PHYSICS DEPARTMENT
PHY 2054
2nd Exam (SF 18–21) 11 July 2006
Name (print, last first):
SOLUTIONS
Signature:
On my honor, I have neither given nor received unauthorized aid on this examination.
YOUR TEST NUMBER IS THE 5DIGIT NUMBER AT THE TOP OF EACH PAGE.
(1)
Code your test number on your answer sheet (use 76–80 for the 5digit number). Code your name on your answer sheet.
DARKEN CIRCLES COMPLETELY. Code your UFID number on your answer sheet.
1
(2) Print your name on this sheet and sign it also.
2
(3) Do all scratch work anywhere on this exam that you like.
Circle your answers on the test form. At the end of the
test, this exam printout is to be turned in. No credit will be given without both answer sheet and printout.
(4)
Blacken the circle of your intended answer completely, using a #2 pencil or blue
or black
ink. Do not make any stray marks
or some answers may be counted as incorrect.
(5) The answers are rounded o
ff
. Choose the closest to exact.
There is no penalty for guessing. If you believe that no correct
answer is listed, leave this item blank!!
(6)
Hand in the answer sheet separately.
Useful (??) Constants:
µ
0 = 4π × 10
−
7
Tm/A
�
0 = 8
.
85 × 10
−
12
C
2
/(Nm
2)
electron charge =
−
1
.
6 × 10
−
19C
electron mass = 9
.
11 × 10
−
31
kg
V=volt
N=newton
J=joule
m=meter
W=watt
µ =
“micro” = 10
−
6
A = ampere
“pico” = 10
−
12
n =
“nano” = 10
−
9
m =
“milli” = 10
−
3
proton mass = 1
.
67 × 10
−
27
kg
c = 3 × 108 m/s
Answer (1) is the correct answer.
The (brief) solution is in bold italics
like this.
[NOTE: ‘RHR’ = “Right Hand Rule”]
1
. Immediately after switch S in the circuit shown is closed, the current through the battery
shown is:
At t=0 , L acts like an open circuit (due to the initial back emf)
so initial
current flows
only through R1 and R2
(1)
Vo/
(
R
1 +
R
2)
(2)
Vo/R
1
(3)
Vo/R
2
(4) 0
2.
In the diagrams, all light bulbs are identical and all emf devices are identical. In
which circuit (I, II, III, IV, V) will the bulbs be dimmest?
(1) IV
(2) I
(3) II
(4) III
(IV) has 2 bulbs in series across a single emf source, hence the lowest current => dimmest bulbs.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
3.
In the circuit shown, the capacitor is initially uncharged and V =
9.0 volts. At time t = 0, switch S is closed. If τ denotes the time
constant, the approximate current through the 3Ω resistor when
t =
τ/
100 is:
(1)1A
(2)1/2A
(3)3/4A
(4)3/8A
(5)3/2A
i= V/R e
t/τ
but t/τ =
(τ/100)/ τ = 1/100
so I = V/R e
100
)
∼
= V/R = 9V/(6 Ω
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Avery
 Physics, Magnetic Field

Click to edit the document details