (a)the average power delivered will match the average power consumed Using the numbers from example, P av = i rms = (355) (00) = 5W (b)at resonance, ε ε the impedance is equal to the resistance, so the rms current is, rms I rms = Z = rms = 0 00 = 550A The average power consumed will be, P av = i rms = (550) (00) = 605W Since the current is a maximum at resonance, the power consumed also is maximum 33-0

11 7 Transformers Transformers are used to convert current to voltage or vice-versa V p Ip N p N s I s Vs Assuming no losses to heat the aw of onservation of Energy requires that the power that comes in must equal the power that goes out, I p V p = I s V s The changing current in the primary coils creates a changing magnetic field in the iron This changing B magnetic field through the secondary coils creates an induced voltage according to Faraday's aw Assuming that the iron keeps the flux the same in both coils, Φ p = Φ s dφ p = dφ s Using Faraday's aw, dφ p = dφ s V p N p = V s N s In summary, I p V p = I s V s V p N p = V s N s Transformers: onservation of Energy Equal Flux If the voltage increases we call it a "step-up" transformer, if it decreases we call it a "step-down" transformer Example 0 : Design a transformer to convert standard 0V to V for a laptop power supply The secondary side has 00 turns and delivers 00mA Find the number of turns needed in the primary and the current the primary will draw The requirement of equal flux through the primary and secondary yields, V p N p = V s N s N p = N s V p V s = 00 0 = 97turns The aw of onservation of Energy requires, V I p V p = I s V s I p = I s s = (00) V p 0 = 09mA 33-

12 hapter 33 - Summary The ircuit Q = Q m cos( +δ) where ω = The ircuit Q = Q m e t cosω d t where ω d = The Definition of Impedance Χ The Impedance of a esistor Χ = The Impedance of a apacitor Χ = ω The Impedance of an Inductor Χ = ω Series ircuit: Impedance Z = esonance Frequency ω o = + ( Χ Χ ) Phase Angle tan φ = Χ Χ Full Wih Half Maximum ω = 3 Average Power onsumed P av = i rms Transformers onservation of Energy I p V p = I s V s Equal Flux V p N p = V s N s 33-