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Unformatted text preview: OLD DOMINION UNIVERSITY PHYSICS 227N/232N, University Physics Spring 2013 Third Examination 8:30 am — 10:20 am Thursday 07 March 2013
Room 142, OCNPS PROFESSOR: Dr. D. C. Cook Answer all 4 questions: Each question is worth different marks.
Full marks will only be given if complete working is shown. All questions should be answered on the examination paper. SI units need to be provided
for all answers.
If more space is required, attach sheets to paper. Text book may not be used. This examination will be 15% of the total course assessment. Honor Pledge Signature Candidate's Nam / I CONSTANTS AND EQUATIONS eo=8.8510’”C2N"m'2 1nC =10"9 C 1pC=10'6 c
k =lr’(41u:o)=9.0109Nm2C'2 Q=cv e= 1.6x 10"9C
Fe=(k q1q2/r2) f V=IR K=‘/2mv2
E=(kq1/r2)f P=VI UE=qV U =kChCI2 /r R=pllA V =kq2/r Volume of sphere: 3 7H3 Circumference of a circle = 27K 2
Surface area of sphere = 4 71:1‘2 Area of a circle = 7121‘ z ’Isz/ 4 (J 1. CAPACITORS (Total 25 marks) Figure 1 shows a network of capacitors in a circuit. The potential difference across terminals ab is
Vab = 220.00V. The values of the capacitances are C1 = C5 = 8.40uF and C2 = C3 = C4 z" (a) Label the value of each capacitor on Figure l. %
<K. M/uc ‘ (L/M .
1 \ U E
E i a W
b c C If 1 ($111
w .... m _ __ . 7'“ 1 l
56 W Li M 5
Figure 1. /: [(3 K/t
(b) What is the equivalent capacitance of the network? Include diagrams be each stage as you reduce the circuit to one equivalent capacitor. «is. ./ Question 1 continued next page. yéz /
(c) Calculate the charge on each capacitor and potential difference across each capacitor. ow
working below. ANSWERSrﬁ = a {
Q L 79;.
b2 Q: = 2. RESISTORS (Total 25 marks) Figure 2 shows a network of resistors in a circuit containing a 30.0 V battery. The fact that the 20.09
resistor is shown to be in a container of water does not affect this problem. 10.0 D. 10.0 (I 30.0 V 5.0 ﬂ {WI'Ithl— .. mum—w" Figure 2.
(a) Calculate the equivalent resistance of the circuit. Include diagrams below for each stage as
you reduce the circuit to one equivalent resistor. ANSWER: u '33» / .' Question 2 continued next page. (7W 3. RESISTANCE and RESISTIVITY (TOTAL 20 Marks) A heater element for a hot water heater is a resistor required to dissipate 250.00 W of power
with an applied voltage of 125.00 V across it. The resistor is to be wound from a spool containing nichrome wire which has a circular crosssection of diameter 1.50 mm d a
resistivity of 1.10 x 10'6 Q.m.
(a) Calculate what length of wire is needed for the heater. pg“ ANSWER (00. m
Zng: \Z§/<11' J77; :\:
Qw—cnm'“ “i : 13:!) (1.5mm 7“ j‘ WO'AM (b) Another identical heater is needed but this time the nichrome wire available has a
diameter of 0.75 mm. Calculate the length of wire needed. ' ' ' ' ' \ ANSWER (Lie: (ham 7'; 5 \95" \ / 'l End question 3. 4. CIRCUIT ANALYSIS (Total 30 marks) ‘ I In the circuit shown in Figure 3, use KirchofPs Laws to calculate: Manama=— rr. mummun" Figure 3 (a) The current ﬂowing through each resistor, (R1 = 2.00 Q, R2 = 4.00 £2, R3 = . Label our currents on the circuit and Show workin below. IR] = 1R2 Question 4 continued next page. (b) The power dissipated by resistors R1 and R2 and the 36.0 V battery 3 ANSWERS: pm:
gram 5 P =
my _
E0 :2, “ 1 PM:
7 rpm 7 “ND/31 (L Q
[ASL / \
(PW? (H “32 7/“— \I
Y4 w T:
2 Q
\1 7; aw, 20W NV
M Q/ /’
EndQ4. End Examination 3 ...
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 Spring '14
 ALEXANDERL.GODUNOV
 Physics

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