TIME OF COMPLETION_______________            NAME_____SOLUTION________________________

 

 

                                           DEPARTMENT OF NATURAL SCIENCES

 

PHYS 1112, Exam 1                                                                                                      Section 1

Version 1                                                                                                            October 4, 2004

Total Weight: 100 points

 

1.         Check your examination for completeness prior to starting.  There are a total of ten (10) problems on seven (7) pages.

 

2.         Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor).

 

3.         You will have 75 minutes to complete the examination.

 

4.         The total weight of the examination is 100 points.

 

5.         There are six (6) multiple choice and four (4) calculation problems. Work five (5) multiple choice problems and all calculation problems.  Show all work; partial credit will be given for correct work shown.

 

6.         If you have any questions during the examination, see your instructor who will

be located in the classroom.

 

7.         Start:               6:00 p.m.

Stop:                7:15 p.m

 

 

 

              PROBLEM

 

                 POINTS

 

                CREDIT

 

                     1-6

 

                     30

 

 

 

                      7

 

                     20

 

 

 

8

 

                     20

 

 

 

                      9

 

                     15

 

 

 

                     10

 

                     15

 

 

 

                 TOTAL

 

                    100

 

 

 

 

 

           PERCENTAGE

 

 

 

CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT.

 

 

 

 

 

  1. Which one of the equations is the correct Kirchhoff’s junction equation for the situation pictured below?

 

    1. I1 = I2 + I3.

 

    1. I2 = I1 + I3.

(6)

    1. I3 = I1 + I2.

 

    1. I1 + I2 + I3 = 0.

 

  1. The unit of electric current, the ampere, is dimensionally equivalent to which of the following?

 

    1. Volt x Ohm.

 

    1. Volt/ Ohm.

(6)

    1. Ohm x meter.

 

    1. Volt/second.

 

 

 

 

  1. When an electric current exists within a conducting wire, which of the following statements is correct?

 

    1. Electric field inside the wire is zero.

(6)

    1. Electric field inside the wire is parallel to the current flow.

 

    1. Electric field inside the wire is anti-parallel to the current flow.

 

    1. Electric field inside the wire is perpendicular to the current flow.

 

 

 

  1. If two parallel, conducting plates have equal positive charge, the electric field lines will

 

    1. Leave one plate and go straight to the other plate.

 

    1. Leave both plates and go to infinity.

(6)

    1. Enter both plates from infinity.

 

    1. None of the above.

 

 

  1. If three 4.00 mF capacitors are connected in parallel, what is the combined capacitance?

 

    1. 12.0 mF.

 

    1. 0.750 mF.

(6)

    1. 8.00 mF.

 

    1. 0.46 mF.

 

 

  1. A repelling force must occur between two charged objects under which condition?

 

    1. Charges are of unlike signs.

 

    1. Charges are of like signs.

(6)

    1. Charges are of equal magnitude.

 

    1. Charges are of unequal magnitude. 
  1. Two charges, Q1 = 6.00 mC and Q2 = - 6.00 mC, are located as shown below.  

    1. Find the total electric field at the origin. Specify both magnitude and direction.

 

E1= ke Q1/r12= (8.99 x 109 Nm2/C2)(6.00 x 10–6C)/(0.150 m)2 = 2.40 x 106 N/C

E2 = ke Q2/r22= (8.99 x 109 Nm2/C2)( 6.00 x 10–6C)/(0.235 m)2 = 0.978 x 106 N/C

 

E1x = (2.40 x 106 N/C) cos(90o) = 0

E1y= (2.40 x 106 N/C) sin(90o) = 2.40 x 106 N/C

 

E2x = (0.978 x 106 N/C) cos (0o) = 0.978 x 106 N/C

E2y = (0.978 x 106 N/C) sin (0o) = 0

 

 

Etotx = 0 N/C + 0.978 x 106 N/C = 0.978 x 106 N/C

Etoty = 2.40 x 106 N/C + 0 N/C = 2.40 x 106 N/C

 

 

Etot =((0.978 x 106 N/C)2 + (2.40 x 106 N/C)2) 1/2 = 2.59 x 106 N/C

 

q = tan-1(Etoty /Etoty)

 

q = 67.8o

 

    1.   What is the magnitude of the total electric force exerted by charges Q1 and Q2 on 2.00 mC charge placed at the origin? What is the direction of the force?

 

F = |q| E = (2.00 x 10-6 C)( 2.59 x 106 N/C) = 5.18 N, directed the same as E.

 

  1. You are given a network of capacitors pictured below. C1 = 3.00 mF, C2 = 5.00 mF, and C3 = 6.00 mF. The applied potential is Vab = 24.0 V.

 

            

a.      Find the equivalent capacitance between points a and b.

 

C12 = C1 + C2  = 3.00 mF + 5.00 mF = 8.00 mF

 

1/C123 = 1/C3 + 1/C12 = 1/(6.00 mF) + 1/(8.00 mF) = 7/(24.0 mF)        C123 = 3.43 mF

 

b.     Calculate charge on the capacitor C3.

 

Q3 = C123 DV= (3.43 mF) (24.0 V) = 82.3 mC

 

c.      How much energy is stored in the capacitor C3?

 

EST = Q32/(2C3)

EST = (82.3 mC)2/(2 (6.00 mF)) = 564 mJ

 

d.     Find the potential difference between points a and d.

 

DV3 = Q3 /C3 = (82.3 mC)/(6.00 mF) = 13.7 V

 

DVad = 24.0 V – 13.7 V = 10.3 V

 

 

 

  1. A carbon resistor is to be used as a thermometer. On a winter day when the temperature is 4.00oC, the resistance of the carbon resistor is 217.3 W. What is the temperature on a spring day when the resistance is 215.8 W ? (Hint: Take the reference temperature T0 to be  4.00oC). The temperature coefficient of resistivity of carbon is a = -0.500 x 10-3(oC)-1.  

 

 R = R0 [1 + a (T – T0)]

 

R / R0 = 1 + a (T – T0)

                   

R / R0 - 1 = a (T – T0)

 

[R / R0 – 1]/ a = (T – T0)

 

[R / R0 – 1]/ a +  T0 = T

 

[(215.8 W) / (217.3 W) – 1]/ (-0.500 x 10-3(oC)-1) +  13.8oC = T

 

T = 17.8 oC

 

10.  Given the network, write equations that would allow you to solve for the currents in        each resistor   if the values of the emfs and resistances were known. Label and indicate your choices for current directions. DO NOT SOLVE THE EQUATIONS.

 

 

 

I4 = I2 + I3

I6 = I5 + I4

I3 = I1 + I6

 

e2+ I2 R6 – I5 R1 + I4 R3= 0

e3 + I3 R5 + I6 R4  + I4 R3 = 0

e1 + I1 R2 – I2 R6  + I3 R5 = 0