Physics 1502: Lecture 23 Today’s Agenda Announcements: –RL - RV - RLC circuits Homework 06: due next Wednesday …Homework 06: due next Wednesday … Maxwell’s equations / AC current

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

Inductors in Series and Parallel In series (like resistors) a b L2L2 L1L1 a b L eq And: a b L2L2 L1L1 a b L eq In parallel (like resistors) And finally:

Summary of E&M 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! 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! Gauss’ Law For Magnetism Faraday’s Law Ampere’s Law

Ampere’s Law is the Culprit! Gauss’ Law: 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 !

Displacement current Remember: EE I in I out changing electric flux

Maxwell’s Displacement Current Can we understand why this “displacement current” has the form it does? 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” I D =  0 (d  E /dt) = the real current I in the wire. circuit

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

L C   R 

Power Production An 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: N S N S Side View N S End View

Power Production An Application of Faraday’s Law Apply Faraday’s Law

Power Production An Application of Faraday’s Law A design schematic Water

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

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? 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! Last time we discovered that a LC circuit was a natural oscillator. L C R

AC Circuits Series LCR Statement of problem: Given  =  m sin  t, find i(t). Everything else will follow. Procedure: start with loop equation? 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. L C   R

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

Lecture 23, ACT 1a Consider a simple AC circuit with a purely resistive load. For this circuit the voltage source is  = 10V sin (2  50(Hz)t) and R = 5 . What is the average current in the circuit?   R A) 10 AB) 5 AC) 2 AD) √2 AE) 0 A

Chapter 28, ACT 1b Consider a simple AC circuit with a purely resistive load. For this circuit the voltage source is  = 10V sin (2  50(Hz)t) and R = 5 . What is the average power in the circuit?   R A) 0 W B) 20 WC) 10 WD) 10 √2 W

RMS Values 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, So Average Power is,

 C Circuit Now consider C: Loop eqn gives:    C  Voltage across C lags current through C by one-quarter cycle (90  ). Is this always true? YES 0 t x mm  m 0 t 0 0  C  m  C  m

Lecture 23, ACT 2 A circuit consisting of capacitor C and voltage source  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? (a) (b) (c) i t i t t i  t

 L Circuit Now consider L: Loop eqn gives:   Voltage across L leads current through L by one- quarter cycle (90  ).   L Yes, yes, but how to remember? 0 t x mm  m 0 t x  m  L  m  L 0 0

Phasors 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 . Recall uniform circular motion: The projections of r (on the vertical y axis) execute sinusoidal oscillation.    R: V in phase with i C: V lags i by 90  L: V leads i by 90   x y y

Phasors for L,C,R i tt  i tt  i tt   t i 0 0 i i 0 Suppose:

Lecture 23, ACT 3 A series LCR circuit driven by emf  =  0 sin  t produces a current i=i m sin(  t-  ). The phasor diagram for the current at t=0 is shown to the right. –At which of the following times is V C, the magnitude of the voltage across the capacitor, a maximum? i  t=0 (a) (b) (c) i t=0 i t=t b i t=t c

Series LCR AC Circuit Consider the circuit shown here: the loop equation gives: Here all unknowns, (i m,  ), must be found from the loop eqn; the initial conditions have been taken care of by taking the emf to be:  m sin  t. To solve this problem graphically, first write down expressions for the voltages across R,C, and L and then plot the appropriate phasor diagram. L C   R Assume a solution of the form:

Phasors: LCR Assume: From these equations, we can draw the phasor diagram to the right.  L C   R  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. Given: 

Phasors: LCR The phasor diagram has been relabeled in terms of the reactances defined from:  L C   R The unknowns (i m,  ) can now be solved for graphically since the vector sum of the voltages V L + V C + V R must sum to the driving emf .

Phasors:LCR  

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. y x  imRimR imXLimXL imXCimXC mm “Full Phasor Diagram” From this diagram, we can also create a triangle which allows us to calculate the impedance Z: 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). “ Impedance Triangle” Z |  R | X L -X C |