Physics 1402: Lecture 23 Today’s Agenda

1. Announcements: RL - RV - RLC circuits Homework 06: due next Wednesday … Maxwell’s equations / AC current Physics 1402: Lecture 23Today’s Agenda

2. X X X X X X X X X Induction Self-Inductance, RL Circuits long solenoid Energy and energy density

3. In series (like resistors) a a a a L1 L2 Leq Leq L1 L2 b b b b And finally: And: Inductors in Series and Parallel • In parallel (like resistors)

4. J. C. Maxwell (~1860) summarized all of the work on electric and magnetic fields into four equations, all of which you now know. However, he realized that the equations of electricity & magnetism as then known (and now known by you) have an inconsistency related to the conservation of charge! Gauss’ Law Faraday’s Law Gauss’ Law For Magnetism Ampere’s Law Summary of E&M I don’t expect you to see that these equations are inconsistent with conservation of charge, but you should see a lack of symmetry here!

5. Gauss’ Law: ! Ampere’s Law is the Culprit! • Symmetry: both E and B obey the same kind of equation (the difference is that magnetic charge does not exist!) • Ampere’s Law and Faraday’s Law: • If Ampere’s Law were correct, the right hand side of Faraday’s Law should be equal to zero -- since no magnetic current. • Therefore(?), maybe there is a problem with Ampere’s Law. • In fact, Maxwell proposes a modification of Ampere’s Law by adding another term (the “displacement” current) to the right hand side of the equation! ie

6. Remember: Iin Iout changing electric flux Displacement current FE

9. Can we understand why this “displacement current” has the form it does? circuit Maxwell’s Displacement Current • Consider applying Ampere’s Law to the current shown in the diagram. • If the surface is chosen as 1, 2 or 4, the enclosed current = I • If the surface is chosen as 3, the enclosed current = 0! (ie there is no current between the plates of the capacitor) Big Idea: The Electric field between the plates changes in time. “displacement current” ID = e0 (dfE/dt)= the real current I in the wire.

10. Maxwell’s Equations • These equations describe all of Electricity and Magnetism. • They are consistent with modern ideas such as relativity. • They even describe light

11. R C L e ~ w AC Current

12. S S S N N N End View Side View Power ProductionAn Application of Faraday’s Law • You all know that we produce power from many sources. For example, hydroelectric power is somehow connected to the release of water from a dam. How does that work? • Let’s start by applying Faraday’s Law to the following configuration:

13. Power ProductionAn Application of Faraday’s Law • Apply Faraday’s Law

14. Water Power ProductionAn Application of Faraday’s Law • A design schematic

15. Add resistance to circuit: R C L e RLC - Circuits • Qualitatively: Oscillations created by L and C are damped (energy dissipation!) • Compare to damped oscillations in classical mechanics:

16. Answer:Yes, if we can supply energy at the rate the resistor dissipates it! How? A sinusoidally varying emf (AC generator) will sustain sinusoidal current oscillations! R + + C L - - • Last time we discovered that a LC circuit was a natural oscillator. AC Circuits - Intro • However, any real attempt to construct a LC circuit must account for the resistance of the inductor. This resistance will cause oscillations to damp out. • Question:Is there any way to modify the circuit above to sustain the oscillations without damping?

17. We could solve this equation in the same manner we did for the LCR damped circuit. Rather than slog through the algebra, we will take a different approach which uses phasors. R C L e ~ AC CircuitsSeries LCR • Statement of problem: • Given e = emsinwt , find i(t). • Everything else will follow. • Procedure: start with loop equation?

18. Before introducing phasors, per se, begin by considering simple circuits with one element (R,C, or L) in addition to the driving emf. R i R e ~ m m / R Note: this is always, always, true… always. 0 0 m m / R t 0 t 0 x eR Circuit • Begin with R: Loop eqn gives: Þ Voltage across R in phase with current through R

19. Consider a simple AC circuit with a purely resistive load. For this circuit the voltage source is e = 10V sin (2p50(Hz)t) and R = 5 W. What is the average current in the circuit? R e ~ A) 10 A B) 5 A C) 2 A D) √2 A E) 0 A Lecture 23, ACT 1a

20. Consider a simple AC circuit with a purely resistive load. For this circuit the voltage source is e = 10V sin (2p50(Hz)t) and R = 5 W. What is the average power in the circuit? R e ~ A) 0 W B) 20 W C) 10 W D) 10 √2 W Chapter 28, ACT 1b

21. Average values for I,V are not that helpful (they are zero). Thus we introduce the idea of the Root of the Mean Squared. In general, RMS Values So Average Power is,

22. Voltage across C lags current through C by one-quarter cycle (90°). C e ~ Cm m Is this always true? YES 0 0 m Cm 0 t t 0 x • Now consider C: Loop eqn gives: eC Circuit Þ Þ

23. A circuit consisting of capacitor C and voltage source e is constructed as shown. The graph shows the voltage presented to the capacitor as a function of time. Which of the following graphs best represents the time dependence of the current i in the circuit? e t i i i (c) (a) (b) t t t Lecture 23 , ACT 2

24. Voltage across L leadscurrent through L by one-quarter cycle (90°). L e ~ m m L 0 0 Yes, yes, but how to remember? m m L 0 t t 0 x x • Now consider L: Loop eqn gives: eL Circuit Þ Þ

25. A phasor is a vector whose magnitude is the maximum value of a quantity (eg V or I) and which rotates counterclockwise in a 2-d plane with angular velocity w. Recall uniform circular motion: y y • R: V in phase with i Þ Phasors • C: V lags i by 90° Þ • L: V leads i by 90° Þ The projections of r (on the vertical y axis) execute sinusoidal oscillation. w x

26. Suppose: i w i 0 wt t w i i 0 wt w i i wt 0 ß Phasors for L,C,R

27. A series LCR circuit driven by emf e = e0sinwt produces a current i=imsin(wt-f). The phasor diagram for the current at t=0 is shown to the right. At which of the following times is VC, the magnitude of the voltage across the capacitor, a maximum? t=0 f i i t=0 t=tc t=tb i i (c) (a) (b) Lecture 23, ACT 3

28. Here all unknowns, (im,f) , must be found from the loop eqn; the initial conditions have been taken care of by taking the emf to be: e = em sinwt. R C L e ~ • Assume a solution of the form: Series LCRAC Circuit • Consider the circuit shown here: the loop equation gives: • To solve this problem graphically, first write down expressions for the voltages across R,C, and L and then plot the appropriate phasor diagram.

29. From these equations, we can draw the phasor diagram to the right. • Given: R C L e ~ Þ w Phasors: LCR • Assume: Þ • This picture corresponds to a snapshot at t=0. The projections of these phasors along the vertical axis are the actual values of the voltages at the given time.

30. w R C L e ~ The unknowns (im,f) can now be solved for graphically since the vector sum of the voltages VL + VC + VR must sum to the driving emf e. Phasors: LCR • The phasor diagram has been relabeled in terms of the reactances defined from: Ohms ->

31. Phasors:LCR Þ ß

32. y x imXL em f imR Z | XL-XC| |  imXC R “Full Phasor Diagram” “ Impedance Triangle” Phasors:Tips • This phasor diagram was drawn as a snapshot of time t=0 with the voltages being given as the projections along the y-axis. • Sometimes, in working problems, it is easier to draw the diagram at a time when the current is along the x-axis (when i=0). From this diagram, we can also create a triangle which allows us to calculate the impedance Z: