as10

4
Physics 15b Assignment #10 By Monday April 18, read Chapter 8 of Purcell, and, if necessary, the review of complex exponentials (complex.pdf) in the handouts section of the website. Q&A questions to be answered on the Physics 15b website before 11pm on Monday, April 18: 10QA-1. To one signifigant figure, what is the answer to problem 8.1 in Purcell? henry henry henry henry None of the above. 10QA-2. Which is the best answer to Problem 8.10 in Purcell? No. Yes, but only if . Yes, but only if Yes, but only if Yes, but only if In addition, there are some survey questions and feedback questions. 1

Upload: trip-adler

Post on 13-Nov-2014

597 views

Category:

Documents


4 download

TRANSCRIPT

Page 1: as10

Physics 15b Assignment #10

By Monday April 18, read Chapter 8 of Purcell, and, if necessary, the review of complexexponentials (complex.pdf) in the handouts section of the website.

Q&A questions to be answered on the Physics 15b website before 11pm on Monday, April18:

10QA-1. To one signifigant figure, what is the answer to problem 8.1 in Purcell?

A : 0 henry

B : 1 henry

C : 3 henry

D: 7 henry

E : None of the above.

10QA-2. Which is the best answer to Problem 8.10 in Purcell?

A : No.

B : Yes, but only if R = 0.

C : Yes, but only if L < CR2

D: Yes, but only if L = CR2

E : Yes, but only if L > CR2

In addition, there are some survey questions and feedback questions.

1

Page 2: as10

Problems due at the beginning of class on Thursday, April 21 —

10.1-a. Do problem 8.4 in Purcell.

In the resonant circuit of the figure the dissipative element is a resistor R0 connectedin parallel, rather than in series, with the LC combination. Work out the equation,analogous to Eq. 2, which applies to this circuit. Find also the conditions on thesolution analogous to those that hold in the series RLC circuit. If a series RLC and aparallel R0LC have the same L, C, and Q, how must R0 be related to R?

Note that Eq. 2 is the differential equation

d2

dt2V +

RLddtV +

1LC

V = 0 (2)

The parallel circuit (p) looks like

�������������������������������������������������������������������������������������������������������������������������������������������������������� C

..................................................................................

.................................................................................. ................................................. .................................................L

................................................................................................................................................................................................................................................R0

.................

.................

.................

.................

.................

.......................................................................................................................................................................................................................................................................................................................................................................................................................................................................

............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

...................................................

.................

.................

.................

(p)

and the series circuit (s) looks like this

�������������������������������������������������������������������������������������������������������������������������������������������������������� C

..................................................................................

.................................................................................. ................................................. .................................................L

................................................................................................................................................................................................................................................R

.................

.................

.................

.................

.................

.......................................................................................................................................................................................................................................................................................................................................................................................................................................................................

..........................................................................................................................................................................................................

...............................................................................................................................................................................................................................................

(s)

10.1-b. Now do the problem in a different way. Even though there is no alternating externalvoltage in the circuits (p) and (s), we can still use the idea of an effective impedance to get theresult more easily. We simply consider the situation in which all voltages and currents in thesystem are proportional to ei!t. We know from the outset that when we do this, ! will have a non-zero imaginary part, because the system is damped. Find the complex impedance Z (for angularfrequency !) for an inductor L, a capacitor C and a resistor R, connected in series. Then find thecomplex impedance Z 0 for an inductor L, a capacitor C and a resistor R0, connected in parallel.Then looking at the series RLC circuit (s) below and at the parallel R0LC circuit (p) above, showhow to use the relation between voltage drop and current in a circuit element to obtain the resultyou got in part a. Hint: If you get stuck, you might start by computing the impedance Z for theseries circuit elements and the admittance Y 0 for the parallel circuit elements. That might give youa useful clue.

2

Page 3: as10

10.2. Do problem 8.15 in Purcell.

Show that the impedance Z at the terminals of each of the two circuits below is

5000 + 16� 10�3!2 � 16i!1 + 16� 10�6!2

Since they present, at any frequency, the identical impedance, the two black boxesare completely equivalent and indistinguishable from the outside. See if you can dis-cover the general rules for constructing the box on the right, given the values of theresistances and capacitance in the box on the left.

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................����������������������������������������������������������������������������

����������������������������������������������������������������������������

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

..............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

1000 ohms

4000 ohms

1 microfarad

Z

........................................................................................

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........................................................................................................................................................................................................................................................................................................................................................................................................

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........................................................................................................................................................

.............................................................................................................................................................................................................................................................................................................................................................................................................................................����������������������������������������������������������������������������

����������������������������������������������������������������������������

........................................................................................

5000 ohms

1250 ohms0.64 microfarads

ZHint: Note by ”See if you can discover the general rules : : :” Purcell is just asking you to solvethe problem symbolically. Thus you should consider the two circuits with general circuit elements(defined below) and find the condition that they be equivalent in any larger circuit.

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................����������������������������������������������������������������������������

����������������������������������������������������������������������������

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

..............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

R1

R01

C1

Z

........................................................................................

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........................................................................................................................................................................................................................................................................................................................................................................................................

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........

........................................................................................................................................................

.............................................................................................................................................................................................................................................................................................................................................................................................................................................����������������������������������������������������������������������������

����������������������������������������������������������������������������

........................................................................................

R02

R2C2

Z

10.3. Here is an interesting AC circuit. Consider a rectangular parallel plate capacitor with thecorners of the plates at

(�a=2;�b=2; s=2) and (�a=2;�b=2;�s=2)

3

Page 4: as10

with the separation between the plates satisfying s� a ; b. Suppose that the surface charge densityon the upper and lower plates is given by

�� sin(�x=a) cos!t

10.3.a. Find the surface current density associated with the movement of this charge. Hint:How can you satisfy charge conservation?

10.3.b. Use the law of induction to find !. Hint: Use the standard approximation for dealingwith a parallel plate capacitor to find the electric and magnetic fields between the plates.

10.3.c. Find the energy stored in the electric field produced by the charge density.

10.3.d. Find the energy stored in the magnetic field produced by this current density.

10.3.e. Compute~r� ~B and

@@t~E

inside the capacitor and find the relation between the two. What is wrong with the arguments yousaw in Chapter 6 that the curl of the magnetic field is related to the current density?

4