Name: Class: _ Date: _ ID: A Electricity Mock Exam Multiple Choice Identify the letter of the choice that best completes the statement or answers the question.. What happens when a rubber rod is rubbed with a piece of fur, giving it a negative charge? a. Protons are removed from the rod. c. Electrons are added to the fur. b. Electrons are added to the rod. d. The fur is left neutral. 2. An attracting force occurs between two charged objects when the charges are of a. unlike signs. c. equal magnitude. b. like signs. d. unequal magnitude. 3. Charge is most easily transferred in a. nonconductors. c. semiconductors. b. conductors. d. insulators. 4. Which statement is the most correct regarding electric insulators? a. Charges within electric insulators do not readily move. b. Electric insulators have high tensile strength. c. Electric charges move freely in electric insulators. d. Electric insulators are good heat conductors. 5. Electric field strength depends on a. charge and distance. c. Coulomb constant and mass. b. charge and mass. d. elementary charge and radius. 6. When a capacitor discharges, a. it must be attached to a battery. b. charges move through the circuit from one plate to the other until both plates are uncharged. c. charges move from one plate to the other until equal and opposite charges accumulate on the two plates. d. it cannot be connected to a material that conducts. 7. A 0.25 µf capacitor is connected to a 9.0 V battery. What is the charge on the capacitor? a..2 0 2 C c. 2.5 0 6 C b. 2.2 0 6 C d. 2.8 0 2 C 8. A 0.50 µf capacitor is connected to a 2 V battery. Use the expression PE = 2 C(ΔV) 2 to determine how much electrical potential energy is stored in the capacitor. a. 3.0 0 6 J c..0 0 5 J b. 6.0 0 6 J d. 3.6 0 5 J 9. How is current affected if the number of charge carriers decreases? a. The current increases. b. The current decreases. c. The current initially decreases and then is gradually restored. d. The current is not affected.
Name: ID: A 0. The current in an electron beam in a cathode-ray tube is 7.0 0 5 A. How much charge hits the screen in 5.0 s? a. 2.8 0 3 C c. 3.5 0 4 C b. 5.6 0 2 C d. 5.3 0 6 C. The amount of charge that moves through the filament of a lightbulb in 2.00 s is 2.67 C. What is the current in the lightbulb? a. 5.34 amps c. 0.835 amps b..33 amps d. 0.47 amps 2. When electrons move through a metal conductor, a. they move in a straight line through the conductor. b. they move in zigzag patterns because of repeated collisions with the vibrating metal atoms. c. the temperature of the conductor decreases. d. they move at the speed of light in a vacuum. 3. What is the potential difference across a 5.0 Ω resistor that carries a current of 5.0 Amps? a..0 0 2 V c. 0.0 V b. 25 V d..0 V 4. A lightbulb has a resistance of 240 Ω when operating at 20 V. What is the current in the lightbulb? a. 2.0 amps c. 0.50 amps b..0 amps d. 0.20 amps 5. Which of the following wires would have the greatest resistance? a. an aluminum wire 0 cm in length and 3 cm in diameter b. an aluminum wire 5 cm in length and 3 cm in diameter c. an aluminum wire 0 cm in length and 5 cm in diameter d. an aluminum wire 5 cm in length and 5 cm in diameter 6. The power ratings on lightbulbs are measures of the a. rate that they give off heat and light. b. voltage they require. c. density of the charge carriers. d. amount of negative charge passing through them. 7. If a 75 W lightbulb operates at a voltage of 20 V, what is the current in the bulb? a. 0.62 amps c..95 0 2 amps b..6 amps d. 9.0 0 3 amps 8. If a lamp has a resistance of 36 Ω when it operates at a power of.00 0 2 W, what is the potential difference across the lamp? a. 7 V c. 25 V b. 36 V d. 220 V 9. Which of the following is the best description of a schematic diagram? a. uses pictures to represent the parts of a circuit b. determines the location of the parts of a circuit c. shows the parts of a circuit and how the parts connect to each other d. shows some of the parts that make up a circuit 2
Name: ID: A 20. If the potential difference across a pair of batteries used to power a flashlight is 6.0 V, what is the potential difference across the flashlight bulb? a. 3.0 V c. 9.0 V b. 6.0 V d. 2 V 2. Three resistors with values of 4.0 Ω, 6.0 Ω, and 8.0 Ω, respectively, are connected in series. What is their equivalent resistance? a. 8 Ω c. 6.0 Ω b. 8.0 Ω d..8 Ω 22. Three resistors connected in series carry currents labeled I, I 2, and I 3, respectively. Which of the following expresses the total current, I t, in the system made up of the three resistors in series? a. It = I + I2 + I3 c. It = I = I2 = I3 b. Ê It = + + ˆ d. It = + + Ê ˆ Ë Á I I 2 I 3 Ë Á I I 2 I 3 23. Three resistors with values of 4.0 Ω, 6.0 Ω, and 0.0 Ω are connected in parallel. What is their equivalent resistance? a. 20.0 Ω c. 6.0 Ω b. 7.3 Ω d..9 Ω 24. What is the equivalent resistance for the resistors in the figure shown above? a. 25 Ω c. 7.5 Ω b. 0.0 Ω d. 5.0 Ω 25. What is the current of the figure shown above? a. 40.0 amps c. 0.0 amps b. 4.00 amps d. 2.0 amps 26. What is the electric force between an electron and a proton that are separated by a distance of.0 0 0 m? Is the force attractive or repulsive? (e =.60 0 9 C, k C = 9.0 0 9 N m 2 /C 2 ) a. +2.3 X 0-8 N c. +2.3 X 0-8 N b. -2.3 X 0-8 N d. -2.3 X 0-8 N 3
Name: ID: A 27. Two point charges having charge values of 2.0 µc and 4.0 µc, respectively, are separated by.5 cm. What is the value of the mutual force between them? (k C = 9.0 0 9 N m 2 /C 2 ) a. +3.2 X 0 2 N c. +4.8 X 0 2 N b. -3.2 X 0 2 N d. -4.8 X 0 2 N Short Answer 28. a. Draw a schematic diagram with the following components: a 00. V battery, a open switch, a 0.0 Ω resistor, a 25.0 Ω resistor, and a light bulb in a series circuit. b. Find the equivalent resistance for the circuit. c. Find the current over the circuit. 29. a. Draw a schematic diagram with the following components: two 30.0 Ω resistors are connected in parallel. This parallel arrangement is connected in series with a 60.0 resistor. The entire circuit is then placed across a 20. V potential difference. b. Find the equivalent resistance over the circuit. c. Find the current over the circuit. 4
Electricity Mock Exam Answer Section MULTIPLE CHOICE. ANS: B PTS: DIF: I OBJ: 6-. 2. ANS: A PTS: DIF: I OBJ: 6-. 3. ANS: B PTS: DIF: I OBJ: 6-.2 4. ANS: A PTS: DIF: I OBJ: 6-.2 5. ANS: A PTS: DIF: I OBJ: 6-3. 6. ANS: B PTS: DIF: I OBJ: 7-2. 7. ANS: B C = 0.25 µf = 0.25 x 0 6 F ΔV = 9.0 V Q = CΔV = (0.25 0 6 F)(9.0 V) = 2.2 0 6 C PTS: DIF: IIIA OBJ: 7-2.2 8. ANS: D C = 0.50 µf = 0.50 x 0 6 F ΔV = 2 V PE electric = 2 C(ΔV) 2 = 2 (0.50 0 6 F)(2 V) 2 = 3.6 0 5 J PTS: DIF: IIIA OBJ: 7-2.3 9. ANS: B PTS: DIF: II OBJ: 7-3. 0. ANS: C I = 7.0 0 5 A Δt = 5.0 s I = ΔQ Δt Rearrange to solve for ΔQ. ΔQ = IΔt = (7.0 0 5 A)(5.0 s) = 3.5 0 4 C PTS: DIF: IIIA OBJ: 7-3.
. ANS: B ΔQ = 2.67 C Δt = 2.00 s I = ΔQ = 2.67 C Δt 2.00 s =.33 A PTS: DIF: IIIA OBJ: 7-3. 2. ANS: B PTS: DIF: I OBJ: 7-3.2 3. ANS: B R = 5.0 Ω I = 5.0 A ΔV = IR = ( 5.0 A) ( 5.0 Ω) = 25 V PTS: DIF: IIIA OBJ: 7-3.3 4. ANS: C ΔV = 20 V R = 240 Ω ΔV = IR Rearrange to solve for I. I = ΔV R = 20 V 240 Ω = 0.50 A PTS: DIF: IIIA OBJ: 7-3.3 5. ANS: A PTS: DIF: I OBJ: 7-3.4 6. ANS: A PTS: DIF: II OBJ: 7-4.2 2
7. ANS: A P = 75 W ΔV = 20 V P = IΔV Rearrange to solve for I. I = P ΔV = 75 W 20 V = 0.62 A PTS: DIF: IIIA OBJ: 7-4.3 8. ANS: A R = 36 Ω P =.00 0 2 W P = (ΔV) 2 R Rearrange to solve for ΔV. ΔV = PR = (.00 0 2 W)(36 Ω) = 7 V PTS: DIF: IIIB OBJ: 7-4.3 9. ANS: C PTS: DIF: I OBJ: 8-. 20. ANS: B PTS: DIF: II OBJ: 8-.3 2. ANS: A R = 4.0 Ω R2 = 6.0 Ω R3 = 8.0 Ω Req = R + R2 + R3 = 4.0 Ω + 6.0 Ω + 8.0 Ω = 8 Ω PTS: DIF: I OBJ: 8-2. 22. ANS: C PTS: DIF: II OBJ: 8-2. 3
23. ANS: D R = 4.0 Ω R2 = 6.0 Ω R3 = 0.0 Ω R eq = R + R 2 + R 3 = = 0.25 R eq Ω + 0.7 Ω + 0.00 Ω = 0.52 Ω R eq = Ω 0.52 =.9 Ω 4.0 Ω + 6.0 Ω + 0.0 Ω PTS: DIF: IIIA OBJ: 8-2.2 24. ANS: B R = 8.0 Ω R2 = 2.0 Ω R3 = 0.0 Ω R4 = 5.0 Ω R,2 = R + R2 = 8.0 Ω + 2.0 Ω = 0.0 Ω = + = R, 2, 3 R, 2 R 3 0.0 Ω + 0.0 Ω = 2 0.0 Ω R, 2, 3 = 0.0 Ω = 5.00 Ω 2 Req = R,2,3 + R4 = 5.00 Ω + 5.0 Ω = 0.0 Ω PTS: DIF: IIIA OBJ: 8-3. 25. ANS: B R = 0.0 Ω V = 40.0 V I = V/R I = 40.0 / 0.0 = 4.00 PTS: 26. ANS: A PTS: 27. ANS: B PTS: 4
SHORT ANSWER 28. ANS: Diagram may vary. b. 35.0 Ω c. I = V/R I = 00./35.0 =2.85 amps PTS: DIF: I OBJ: 6-3.2 29. ANS: a. Diagrams may vary. b. R eq = 5.0Ω + 60.0Ω = 75Ω c. I = V/R I = 20. V / 75 Ω I =.60 amps PTS: 5