Final Exam Spring 2012

Final Exam Spring 2017

Covers Circuits
Covers Optics

Students who do well on tests are those who test themselves ahead of time. The topics covered on this final are topics on which students should have tested themselves over a month ago (for circuits) and now (for optics). Tools were given to perform this self-testing and practice. Nonetheless, yet another practice test has is here for students. This brand new practice test covers circuits and optics.

Students have been able to test themselves with other tools that I provided throughout the year. (Those who did things the way I instructed and mastered topics as we did them won’t forget those things, don’t need to restudy stressfully for this final test, and they can lightly review.)

The rest of this note is for anyone who didn’t master BOTH circuits and optics before and is attempting to do so now:

1)At the time of studying circuits, I repeatedly told all students how to test themselves. I told students to test themselves by solving for every current and every voltage for any of the circuit diagrams at the end of Chapter 20 of the textbook. I showed students how all answers can be self-checked for correctness without needing answers in the back of the book. Any student who has mastered circuits walks away from a circuit problem having checked every prediction of current and voltage and knows that they all add up correctly. The student has no doubt that the answers are correct, because they confirm K1 and K2.

2)When studying optics, I show how algebraic solutions can be checked against diagram solutions for the same problem. I’ve done this at least once in class. Ray diagramming is old by now. Algebraic formula usage in optics is slightly new. The main ideas are:

diagramming techniques

algebraic formula techniques that use both p-1 + q-1 = f-1 and M = -q/p to predict the image’s location and characteristics. (You have to be very careful about sign for q and f)

The aware student who did all the ray diagramming I assigned knows that there are 4 unique spherical mirror situations and 4 unique thin lens situations, and this student can easily ray-trace all of them.

Students are hereby clearly told the objectives of the optics unit before the final: 1) Anything related to Snell’s Law; 2) Full characterization of images through ray tracing; 3) Full characterization of images through algebra.

For help with the algebraic part of Optics, I have another practice posted called Optics Practice Test. That one only deals with Optics, no Circuits.

DC Circuits:

Directions: Each of the questions or incomplete statements below is followed by four or five suggested answers or completions. There is only one correct choice. Select the one that is best in each case.

1.If 15 coulombs pass a point in a circuit during 30 seconds, the current is

A. 2 amperes B. 500 milliamperes C. 30 amperes D. 300 milliamperes

2.If 3.12 x 1019 electrons pass a point in a circuit during each second, the current in amperes is about

A. 0.5 amperes B. 2.0 amperes C. 5.0 amperes

D. 7.0 amperesE. 10.0 amperes

3.Microwave ovens operate at about 600 watts. If a mean little kid puts his little brother's pet Tamagotchi in the microwave oven for 10 minutes, how much electrical energy does the oven consume when warming up the toad?

A. 3.6 x 105 joules B. 1 joule C. 6000 joules D. 3600 joules E. none of these

4.If the cost of electricity is 10 cents per kilowatt-hour, the cost of leaving a 100 watt light bulb on for 10 hours is closest to (Recall that 3600 seconds = 1 hour)

A. a penny B. a dime C. a quarter D. 50 cents E. a dollar

The following circuit diagram and information relates to questions 4 through 7.

The circuit at the right consists of a battery, three resistors (R1 > R2 > R3), and 6 meters. The meters are either voltmeters or ammeters and are connected correctly.

5.Which meters are measuring potential difference?

A. M3 and M5D. M1, M2 and M6

B. M4 and M6E. M1, M2, M3, and M5

C. M1, M2, and M4

6.Which meters are measuring current? (Choose answers from those in question #5)

7.Which meters will read the same?

A. M3 and M5 only B. M1 and M2 only C. M4 and M6 only

D. M1, M2, and M4 onlyE. none of them

8.What is the equivalent resistance, Req, for the circuit shown at the right?

A. 22  B. 160  C. 200  D. 240 

E. none of these

9.Which of the following statement(s) is/are true for the circuit shown at the right?

i.The voltage across each resistor is 10 volts.

ii.The current flowing through the 20  resistor is half the current flowing through the 10  resistor.

iii. The power dissipated in the 10  is greater than that dissipated in either the 20 or the 60  resistor.

A. i & ii only B. i & iii only C. ii & iii only

D. all of the statements are true

E. none of the statements are true

10. The equivalent resistance for the circuit shown in question #9 is

A.  B. 6  C. 9  D. 90  E. none of these

The diagram and information below relate to questions 11 through 20.

A circuit consisting of 3 resistors, an ammeter, a voltmeter, and a DC power supply is set up as shown in the diagram at the right. The following quantities are known about the circuit:

Vps = 12 volts = the voltage across the power supply

I = 30 milliamperes = the current measured

by meter A

V3 = 9 volts = the voltage measured by meter V

R1 = 100 ohmsR3 = 450 ohms

11. As stated above Vps = 12 volts. This means that when 0.5 coulomb of charge passes through the power supply it has its potential energy

A. increased by 12 voltsD. decreased by 6 joules

B. increased by 24 joulesE. none of these

C. decreased by 12 joules

12. As 1 elementary charge passes through R3 its potential energy

A. increases by 1.9 x 10-18 joulesD. increases by 4.8 x 10-19 joules

B. decreases by 1.4 x 10-18 joulesE. decreases by 5.6 x 10+19 joules

C. decreases by 9 joules

13. The amount of charge passing through meter A in 2 seconds is

A. 0.3 coulomb B. 30 coulombs C. 60 coulombs

D. 0.06 coulombE. none of these

14. The number of elementary charges passing through meter A in 1 second is about

A. 1.9 x 10+20 B. 4.8 x 10-18 C. 2.1 x 10+20 D. 2.1 x 10+17 E. 1.9 x 10+17

15. The power output of the power supply is

A. 0.36 watts B. 360 watts C. 0.4 watts D. 400 watts E. none of these

16. The potential difference across resistor R2 is

A. 3 volts B. 4 volts C. 6 volts D. 9 volts E. none of these

17. The current flowing through resistor R3 is

A. 30 milliamps B. 10 milliamps C. 0.05 amperess

D. 0.02 amperesE. none of these

18. The current flowing through resistor R1 is

A. 30 milliamps B. 10 milliamps C. 0.05 amperess

D. 0.02 amperesE. none of these

19. The current flowing through resistor R2 is

A. 30 milliamps B. 10 milliamps C. 0.05 amperess

D. 0.02 amperesE. none of these

20. The energy dissipated as heat in resistor R1 in 3 seconds is

A. 0.09 joules B. 0.03 joules C. 9 joules D. 270 joules E. 0.27 joules

Geometric Optics:

1.Light waves are thought to be transverse waves because

A.they can be focused by a lens.

B.they exhibit interference effects.

C.they undergo refraction when passing from one medium to another.

D.they diffract around objects.

E. they can be polarized. Don’t worry if you don’t know this one. But look it up.

2.The property of light waves that is associated with color is

A. their amplitude B. their frequency C. their intensity

D. their speedE. the fact that light can be polarized

3.As shown in the diagram at the right, a beam of light reflects between two plane mirrors which are mounted perpendicular to one another. The beam strikes the first with an incident angle of 35o and reflects toward the other mirror. The reflected beam then strikes the second mirror at point P with an incident angle i. The angle of reflection, r, at point P will be

A. 35o B. 55o C. 70o D. 90o E. none of these

4.If light passes from air into glass, the index of refraction is found to be 1.5. If the speed of light in air is 3 x 108 m/s, the speed of light in the glass is about

A. 1 x 108 m/s B. 1.5 x 108 m/s C. 2 x 108 m/s D. 3 x 108 m/s E. 4.5 x 108 m/s

5.The drawing at the right represents light rays from a very distant source entering a plano-convex lens. The lens is made of glass and is submerged in a liquid having a greater optical density than glass. Which of the following best represents the path of the upper light ray as it passes through the lens?

The following diagram and information relate to Questions 6, 7, and 8.

In the diagram at the right AB is the boundary between

medium I and medium II. MN is the normal to the boundary at point P.

6.If HP is an incident monochromatic (one color) light ray and PQ is the refracted ray, which of the following statement(s) is/are true?

i.i is greater than r.

ii.The index of refraction for light passing from medium I into medium II is greater than 1.

iii.Medium I is of lower optical density than medium II.

iv.The wavelength of the light increases as it crosses the boundary.

v.The speed of the light increases as it crosses the boundary.

A. i, ii, & iii only B. iii, iv, & v only C. i, iii, & iv only D. iv & v only

E. none of these combinations are correct

7.Assume medium I is air and medium II is water. If KP is the incident ray, the refracted ray is most represented by ray

A. PU B. PV C. PT D. PG E. PR

8.Assume medium I is air and medium II is glass. If VP is the incident ray and angle VPN is greater than the critical angle, the ray leaving point P will most likely be

A. PK B. PJ C. PF D. PQ E. PB

9.A light ray is incident along the normal to side AC of a right angle glass prism as shown at the right. The index of refraction for a ray passing from air into the glass is 1.5. Which one of the

following statements is true?

A. The ray is totally reflected from side AB with a reflected

angle greater than 10o.

B. The ray emerges from side BC with an angle of refraction

greaterthan 45o.

C. The ray emerges from side BC with an angle of refraction less than 45o.

D. The ray reflects back along the same path it entered.

E. The ray is totally reflected from side BC and emerges from side AB along its normal.

10. For reflected rays of light to be parallel to the principal axis of a concave mirror, a small source of light should be placed

A. as close to the mirror as possible.

B. at the center of curvature of the mirror.

C. half way between the center of curvature and the principal focus of the mirror.

D. at the principal focus of the mirror.

E. at a point half way between the principal focus and the mirror.

11. An object is placed in front of a concave mirror as shown at the right. The center of curvature is C and the principal focus is at F. The image formed is

A. erect and larger than the object.

B. erect and the same size as the object.

C. erect and smaller than the object.

D. inverted and larger than the object.

E. inverted and smaller than the object.

12. An object is located at the center of curvature of a concave mirror of 15 cm radius. The distance from the mirror to the image formed is

A. 30 cm B. 22.5 cm C. 15 cm D. 7.5 cm E. none of these

13. The image formed of an object due to reflection from a convex mirror is

A. real, erect, and larger than the object.

B. real, erect, and smaller than the object.

C. real, inverted, and smaller than the object.

D. virtual, inverted, and larger than the object.

E. virtual, erect, and smaller than the object.

14. An object is placed 20 centimeters in front of a convex mirror forms an image 10 cm behind the mirror. What is the focal length of the mirror?

A. -20/3 cm B. -10 cm C. 10 cm D. 20 cm E. none of these

15. If the height of the object in question #14 is 2 cm, what is the height of the image?

A. 0.5 cm B. 1.0 cm C. 1.5 cm D. 2.0 cm E. none of these

16. The diagram at the right shows an object O, an image I, and two rays from the object. Which of the following will when placed on the line PQ will produce the image shown?

A. A plane mirror D. A converging lens

B. A concave mirror E. A diverging lens

C. A convex mirror

17. The diagram at the right shows a converging lens, its principal foci, F1 and F2, located on its principal axis, and the following light ray segments.

Ray W is parallel to the principal axis and strikes the lens near its top.

Ray X passes through F1 and strikes the lens near its top.

Ray Y passes through F1 and travels along the principal axis.

Ray Z is parallel to ray X and passes through the center of the lens.

Which rays pass through F2?

A. X and Y B. X and Z C. W and X D. W and Y E. Y and Z

The diagram and information below relate to Questions 18, 19, and 20.

The diagram at the right represents a converging lens having a focal length of 20 cm. Along the principal axis the principal foci are labeled F and F' and the secondary foci 2F and 2F'.

18. Which type of image cannot be formed by this lens?

A. A real image smaller than the object

B. A real image larger than the object

C. A virtual image smaller than the object

D. A virtual image larger than the object

19. If the object is 30 cm from the lens, the distance from the lens to the image will be

A. 10 cm B. 12 cm C. 50 cm D. 60 cm E. none of these

20. An object 12 cm tall is placed 40 cm from the lens. The height of the image formed is

A. 6 cm B. 12 cm C. 24 cm D. 48 cm E. none of these

DC Circuits:

Key: The following is more a set of hints instead of a list of answers. It’s to help you study based on concepts.

1.Q = I(Δt) by definition.

2.If a person knows that C = (A)(s) and that every 1 e equals 1.6 x 10-19C, and applies these facts, then this is just a simple unit manipulation question.

3.More unit manipulation: W = J/s, so J = (W)(s), meaning Energy = (Power)(time), so when people don’t know simple definitions, they miss out on how simple things are.

Energy = (600 J/s)(1 hr) = (600 J/s)(3600 s) = answer

How easy it is for those who write units and expect them to cancel correctly to match the units of the final answer quantity.

4.If the cost of electricity is 10 cents per kilowatt-hour, the cost of leaving a 100 Watt light bulb on for 10 hr is closest to (Recall that 3600 s = 1 hr)

More units: If a person cares about dimensions, then they will appreciate that Watt-s is in the same dimension category as kilowatt-hour, and that dimension category is ______.

If at this point, you didn’t fill in the last blank with the word “energy” or “work”, then the source of difficulty is not mathematical; it’s verbal. But I’ve heavily stressed dimensions all year, and the people who caught on developed simpler methods.

Just like #3, the amount of energy in the problem is really easy to calculate:

Energy = Pt = (100 J/s)(36,000 s) = 3.6 x 106J or it equals Pt = (1000 W)(1 hr) = 1000 W-hr

The money cost is charged per unit ENERGY and NOT per unit power, so it should be simple to know that calculating the energy therefore calculates the money. The energy has been calculated; it is either 3.6 x 106J or 1000 W-hr or 1 kW-hr, depending on you feel like stating it. Use any of these facts to know how much that much energy costs if the price is 10 cents per kW-hr. This was a vocabulary question, and it reminds you that energy is related to circuits through P = I(ΔV), where the energy used = Pt which would equal current times voltage times time.

There was no P = I(ΔV) on these two particular problems #3 and #4, but Power is obviously closely related to energy and money, so P = I(ΔV) could be related to things like this on the real test.

9.A person should quickly calculate all FOUR currents in the circuit before attempting to answer any of what is asked for #9. Calculating all FOUR of those currents should take about 20 seconds.

A circuit consisting of 3 resistors, an ammeter, a voltmeter, and a DC power supply is set up as shown in the diagram at the right. The following quantities are known about the circuit:

Vps = 12 volts = the voltage across the power supply

I = 30 milliamperes = the current measured

by meter A

V3 = 9 volts = the voltage measured by meter V

R1 = 100 ohmsR3 = 450 ohms

11.As stated above Vps = 12 volts. This means that when 0.5 coulomb of charge passes through the power supply it has its potential energy

By definition, J = (Coulomb)(Volt)

12.As 1 elementary charge passes through R3 its potential energy decreases by how much?

By definition, J = (Coulomb)(Volt)

And in this case the charge value is 1.6 x 10-19C. So the energy to be solved for is that much charge multiplied by the voltage. So what is the voltage across R3? You are NOT supposed to expect that it is the power supply voltage.

13.The amount of charge passing through meter A in 2 seconds is

Q = I(Δt)

14.The number of elementary charges passing through meter A in 1 second is about

The Coulomb value is half the answer to #13. To convert the Coulomb value to number of e, one knows how to use 1 e = 1.6 x 10-19C.

16.The potential difference across resistor R2 is

This is a sophisticated question where a person is expected to know the loop rule to relate the four important voltages of this circuit to each other.

More challenging questions would be things like the following:

How much energy is turned into heat as one elementary charge is forced across R2?
How much heat is created when current flows through R1 for 20 seconds.

To answer either of those, one would have to start by being crystal clear on the idea the voltage across R2 has to be 9 V and that across R1 has to be 3 V. The easiest way to do the heat-for-20-seconds question is to know that 3 V is the voltage and to multiply that by the proper current. This gets the person power in Watts. That times 20 s yields heat in J. It would be wise to answer that question for practice now:

18.The current flowing through resistor R1 is30

20.The energy dissipated as heat in resistor R1 in 3 seconds is

I don’t know the answer to #20, but I do know that the answer to the heat-for-20-seconds question above is 1.8 J. I talked you through that one, so I hope you did it on your own in the space above. Will you do #20 on your own? The key’s on another paper.

Geometric Optics:

4.We didn’t have much time for this, but the speed in a medium is the vacuum speed of light divided by the index of refraction. Index of refraction is symbolized with the letter n, and is always greater than 1.