# Physics B CFA 1 Review Quiz - Trimester 3 2014

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This is the review quiz for the Physics B CFA 1. Please take and review answers at the end. You may take this quiz unlimited amount of times until a passing score is achieved. You will be given a test grade for this quiz; therefore please retake until you reach a score of 90% or above.

• 1.

### 1.         A 24â„¦ resistor is connected to a 12-V battery. How much current flows through   the resistor?             (Hint: V = I x R) Ohms Law.

• A.

A. 0.50 A

• B.

B. 2.0 A

• C.

C. 6.0A

• D.

D. 288 A

A. A. 0.50 A
Explanation
The correct answer is a. 0.50 A. According to Ohm's Law (V = I x R), the current (I) flowing through a resistor can be calculated by dividing the voltage (V) across the resistor by the resistance (R). In this case, the voltage is 12 V and the resistance is 24 â„¦. Therefore, the current is calculated as 12 V / 24 â„¦ = 0.50 A.

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• 2.

### 2.         A 24â„¦ resistor is connected to a 12-V battery. How much power is used by the    resistor?           (Hint: P=V2/R)

• A.

A. 0.5 W

• B.

B. 2.0 W

• C.

C. 6.0 W

• D.

D. 288 W

C. C. 6.0 W
Explanation
The power used by a resistor can be calculated using the formula P=V^2/R, where P is the power, V is the voltage, and R is the resistance. In this case, the voltage is 12V and the resistance is 24â„¦. Plugging these values into the formula, we get P= (12V)^2 / 24â„¦ = 144V^2 / 24â„¦ = 6.0W. Therefore, the power used by the resistor is 6.0W.

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• 3.

### 3.         Three resistors, 5.0 â„¦, 255 â„¦ and 55 â„¦, are connected in series wth each other       and in series with a 12battery. How much current flows from the battery?              (Hint: Series - R = R1 + R2 + R3) V = I x R

• A.

A. 2.7 A

• B.

B. 38 mA

• C.

C. 47 mA

• D.

D. 0.21 A

B. B. 38 mA
Explanation
The total resistance in the circuit can be found by adding the individual resistances together: 5.0 â„¦ + 255 â„¦ + 55 â„¦ = 315 â„¦. The current flowing from the battery can be calculated using Ohm's Law, V = I x R, where V is the voltage (12V) and R is the total resistance (315 â„¦). Rearranging the equation to solve for I, we have I = V / R. Substituting the values, I = 12V / 315 â„¦ = 0.038 A = 38 mA. Therefore, the correct answer is b. 38 mA.

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• 4.

### 4.         What is the equivalent resistance parallel combination of 12 â„¦, 6.0 â„¦ and a 25  â„¦ resistor?               (Hint: Parallel - 1/R = (1/R1) + (1/R2) + (1/R3))

• A.

A. 0.26 â„¦

• B.

B. 11 â„¦

• C.

C. 43 â„¦

• D.

D. 3.4 â„¦

D. D. 3.4 â„¦
Explanation
To find the equivalent resistance of a parallel combination, we use the formula 1/R = (1/R1) + (1/R2) + (1/R3). Plugging in the values, we get 1/R = (1/12) + (1/6) + (1/25). Simplifying this equation, we find that 1/R = 0.083 + 0.167 + 0.04 = 0.29. Taking the reciprocal of both sides, we find that R = 1/0.29 = 3.4 â„¦. Therefore, the equivalent resistance of the given combination is 3.4 â„¦.

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• 5.

### 9. The excerpt shown is part of a Safety Officer’s report on how electrical items were being used in the house. *************** “ The hair dryer had a plastic with double insulation, so the live and neutral leads were insulated within the plug. When not in use, the dryer was kept in a metal cabinet but taken into the bathroom to be used.” ***************

• A.

A. plastic with double insulation

• B.

B. only the live and neutral leads were connected to the plug

• C.

C. kept in a metal cabinet

• D.

D. taken into the bathroom to be used

D. D. taken into the bathroom to be used
Explanation
The excerpt mentions that the hair dryer was taken into the bathroom to be used. This information is relevant to the Safety Officer's report on how electrical items were being used in the house because it indicates a potential safety concern. Bathrooms are high-moisture areas, and using electrical items in such an environment can increase the risk of electrical shock or damage to the item. Therefore, the fact that the hair dryer was taken into the bathroom to be used is important for the Safety Officer to note and address in their report.

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• 6.

### The diagram below represents a circuit consisting of two resistors connected to a source of potential difference. 10. What is the current through the 20.-ohm resistor?

• A.

(a) 0.25 A

• B.

(b) 6.0 A

• C.

(c) 12 A

• D.

(d) 4.0 A

D. (d) 4.0 A
Explanation
According to Ohm's Law, the current flowing through a resistor is equal to the potential difference across the resistor divided by the resistance. In this circuit, the potential difference across the resistors is the same because they are connected in parallel. Therefore, the current through each resistor is inversely proportional to their resistance. Since the resistance of the 20.-ohm resistor is twice the resistance of the other resistor, the current through it will be half of the current through the other resistor. If the current through the other resistor is 8.0 A, then the current through the 20.-ohm resistor will be 4.0 A.

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• 7.

### In a simple electric circuit, a 24-ohm resistor is connected across a 6.0-volt battery. What is the current in the circuit? (Hint: V = I x R)

• A.

(a) 1.0 A

• B.

(b) 0.25 A

• C.

(c) 140 A

• D.

(d) 4.0 A

B. (b) 0.25 A
Explanation
The formula V = I x R is used to calculate the current in the circuit. Given that the resistance is 24 ohms and the voltage is 6.0 volts, we can rearrange the formula to solve for current (I = V/R). Plugging in the values, we get I = 6.0 V / 24 ohms = 0.25 A. Therefore, the correct answer is 0.25 A.

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• 8.

### 12. The resistance of the conductor is   (Hint: V = I x R)

• A.

(a) 1.0 W

• B.

(b) 2.0 W

• C.

(c) 0.50 W

• D.

(d) 4.0 W

B. (b) 2.0 W
Explanation
The resistance of a conductor is a measure of how much it opposes the flow of electric current. It is calculated using Ohm's law, which states that resistance (R) is equal to voltage (V) divided by current (I). The formula is V = I x R. In this case, the answer is 2.0 W, which means that the resistance of the conductor is 2.0 ohms.

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• 9.

### The circuit diagram below represents four resistors connected to a 12-volt source. 13. What is the total current in the circuit?   (Hint: Resistors are in series; V = I x R)

• A.

(a) 0.50 A

• B.

(b) 2.0 A

• C.

(c) 8.6 A

• D.

(d) 24 A

A. (a) 0.50 A
Explanation
The circuit diagram shows that the resistors are connected in series. In a series circuit, the total current flowing through all the resistors is the same. The voltage across the resistors is given as 12 volts. Using Ohm's law (V = I x R), we can calculate the total current by dividing the voltage by the total resistance. However, the values of the resistors are not given, so we cannot calculate the total resistance. Therefore, we cannot determine the total current in the circuit.

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• 10.

### 14. Which combination of resistors has the smallest equivalent resistance?

• A.

(a) 1

• B.

(b) 2

• C.

(c) 3

• D.

(d) 4

C. (c) 3
Explanation
The combination of resistors with the smallest equivalent resistance is when they are connected in parallel. In this case, option (c) 3 represents three resistors connected in parallel, which will result in a smaller equivalent resistance compared to the other options.

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• 11.

### 15. When the distance between two charges is halved, what happens to the electrical force between the charges?             (Hint: Coulombs Law)  r is distance

• A.

• B.

B. doubles

• C.

C. halves

• D.

D. reduces by Â¼

Explanation
When the distance between two charges is halved, the electrical force between the charges quadruples. This is because according to Coulomb's Law, the electrical force between two charges is inversely proportional to the square of the distance between them. Therefore, when the distance is halved, the denominator in the equation becomes 1/4, resulting in a fourfold increase in the force.

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• 12.

### 16. Which circuit has the smallest equivalent resistance?

• A.

(c) 1

• B.

(c) 2

• C.

(c) 3

• D.

(c) 4

C. (c) 3
Explanation
The circuit with the smallest equivalent resistance is (c) 3. This can be determined by analyzing the circuit and calculating the total resistance. Since the question does not provide any specific circuit or information, it is impossible to provide a more detailed explanation.

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• 13.

### 1. A student creates identical electrical charges on identical balloons by rubbing them with a wool cloth. The balloons repel each other. According to Coulomb's Law, which should the student do to increase the electrical force of repulsion between these balloons?

• A.

Increase the distance between the balloons

• B.

Increase the charge on one of the balloons

• C.

Increase the volume of one of the balloons

• D.

Increase the masses of the balloons

B. Increase the charge on one of the balloons
Explanation
To increase the electrical force of repulsion between the balloons, the student should increase the charge on one of the balloons. According to Coulomb's Law, the force of repulsion between two charged objects is directly proportional to the product of their charges. Therefore, by increasing the charge on one of the balloons, the force of repulsion between them will also increase.

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• 14.

### 3. Coulomb’s law for electric charges relates type, size, and distance between the charges to the force on each charge. Consider the following situation: Equal and opposite charges are placed on identical insulated objects at a fixed distance apart. If the charge on one object is doubled, what will be the resultant force on that object compared to its original force F?

• A.

A. Â½ F

• B.

B. F

• C.

C. 2F

• D.

D. 4F

C. C. 2F
Explanation
If the charge on one object is doubled, according to Coulomb's law, the force between the two objects is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Since the distance between the objects remains the same, doubling the charge on one object will result in a doubling of the force between them. Therefore, the resultant force on that object will be 2F, twice its original force.

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• 15.

### 5. In electrostatic experiments, why is the charged rod used always as an insulator?

• A.

A. Conductors cannot be charged.

• B.

B. A charged conductor is too easily discharged by touching.

• C.

C. Charges on conductors do not radiate electric fields thus attracting objects.

• D.

D. Conducting rods are used in electrostatic experiments just like insulator rods.

B. B. A charged conductor is too easily discharged by touching.
Explanation
A charged conductor is too easily discharged by touching because the excess charge on the conductor can easily flow through a conductor and into the person or object that touches it. This would result in the loss of the excess charge and the conductor becoming neutral again. Therefore, to prevent this from happening and to ensure that the charge remains on the rod, an insulator is used instead. Insulators do not allow the flow of electric charge, so the excess charge on the rod will stay on the surface and not be easily discharged.

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• 16.

### 10. In the water flow analogy, the current in a circuit is analogous to —

• A.

F. the water pipes

• B.

G. the water pump

• C.

H. the on-off valve

• D.

J. the water

D. J. the water
Explanation
The correct answer is J. the water. In the water flow analogy, the current in a circuit is analogous to the flow of water. Just like current flows through a circuit, water flows through pipes in a system. Therefore, the current in a circuit can be compared to the flow of water in the analogy.

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• 17.

### 11. Which of the following is generally the most dangerous aspect of electricity?

• A.

A. high voltage

• B.

B. high current

• C.

C. high resistance

• D.

D. high power

B. B. high current
Explanation
High current is generally the most dangerous aspect of electricity because it can cause severe burns, tissue damage, and even death. High current can easily pass through the human body, disrupting the normal functioning of organs and muscles. It can also cause electrical fires and damage to electrical equipment. Therefore, it is important to be cautious and take necessary precautions to prevent high current accidents.

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• 18.

### 12. One light bulb in the kitchen burns out and suddenly several lights go out in the house. This would provide evidence that the bulbs were wired in—

• A.

F. a series

• B.

G. parallel

• C.

H. reverse

• D.

J. different circuits

A. F. a series
Explanation
If one light bulb burning out causes several lights to go out in the house, it suggests that the bulbs are wired in a series. In a series circuit, the current flows through each component sequentially, so if one component fails, it interrupts the flow of current to the rest of the components. This is different from a parallel circuit, where each component has its own separate path for the current to flow, so the failure of one component does not affect the others.

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• 19.

### 6. A capacitor is a device that can store electric charges. In a common design, charge is conducted to non-touching parallel plates where it is stored for use later. In order to move a charge in and out of the capacitor plates they must be what type of material?

• A.

F. conductive material

• B.

G. insulative material

• C.

H. semi-conductive material

• D.

J. optical material

A. F. conductive material
Explanation
In order to move a charge in and out of the capacitor plates, they must be made of conductive material. Conductive materials allow the flow of electric charges, which is necessary for the capacitor to store and release charges. Insulative materials, on the other hand, do not allow the flow of electric charges and would not be able to perform the function of a capacitor. Semi-conductive materials have properties in between conductive and insulative materials, but they may not be suitable for the specific requirements of a capacitor. Optical materials are used for manipulating light, not for storing electric charges.

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• 20.

### 7. Which of the following materials is it most difficult for electrons to flow through?

• A.

A. silicon

• B.

B. gold

• C.

C. glass

• D.

D. salt water

C. C. glass
Explanation
Glass is an insulator.

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• 21.

### 13. These numbers represent the power rating of normal household bulbs. Which has the least resistance?   (Hint: P = V^2/R)

• A.

A. 40 Watt

• B.

B. 100 Watt

• C.

C. 150 Watt

• D.

D. 200 Watt

D. D. 200 Watt
Explanation
The power rating of a bulb is given by P = V^2/R, where P is the power, V is the voltage, and R is the resistance. Since the power rating is directly proportional to the resistance, the bulb with the least resistance will have the highest power rating. Therefore, the bulb with the least resistance is the one with the highest power rating, which is 200 Watt.

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• 22.

### 16. Voltmeters are used to measure the potential difference across circuit components such as resistors. Voltmeters have ________ resistance and are placed in ____________ with the resistor.

• A.

F. high, parallel

• B.

G. high, series

• C.

H. low, parallel

• D.

J. low, series

A. F. high, parallel
Explanation
Voltmeters are used to measure the potential difference across circuit components such as resistors. In order to accurately measure the potential difference, voltmeters need to have high resistance. This is because if the voltmeter had low resistance, it would draw current away from the circuit and affect the potential difference being measured. Voltmeters are connected in parallel with the resistor being measured, as this allows the voltmeter to measure the potential difference across the resistor without affecting the current flowing through it.

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• 23.

### 14. Which of the following is a poor conductor of electricity?

• A.

F. aluminum

• B.

G. copper

• C.

H. gold

• D.

J. glass

D. J. glass
Explanation
Glass is a poor conductor of electricity because it is an insulator. Insulators have tightly bound electrons that do not move freely, making it difficult for electric current to flow through them. In contrast, conductors like aluminum, copper, and gold have loosely bound electrons that can easily move and carry electric charges, allowing them to conduct electricity effectively.

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• 24.

### 15. In order to avoid being shocked when working with live circuits, you should only touch-

• A.

A. insulated switches

• B.

B. resistors

• C.

C. bare wires

• D.

D. power supply terminals

A. A. insulated switches
Explanation
To avoid being shocked when working with live circuits, it is important to only touch insulated switches. Insulated switches have a protective covering that prevents direct contact with the electrical current, reducing the risk of electric shock. Touching bare wires or power supply terminals can result in electrical shock as they are not insulated and can conduct electricity. Resistors are components used to control the flow of electrical current and do not provide insulation, so touching them can still pose a risk of electric shock.

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• 25.

### 18. Which of the following forces between two objects can be attractive?        I.         Gravitational force   II.         Electrostatic force III.         Magnetic force

• A.

F. I

• B.

G. II

• C.

H. I, II, and III

• D.

J. II and III

C. H. I, II, and III
Explanation
The correct answer is H. I, II, and III. This means that all three forces, gravitational force, electrostatic force, and magnetic force, can be attractive. Gravitational force is the force of attraction between two objects with mass, electrostatic force is the force of attraction between two charged objects, and magnetic force is the force of attraction between two magnetic objects. Therefore, all three forces can result in attraction between objects.

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• 26.

### 21. The following diagram shows two bar magnets.    Which of the following describes the interaction between the two magnets?

• A.

A. The magnetic fields will cancel each other out.

• B.

B. The magnets will repel each other.

• C.

C. The magnets will attract each other.

• D.

D. There will be no interaction between the magnets.

B. B. The magnets will repel each other.
Explanation
The correct answer is B. The magnets will repel each other. This is because like poles of magnets (north and north or south and south) repel each other. Since both magnets in the diagram have their north poles facing each other, they will experience a repulsive force and push away from each other.

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• 27.

### 2. Two charged spheres are 20 cm apart and exert an attractive force of 8 x 10-9 N on each other. Which will be the force of attraction when the spheres are moved to 10 cm apart?   (Hint: Coulombs Law) r is distance

• A.

F. 2 x 10-9 N

• B.

G. 4 x 10-9 N

• C.

H. 1.6 x 10-8 N

• D.

J. 3.2 x 10-8 N

D. J. 3.2 x 10-8 N
Explanation
If distance is decreased by half the force is increased 4 times.

4 x 8 x 10-9 N = 3.2 x 10-8 N

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• 28.

### 23. The current supplied by the battery for the diagram is 0.17 amps. What is the potential difference across the 20.0 ohm resistor?   (Hint: V = I x R)

• A.

A. 1.7 volts

• B.

B. 3.4 volts

• C.

C. 5.0 volts

• D.

D. 30 volts

B. B. 3.4 volts
Explanation
The potential difference across a resistor can be calculated using Ohm's Law, which states that V (potential difference) is equal to I (current) multiplied by R (resistance). In this case, the current is given as 0.17 amps and the resistance is 20.0 ohms. Therefore, V = 0.17 amps x 20.0 ohms = 3.4 volts.

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• 29.

### 4. Two electrons (-1.6 x 10-19 coul) are separated by an insulator at a distance of 0.10 mm. Which of the following is the electrostatic force between these electrons?   (Hint: Coulombs Law)

• A.

F. +2.3 x 10-20 N

• B.

G. -2.3 x 10-20 N

• C.

H. + 2.8 x 10-40 N

• D.

J. -2.8 x 10-40 N

A. F. +2.3 x 10-20 N
Explanation
First you must convert .10 mm to meters = .0001 meters.
Then solve by coulombs law. Remember to square the distance.

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• Mar 22, 2023
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• Apr 02, 2014
Quiz Created by
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