EM Oscillations. David J. Starling Penn State Hazleton PHYS 212

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1 I ve got an oscillating fan at my house. The fan goes back and forth. It looks like the fan is saying No. So I like to ask it questions that a fan would say no to. Do you keep my hair in place? Do you keep my documents in order? Do you have 3 settings? Liar! My fan... lied to me. Now I will pull the pin up. Now you ain t sayin [nothin ]. -Mitch Hedberg David J. Starling Penn State Hazleton PHYS 212

2 We have seen circuits that grow/decay exponentially: RC circuit: q(t) = CV(1 e t/τ C) or q(t) = CVe t/τ C RL circuit: i(t) = (V/R)(1 e t/τ L ) or i(t) = (V/R)e t/τ L But what if we put a charged capacitor together with an inductor? Here, there is no energy dissipation!

3 Capacitors and inductors store energy: U E = q2 2C U B = Li2 2 P = iv = i(ldi/dt) U B = Pdt = Li 2 /2 Initially, all the energy is stored in the capacitor and no current flows Eventually, the capacitor discharges and current flows The energy is transfered from the capacitor to the inductor... and then back!

4

5 We can measure the voltage across the capacitor V = q/c the current in the circuit i We find: The voltage lags by 90

6 An analogy: block on a spring Potential energy is like the energy in a capacitor U U E Kinetic energy is like the energy in the inductor K U B ω = k/m 1/LC

7 Let s look more closely at this LC circuit: The total energy: U = Li2 2 + q2 2C = const Therefore, du dt d 2 q dt 2 + q LC = 0 = d ( ) Li 2 dt 2 + q2 2C = Li di dt + q dq C dt = L dq d 2 q dt dt 2 + q dq C dt = 0 Solution: q(t) = Q cos(ωt + φ), i(t) = I sin(ωt + φ) with I = ωq and ω = 1/LC

8 For an LC circuit: q(t) = Q cos(ωt + φ) (1) i(t) = I sin(ωt + φ) (2) Therefore, the energy stored is: U E = q2 2C = Q2 2C cos2 (ωt + φ) U B = Li2 2 = 1 2 LI2 sin 2 (ωt + φ) = Q2 2C sin2 (ωt + φ) [note: I 2 = (ωq) 2 = Q 2 /LC] Adding them up, we get U = U E + U B = Q2 2C [sin2 (ωt + φ) + cos 2 (ωt + φ)] = Q2 2C

9 For an LC circuit: q(t) = Q cos(ωt + φ) i(t) = I sin(ωt + φ) U E = Q2 2C cos2 (ωt + φ) U B = Q2 2C sin2 (ωt + φ) U total = Q2 2C

10 What if we throw a resistor in the mix? How does the previous analysis change? Energy is no longer constant, but decreases over time The power dissipated by a resistor is just P = iv = i 2 R Therefore, du dt = Li di dt + q dq C dt = i2 R (3)

11 The equation for an RLC circuit: Li di dt + q dq C dt = i2 R Li di dt + q C i = i2 R L d2 q dt 2 + q C + ir = 0 L d2 q dt 2 + Rdq dt + q C = 0 The solution: q(t) = Qe Rt/2L cos(ω t + φ) Frequency: ω = ω 2 (R/2L) 2 ω = 1/LC

12 For an RLC circuit: q(t) = Qe Rt/2L cos(ω t + φ) U E (t) = q2 2C = Q2 2C e Rt/L cos 2 (ω t + φ) The energy in the circuit dissipates over a time with a time constant τ L = L/R

13 Lecture Question 15.2 The current in an oscillating LC circuit is zero. Which one of the following statements is true? (a) The charge on the capacitor is equal to zero coulombs. (b) Charge is moving through the inductor. (c) The energy is equally shared between the electric and magnetic fields. (d) The energy in the electric field is maximized. (e) The energy in the magnetic field is maximized.

14 Generators create an oscillating emf: As the flux varies, so does the enduced emf E E(t) = E sin(ω d t) Therefore, i(t) = I sin(ω d t φ)

15 A general RLC circuit with an oscillating emf: The oscillations in the circuit are at frequency ω d This is true even if ω d ω The current moves back and forth through each element

16 Resistive load: For this case, E ir = 0 i = E/R = E m sin(ω d t)/r In general, we expect i(t) = I sin(ω d t φ) For a resistive load, there is no phase delay (φ = 0 )

17 We can graph the applied voltage and the resulting current: On the right, we have a phasor diagram The angle of the phasor gives the argument of the sine The length of the phasor is the max value of v or i The projection onto the y-axis gives the instantaneous v or i

18 Capacitive load: For this case, q = Cv = CV sin(ω d t) i = dq/dt = ω d CV cos(ω d t) = (V/X C ) cos(ω d t) The Capacitive Reactance: X C = 1/ω d C X C has units of ohms and V = IX C.

19 We can graph the applied voltage and the resulting current: On the right, we have a phasor diagram Notice how the current leads the voltage by 90 q = CV sin(ω d t) i = (V/X C ) sin(ω d t + 90 )

20 Inductive load: For this case, v = Ldi/dt = V sin(ω d t) di dt = (V/L) sin(ω dt)

21 Solving for i(t), di = (V/L) sin(ω d t)dt i(t) = V sin(ω d t)dt L = V ω d L cos(ω dt) = V X L sin(ω d t 90 ) The Inductive Reactance: X L = ω d L X L has units of ohms and V = IX L.

22 We can graph the applied voltage and the resulting current: On the right, we have a phasor diagram Notice how the current lags the voltage by 90 v = V sin(ω d t) i = (V/X L ) sin(ω d t 90 )

23 Summary of Simple Circuits

24 Putting them together: Each component operates as before: resistor: i(t) and v R (t) are in phase capacitor: i(t) leads v C (t) by 90 inductor: i(t) lags v L (t) by 90

25 The result: E is equal to the vector sum of the three potential differences

26 The vectors: E 2 m = V 2 R + (V L V C ) 2 = I 2 [R 2 + (X L X C ) 2 I = E m R 2 + (X L X C ) 2 I = E m /Z Impedence: Z = R 2 + (X L X C ) 2

27 We can now find the max current. But what is the phase between the applied voltage and the current? tan(φ) = V L V C V R = IX L IX C IR = X L X C R

28 tan(φ) = X L X C R

29 Notice that the max current depends on the impedence: I = E m R 2 + (X L X C ) 2 = E m /Z For a fixed R, I is max if X L = X C, so ω d L = 1/ω d C ω d = 1/ LC But if X L = X C, then φ = 0. This is called resonance!

30

31 Lecture Question 15.4 An inductor circuit operates at a frequency f = 120 Hz. The peak voltage is 120 V and the peak current through the inductor is 2.0 A. What is the inductance of the inductor in the circuit? (a) H (b) H (c) H (d) H (e) H

32 The power dissipated in the RLC circuit leaves through the resistor only: P = i 2 R = [I sin(ω d t φ)] 2 R = I 2 R sin 2 (ω d t φ) The average power is just P avg = I 2 R/2 Define: I rms = I/ 2 (root-mean-squared) This gives P avg = I 2 rmsr

33 A transformer changes one AC voltage to another AC voltage: A primary coil is wrapped around an iron core An emf is induced in a secondary coil also on the core The relationship between the two voltages is V s = V p N s N p

34 V s = V p N s N p Step-up transformer: N s > N p Step-down transformer: N p > N s Energy is conserved: I p V p = I s V s I s = I p N p /N s

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