Electromagnetic Oscillations and Currents. March 23, 2014 Chapter 30 1

Transcription

1 Electromagnetic Oscillations and Currents March 23, 2014 Chapter 30 1

2 Driven LC Circuit! The voltage V can be thought of as the projection of the vertical axis of the phasor V m representing the time-varying emf in the circuit! The voltage phasors must sum as vectors! Impedance Z = 2 + ( X L X C ) 2! The impedance of a circuit depends on the frequency of the time-varying emf! Impedance of an AC circuit has unit Ω March 23, 2014 Chapter 30 2

3 ! Time-varying emf esistance and eactance V emf! Time-varying emf V with resistor = V max sinωt i = V = I sinωt esistance! Time-varying emf V C with capacitor X C = 1 ωc i C = V C X C sin(ωt + 90 ) Capacitive eactance X C! Time-varying emf V L with inductor X L = ω L i L = V L sin( ωt 90 ) X L Inductive eactance X L October 22, 2008 Physics for Scientists & Engineers 2, Lecture 33 4

4 Driven LC Circuit! Phase constant ϕ is the phase difference between the voltage phasors V and V m φ = tan 1 V L V C V! The current is then ( ) i = I m sin ωt φ = tan 1! emember that the voltage across all components is V = V emf ( t) = V m sin( ωt) X L X C March 23, 2014 Chapter 30 5

5 Driven LC Circuit! Thus we have three conditions for an alternating current circuit For X L > X C, ϕ is positive, and the current in the circuit will lag behind the voltage in the circuit This circuit will be similar to a circuit with only an inductor, except that the phase constant is not necessarily π/2 For X L < X C, ϕ is negative, and the current in the circuit will lead the voltage in the circuit This circuit will be similar to a circuit with only a capacitor, except that the phase constant is not necessarily -π/2 For X L = X C, ϕ is zero, and the current in the circuit will be in phase with the voltage in the circuit This circuit is similar to a circuit with only a resistance When ϕ = 0 we say that the circuit is in resonance March 23, 2014 Chapter 30 6

6 Driven LC Circuit X L > X C Inductive X C > X L Capacitive X L = X C esonance March 23, 2014 Chapter 30 7

7 Driven LC Circuit! The current amplitude, I m, in the series LC circuit depends on the frequency of the time-varying emf, as well as on L and C! The maximum current occurs when ωl 1 ωc = 0! This corresponds to ϕ = 0 and X L = X C! The angular frequency, ω 0, at which the maximum current occurs, is called the resonant angular frequency ω 0 = 1 LC March 23, 2014 Chapter 30 8

8 esonant Behavior of LC Circuit! The resonant behavior of an LC circuit resembles the response of a damped oscillator! Here we show the calculated maximum current as a function of the ratio of the angular frequency of the time varying emf divided by the resonant angular frequency, for a circuit with V m = 7.5 V, L = 8.2 mh, C = 100 μf, and three resistances! As the resistance is lowered, the maximum current at the resonant angular frequency increases and there is a more pronounced resonant peak March 23, 2014 Chapter 30 9

9 Energy and Power! When an LC circuit operates, some of the energy is stored in the electric field of the capacitor, some of the energy is stored in the magnetic field of the inductor, and some energy is dissipated in the form of heat in the resistor! The sum of energy stored in the capacitor and inductor does not change in steady state operation! Therefore the energy transferred from the source of emf to the circuit is transferred to the resistor! The rate at which energy is dissipated in the resistor is the power P given by P = i 2! The average power is given by P = 1 2 I 2 ( ) ( ) 2 = I sin ωt φ = I 2 2 ( ) = I 2 sin 2 ωt φ March 23, 2014 Chapter 30 10

10 Energy and Power! We define the root-mean-square (rms) current to be I rms = I 2! So we can write the average power as 2 P = I rms! We can make similar definitions for other quantities V rms = V m 2! The currents and voltages measured by a alternating current ammeter or voltmeter are rms values! For example, we normally say that the voltage in the wall socket is 110 V! This rms voltage would correspond to a maximum voltage V 156 V March 23, 2014 Chapter 30 11

11 Energy and Power! We can then re-write our formula for the current as I rms = V rms Z! Which allows us to express the average power dissipated as 2 P = I rms! We can relate the phase constant to the ratio of the maximum value of the voltage across the resistor divided by the maximum value of the time-varying emf cosφ = V V m = = V rms Z = V rms 2 + ωl 1 2 ωc I IZ I rms = Z = I rms V rms Z March 23, 2014 Chapter 30 12

12 Energy and Power! We can then express the power in the circuit as P = I rms V rms cosφ! We can see that the maximum power is dissipated when ϕ = 0! We call cosϕ the power factor! The maximum power is dissipated in the circuit when the frequency of the time-varying emf matches the resonant frequency of the circuit! We define the quality factor of a series LC circuit as Q = ω 0 L = 1 L C! The quality factor is the ratio of the total energy stored in the system divided by the energy per cycle March 23, 2014 Chapter 30 13

13 Transformers! When using or generating electrical power, high currents and low voltages are desirable for convenience and safety! When transmitting electric power, high voltages and low currents are desirable, P=VI The power loss in the transmission wires goes as P = I2 Assume we have 500 MW of power to transmit If we transmit at 750 kv, the current would be 667 A If the resistance of the power lines is 200 Ω, the power dissipated in the power lines is 89 MW 18% loss Suppose we transmit at 375 kv instead 75% loss! The ability to raise and lower alternating voltages would be very useful in everyday life March 23, 2014 Chapter 30 16

14 Transformers! To transform alternating currents and voltages from high to low one uses a transformer! A transformer consists of two sets of coils wrapped around an iron core! Consider the primary windings with N P turns connected to a source of emf V emf = V max sinωt! We can assume that the primary windings act as an inductor! The current is out of phase with the voltage and no power is delivered to the transformer! A transformer that takes voltages from lower to higher is a step-up transformer and a transformer that takes voltages from higher to lower is a step-down transformer March 23, 2014 Chapter 30 17

15 Transformers! Now consider the second coil with N S turns! The time-varying emf in the primary coil induces a time-varying magnetic field in the iron core! This core passes through the secondary coil! Thus a time-varying voltage is induced in the secondary coil described by Faraday s Law V emf = N dφ B dt! Because both the primary and secondary coils experience the same changing magnetic field we can write dφ V S = N B S dt dφ V P = N B P dt March 23, 2014 Chapter 30 18

16 Transformers! Dividing these two equations gives us V P N P = V S N S V S = V P N S N P! The transformer changes the voltage of the primary circuit to a secondary voltage, given by the ratio of the number turns in the secondary coil divided by the number of turns in the primary coil! If we now connect a resistor across the secondary windings, a current will begin to flow through the secondary coil! The power in the secondary circuit is then P S = I S V S March 23, 2014 Chapter 30 19

17 Transformers! This current will induce a time-varying magnetic field that will induce an emf in the primary coil! The emf source then will produce enough current I P to maintain the original emf! This induced current will not be 90 out of phase with the emf thus power can be transmitted to the transformer! Energy conservation tells that the power produced by the emf source in the primary coil will be transferred to the secondary coil so we can write P P = I P V P = P S = I S V S! The current in the secondary circuit is then I S = I P V P V S = I P N P N S March 23, 2014 Chapter 30 20

18 Transformers! When the secondary circuit begins to draw current, then current must be supplied to the primary circuit! We can define the current in the secondary circuit using V S = I S! We can then write the primary current as I P = N S I S = N S V S N P N P = N S N P N V S P! With an effective primary resistance of P = V P N = V P P I P N S 2 = V P N P N S! Note that these equations assume no losses in the transformers and that the load is purely resistive N P 2 1 = N S N P 2 VP March 23, 2014 Chapter 30 21