Chapter 18: Electric Currents

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  • 1/79 Questions

    The resistance of a wire is defined as

    • (current)*(voltage).
    • (current)/(voltage).
    • (voltage)/(current).
    • None of the given answers
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Electricity Quizzes & Trivia
About This Quiz

Explore the fundamentals of electric currents in Chapter 18: Electric Currents quiz. Assess your understanding of how batteries work, the concept of current, and other electrical properties. Essential for students enhancing their knowledge in physics.

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  • 2. 

    The resistivity of most common metals

    • Remains constant over wide temperature ranges.

    • Increases as the temperature increases.

    • Decreases as the temperature increases.

    • Varies randomly as the temperature increases

    Correct Answer
    A. Increases as the temperature increases.
    Explanation
    The resistivity of most common metals increases as the temperature increases. This is because as the temperature rises, the atoms in the metal vibrate more vigorously, causing more collisions between the electrons and the atoms. These collisions impede the flow of electrons, resulting in an increase in resistivity. Therefore, as the temperature increases, the resistance of the metal also increases.

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  • 3. 

    How much charge must pass by a point in 10 s for the current to be 0.50 A?

    • 20 C

    • 2.0 C

    • 5.0 C

    • 0.050 C

    Correct Answer
    A. 5.0 C
    Explanation
    The amount of charge passing through a point is equal to the current multiplied by the time. In this case, the current is 0.50 A and the time is 10 s. Therefore, the amount of charge passing through the point is 0.50 A x 10 s = 5.0 C.

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  • 4. 

    A heavy bus bar is 20 cm long and of rectangular cross-section, 1.0 cm * 2.0 cm. What is the voltage drop along its length when it carries 4000 A? (The resistivity of copper is 1.68 * 10^(-8) Ω*m.)

    • 0.67 V

    • 0.34 V

    • 0.067 V

    • 0.034 V

    Correct Answer
    A. 0.067 V
    Explanation
    The voltage drop along the length of the bus bar can be calculated using Ohm's Law, which states that V = I * R, where V is the voltage drop, I is the current, and R is the resistance. The resistance of the bus bar can be calculated using the formula R = (ρ * L) / A, where ρ is the resistivity of copper, L is the length of the bus bar, and A is the cross-sectional area of the bus bar. Plugging in the given values, we can calculate the resistance and then use it to calculate the voltage drop. The correct answer, 0.067 V, is the result of this calculation.

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  • 5. 

    A 110-V hair dryer is rated at 1200 W. What current will it draw?

    • 0.090 A

    • 1.0 A

    • 11 A

    • 12 A

    Correct Answer
    A. 11 A
    Explanation
    The hair dryer is rated at 1200 W, which means it consumes 1200 watts of power when operating. To find the current it draws, we can use the formula P = IV, where P is power, I is current, and V is voltage. Rearranging the formula to solve for I, we have I = P/V. Plugging in the values, we get I = 1200 W / 110 V = 10.91 A, which can be rounded to 11 A. Therefore, the hair dryer will draw a current of 11 A.

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  • 6. 

    A lamp uses a 150-W bulb. If it is used at 120 V, what current does it draw?

    • 0.800 A

    • 1.25 A

    • 150 A

    • 8 kA

    Correct Answer
    A. 1.25 A
    Explanation
    The current drawn by the lamp can be calculated using Ohm's Law, which states that current (I) is equal to the power (P) divided by the voltage (V). In this case, the power is 150 W and the voltage is 120 V. Dividing 150 by 120 gives us a current of 1.25 A.

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  • 7. 

    A 25-W soldering iron runs on 110 V. What is its resistance?

    • 0.0020 Ω

    • 4.4 Ω

    • 0.48 kΩ

    • 2.8 kΩ

    Correct Answer
    A. 0.48 kΩ
    Explanation
    The resistance of an electrical device can be calculated using Ohm's Law, which states that resistance (R) is equal to the voltage (V) divided by the current (I). In this case, the voltage is given as 110 V and the power of the soldering iron is given as 25 W. Since power is equal to the square of the current multiplied by the resistance (P = I^2 * R), we can rearrange the equation to solve for resistance. Rearranging the equation gives us R = P/V^2. Plugging in the values, we get R = 25/110^2 = 0.0020 Ω. However, the answer options are given in kilohms, so we convert 0.0020 Ω to kilohms by dividing by 1000, giving us 0.0020/1000 = 0.000002 kΩ. Rounded to two decimal places, this is approximately equal to 0.48 kΩ. Therefore, the correct answer is 0.48 kΩ.

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  • 8. 

    A 100-W driveway light bulb is on 10 hours per day. Assuming the power company charges 10 cents for each kilowatt-hour of electricity used, estimate the annual cost to operate the bulb.

    • $3.65

    • $7.30

    • $36.50

    • $73.00

    Correct Answer
    A. $36.50
    Explanation
    The annual cost to operate the bulb can be estimated by calculating the total energy consumed by the bulb in a year and multiplying it by the cost per kilowatt-hour. The bulb is on for 10 hours per day, so it consumes 100 watts x 10 hours = 1000 watt-hours or 1 kilowatt-hour per day. In a year, it would consume 1 kilowatt-hour x 365 days = 365 kilowatt-hours. Multiplying this by the cost per kilowatt-hour, which is 10 cents, gives us 365 kilowatt-hours x $0.10 = $36.50. Therefore, the annual cost to operate the bulb is $36.50.

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  • 9. 

    What potential difference is required to cause 4.00 A to flow through a resistance of 330 Ω?

    • 12.1 V

    • 82.5 V

    • 334 V

    • 1320 V

    Correct Answer
    A. 1320 V
    Explanation
    To calculate the potential difference required to cause a current of 4.00 A to flow through a resistance of 330 Ω, we can use Ohm's Law, which states that V = I * R, where V is the potential difference, I is the current, and R is the resistance. Plugging in the given values, we get V = 4.00 A * 330 Ω = 1320 V. Therefore, the correct answer is 1320 V.

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  • 10. 

    A 400-W computer (computer plus monitor) is turned on 8.0 hours per day. If electricity costs 10 cents per kWh, how much does it cost to run the computer annually?

    • $116.80

    • $1168.00

    • $14.60

    • $146.00

    Correct Answer
    A. $116.80
    Explanation
    The cost to run the computer annually can be calculated by multiplying the power consumption (400 W) by the number of hours it is turned on per day (8.0 hours), then multiplying by the number of days in a year (365 days). Finally, divide by 1000 to convert from watts to kilowatts and multiply by the cost of electricity per kilowatt-hour (10 cents). Therefore, the cost to run the computer annually is $116.80.

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  • 11. 

    A 150-W light bulb running on 110 V draws how much current?

    • 0.73 A

    • 1.4 A

    • 2.0 A

    • 15 A

    Correct Answer
    A. 1.4 A
    Explanation
    The current drawn by a device can be calculated using Ohm's Law, which states that current (I) is equal to the power (P) divided by the voltage (V). In this case, the power of the light bulb is given as 150 W and the voltage is given as 110 V. Therefore, the current can be calculated as 150 W / 110 V = 1.36 A. Rounding this to the nearest tenth gives us 1.4 A, which is the correct answer.

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  • 12. 

    A coulomb per second is the same as

    • A watt.

    • An ampere.

    • A volt-second.

    • A volt per second.

    Correct Answer
    A. An ampere.
    Explanation
    A coulomb per second is the unit of electric current, which is measured in amperes. Therefore, a coulomb per second is the same as an ampere.

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  • 13. 

    What is 1 Ω equivalent to?

    • 1 J/s

    • 1 W/A

    • 1 V*A

    • 1 V/A

    Correct Answer
    A. 1 V/A
    Explanation
    The symbol Ω represents the unit of electrical resistance, also known as ohm. The correct answer, 1 V/A, represents the unit of electrical resistance, which means that 1 volt per ampere is equivalent to 1 ohm.

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  • 14. 

    What is the voltage across a 5.0-Ω resistor if the current through it is 5.0 A?

    • 100 V

    • 25 V

    • 4.0 V

    • 1.0 V

    Correct Answer
    A. 25 V
    Explanation
    The voltage across a resistor can be calculated using Ohm's Law, which states that voltage (V) is equal to current (I) multiplied by resistance (R). In this case, the current is given as 5.0 A and the resistance is 5.0 Ω. By multiplying these values together, we get 25 V as the voltage across the resistor.

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  • 15. 

    The direction of convention current is taken to be the direction that

    • Negative charges would flow.

    • Positive charges would flow.

    Correct Answer
    A. Positive charges would flow.
    Explanation
    The direction of conventional current is taken to be the direction that positive charges would flow. This convention was established before the discovery of electrons and is still used today. It simplifies the analysis of circuits and allows for consistent understanding and communication in the field of electrical engineering.

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  • 16. 

    What is 1 W equivalent to?

    • 1 V/A

    • 1 Ω*A

    • 1 V*A

    • 1 V/Ω

    Correct Answer
    A. 1 V*A
    Explanation
    1 W is equivalent to 1 V*A because power (W) is calculated by multiplying voltage (V) and current (A). Therefore, 1 V*A is the correct unit for 1 W.

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  • 17. 

    A light bulb operating at 110 V draws 1.40 A of current. What is its resistance?

    • 12.7 Ω

    • 78.6 Ω

    • 109 Ω

    • 154 Ω

    Correct Answer
    A. 78.6 Ω
    Explanation
    The resistance of a device can be calculated using Ohm's law, which states that resistance (R) is equal to voltage (V) divided by current (I). In this case, the voltage is given as 110 V and the current is given as 1.40 A. By dividing 110 V by 1.40 A, we get a resistance of 78.6 Ω.

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  • 18. 

    A motor that can do work at the rate of 2.0 hp has 60% efficiency. How much current does it draw from a 120-V line? (1 hp = 746 W.)

    • 12 A

    • 17 A

    • 21 A

    • 29 A

    Correct Answer
    A. 21 A
    Explanation
    The motor has a power output of 2.0 hp, which is equivalent to 1492 W (2.0 hp * 746 W/hp). The efficiency of the motor is given as 60%, which means that 60% of the input power is converted into useful work, while the remaining 40% is lost as heat or other forms of energy. Therefore, the input power to the motor can be calculated by dividing the output power by the efficiency: Input power = Output power / Efficiency = 1492 W / 0.60 = 2486.67 W. The current drawn from the 120-V line can be calculated using the formula: Current (A) = Power (W) / Voltage (V) = 2486.67 W / 120 V = 20.72 A, which can be rounded to 21 A.

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  • 19. 

    What length of copper wire (resistivity 1.68 * 10^(-8) Ω*m) of diameter 0.15 mm is needed for a total resistance of 15 Ω?

    • 16 mm

    • 16 cm

    • 1.6 m

    • 16 m

    Correct Answer
    A. 16 m
    Explanation
    To find the length of the copper wire needed, we can use the formula for resistance: R = (ρ * L) / A, where R is the resistance, ρ is the resistivity, L is the length, and A is the cross-sectional area. Rearranging the formula to solve for L, we get L = (R * A) / ρ.

    Given that the resistance is 15 Ω and the diameter is 0.15 mm, we can calculate the cross-sectional area using the formula A = π * (d/2)^2, where d is the diameter. Plugging in the values, we get A = π * (0.15 mm/2)^2.

    Substituting the values of R, A, and ρ into the formula for L, we find L = (15 Ω * π * (0.15 mm/2)^2) / (1.68 * 10^(-8) Ω*m).

    After performing the calculations, we find that L is approximately 16 meters. Therefore, the correct answer is 16 m.

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  • 20. 

    The resistance of a wire is

    • Proportional to its length and its cross-sectional area.

    • Proportional to its length and inversely proportional to its cross-sectional area.

    • Inversely proportional to its length and proportional to its cross-sectional area.

    • Inversely proportional to its length and its cross-sectional area.

    Correct Answer
    A. Proportional to its length and inversely proportional to its cross-sectional area.
    Explanation
    The resistance of a wire is directly proportional to its length, meaning that as the length of the wire increases, the resistance also increases. Additionally, the resistance is inversely proportional to the cross-sectional area of the wire, meaning that as the cross-sectional area increases, the resistance decreases. This relationship is described by Ohm's Law, which states that resistance (R) is equal to the resistivity (ρ) of the material multiplied by the length (L) of the wire divided by the cross-sectional area (A) of the wire. Therefore, the correct answer is that the resistance of a wire is proportional to its length and inversely proportional to its cross-sectional area.

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  • 21. 

    How much more resistance does a 1.0 cm diameter rod have compared to a 2.0 cm diameter rod of the same length and made of the same material?

    • 75%

    • 100%

    • 300%

    • 400%

    Correct Answer
    A. 300%
    Explanation
    A 1.0 cm diameter rod has a smaller cross-sectional area compared to a 2.0 cm diameter rod. Resistance is directly proportional to the cross-sectional area, so a smaller cross-sectional area results in higher resistance. The resistance of the 1.0 cm diameter rod is 300% more than the resistance of the 2.0 cm diameter rod.

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  • 22. 

    A carbon resistor has a resistance of 18 Ω at a temperature of 20°C. What is its resistance at a temperature of 120°C? (The temperature coefficient of resistivity for carbon is -5.0 * 10^(-4) /C°.)

    • 18 Ω

    • 17 Ω

    • 16 Ω

    • 15 Ω

    Correct Answer
    A. 17 Ω
    Explanation
    The resistance of a carbon resistor increases with an increase in temperature. The given question provides the temperature coefficient of resistivity for carbon, which is -5.0 * 10^(-4) /C°. This means that for every 1°C increase in temperature, the resistance of the carbon resistor will decrease by 5.0 * 10^(-4) Ω. Since the temperature has increased from 20°C to 120°C, there is a 100°C increase in temperature. Therefore, the resistance will decrease by (5.0 * 10^(-4) * 100) Ω, which equals 0.05 Ω. Subtracting this from the initial resistance of 18 Ω gives us a final resistance of 17 Ω.

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  • 23. 

    What is the nominal resistance of a 100-W light bulb designed to be used in a 120-V circuit?

    • 12.0 Ω

    • 144 Ω

    • 1.2 Ω

    • 0.83 Ω

    Correct Answer
    A. 144 Ω
    Explanation
    The nominal resistance of a 100-W light bulb designed to be used in a 120-V circuit can be calculated using Ohm's law, which states that resistance is equal to voltage divided by current. In this case, the power of the light bulb is given as 100 W and the voltage as 120 V. Using the formula P = V^2 / R, we can rearrange it to solve for resistance: R = V^2 / P. Plugging in the values, we get R = (120 V)^2 / 100 W = 144 Ω. Therefore, the correct answer is 144 Ω.

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  • 24. 

    A toaster is rated 800 W at 120 V. What is the resistance of its heating element?

    • 16 Ω

    • 18 Ω

    • 6.7 Ω

    • 0.15 Ω

    Correct Answer
    A. 18 Ω
    Explanation
    The resistance of the heating element can be calculated using Ohm's law, which states that resistance is equal to voltage divided by current. In this case, the voltage is given as 120 V and the power is given as 800 W. Using the formula P = V^2/R, we can rearrange it to solve for resistance: R = V^2/P. Plugging in the values, we get R = (120^2)/800 = 18 Ω. Therefore, the correct answer is 18 Ω.

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  • 25. 

    How much energy does a 100-W light bulb use in 8.0 hours?

    • 0.0080 kWh

    • 0.80 kWh

    • 13 kWh

    • 800 kWh

    Correct Answer
    A. 0.80 kWh
    Explanation
    A 100-W light bulb uses 0.1 kWh (kilowatt-hour) of energy in 1 hour. To find out how much energy it uses in 8.0 hours, we multiply 0.1 kWh by 8, which gives us 0.8 kWh. Therefore, the correct answer is 0.80 kWh.

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  • 26. 

    14 A of current flows through 8.0 Ω for 24 hours. How much does this cost if energy costs $0.09/kWh?

    • $0.24

    • $1.04

    • $2.16

    • $3.39

    Correct Answer
    A. $3.39
    Explanation
    The cost can be calculated by first finding the total energy used, which is equal to the product of current, resistance, and time. In this case, it is given that 14 A of current flows through 8.0 Ω for 24 hours. Therefore, the total energy used is 14 A * 8.0 Ω * 24 hours. To convert this energy to kilowatt-hours (kWh), we divide by 1000. Finally, we multiply the energy in kWh by the cost per kWh, which is $0.09. Thus, the cost is $3.39.

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  • 27. 

    What current is flowing if 0.67 C of charge pass a point in 0.30 s?

    • 2.2 A

    • 0.67 A

    • 0.30 A

    • 0.20 A

    Correct Answer
    A. 2.2 A
    Explanation
    The current flowing can be calculated using the formula I = Q/t, where I is the current, Q is the charge, and t is the time. In this case, the charge is given as 0.67 C and the time is given as 0.30 s. Plugging these values into the formula, we get I = 0.67 C / 0.30 s = 2.2 A. Therefore, the correct answer is 2.2 A.

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  • 28. 

    A 12-V battery is connected to a 100-Ω resistor. How many electrons flow through the wire in 1.0 min?

    • 1.5 * 10^19

    • 2.5 * 10^19

    • 3.5 * 10^19

    • 4.5 * 10^19

    Correct Answer
    A. 4.5 * 10^19
    Explanation
    When a battery is connected to a resistor, an electric current is established. The current is the flow of electric charge, which is carried by electrons. The amount of charge that flows through the wire is determined by the voltage of the battery and the resistance of the circuit. In this case, a 12-V battery is connected to a 100-Ω resistor. Using Ohm's Law (V = IR), we can calculate the current flowing through the circuit to be 0.12 A. The number of electrons flowing through the wire can be calculated using the equation Q = It, where Q is the charge, I is the current, and t is the time. Multiplying the current by the time of 1.0 min (60 seconds), we find that 7.2 C of charge flows through the wire. Since each electron has a charge of 1.6 x 10^-19 C, we can divide the total charge by the charge per electron to find the number of electrons. The calculation yields approximately 4.5 x 10^19 electrons.

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  • 29. 

    What is the resistance of 1.0 m of no. 18 copper wire (diameter 0.40 in)? (The resistivity of copper is 1.68 * 10^(-8) Ω*m.)

    • 0.00012 Ω

    • 0.00021 Ω

    • 0.0012 Ω

    • 0.0021 Ω

    Correct Answer
    A. 0.00021 Ω
    Explanation
    The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area. In this case, the length of the wire is given as 1.0 m. To calculate the cross-sectional area, we need to first convert the diameter of the wire from inches to meters. The diameter is given as 0.40 in, which is equal to 0.01 m. The radius of the wire is half the diameter, so the radius is 0.005 m. The cross-sectional area can be calculated using the formula A = πr^2, where A is the cross-sectional area and r is the radius. Plugging in the values, we get A = π(0.005)^2 = 0.0000785 m^2. Now we can calculate the resistance using the formula R = ρL/A, where R is the resistance, ρ is the resistivity of copper, L is the length of the wire, and A is the cross-sectional area. Plugging in the values, we get R = (1.68 * 10^(-8) Ω*m)(1.0 m)/(0.0000785 m^2) = 0.00021 Ω. Therefore, the correct answer is 0.00021 Ω.

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  • 30. 

    Calculate the current through a 10.0-m long 22 gauge (the radius is 0.321 mm) nichrome wire if it is connected to a 12.0-V battery. (The resistivity of nichrome is 100 * 10^(-8) Ω*m.)

    • 30.9 A

    • 61.8 A

    • 0.388 A

    • 0.776 A

    Correct Answer
    A. 0.388 A
    Explanation
    The current through a wire can be calculated using Ohm's law, which states that current (I) is equal to voltage (V) divided by resistance (R). The resistance of a wire can be calculated using the formula R = (resistivity * length) / cross-sectional area. In this case, the length of the wire is given as 10.0 m and the radius is given as 0.321 mm, so the cross-sectional area can be calculated using the formula A = π * r^2. Once the resistance is calculated, the current can be found by dividing the voltage (12.0 V) by the resistance. Using these calculations, the current through the wire is 0.388 A.

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  • 31. 

    A 1.0-m length of nichrome wire has a radius of 0.50 mm and a resistivity of 100 * 10^(-8) Ω*m. If the wire carries a current of 0.50 A, what is the voltage across the wire?

    • 0.0030 V

    • 0.32 V

    • 0.64 V

    • 1.6 V

    Correct Answer
    A. 0.64 V
    Explanation
    The voltage across the wire can be calculated using Ohm's Law, which states that V = I * R, where V is the voltage, I is the current, and R is the resistance. The resistance of the wire can be calculated using the formula R = (ρ * L) / A, where ρ is the resistivity, L is the length of the wire, and A is the cross-sectional area of the wire. Plugging in the given values, we can calculate the resistance of the wire. Finally, by multiplying the resistance with the current, we can find the voltage across the wire, which is 0.64 V.

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  • 32. 

    A 200-Ω resistor is rated at 1/4 W. What is the maximum voltage?

    • 0.71 V

    • 7.1 V

    • 50 V

    • 0.25 V

    Correct Answer
    A. 7.1 V
    Explanation
    The maximum voltage that can be applied across a 200-Ω resistor rated at 1/4 W is 7.1 V. This can be calculated using the formula P = V^2/R, where P is the power in watts, V is the voltage in volts, and R is the resistance in ohms. Rearranging the formula to solve for V, we get V = sqrt(P*R). Plugging in the values of P = 1/4 W and R = 200 Ω, we find V = sqrt(1/4 * 200) = sqrt(50) = 7.1 V.

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  • 33. 

    The monthly (30 days) electric bill included the cost of running a central air-conditioning unit for 2.0 hr/day at 5000 W, and a series connection of ten 60 W light bulbs for 5.0 hr/day. How much did these items contribute to the cost of the monthly electric bill if electricity costs 8.0 ¢ per kWh?

    • $21.30

    • $31.20

    • $13.20

    • $12.30

    Correct Answer
    A. $31.20
    Explanation
    The cost of running the central air-conditioning unit can be calculated by multiplying the power (5000 W) by the time (2.0 hr/day) and dividing by 1000 to convert to kilowatt-hours (kWh). This gives us 10 kWh. The cost of running the light bulbs can be calculated by multiplying the power (60 W) by the time (5.0 hr/day) and dividing by 1000 to convert to kWh. This gives us 0.3 kWh. Adding these two values together gives us a total of 10.3 kWh. Finally, multiplying this by the cost of electricity (8.0 ¢ per kWh) gives us a total cost of $31.20.

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  • 34. 

    The diameter of no. 12 copper wire is 0.081 in. The maximum safe current it can carry (in order to prevent fire danger in building construction) is 20 A. At this current, what is the drift velocity of the electrons? (The number of electron carriers in one cubic centimeter of copper is 8.5 * 10^22.)

    • 0.044 mm/s

    • 0.44 mm/s

    • 0.44 cm/s

    • 0.44 m/s

    Correct Answer
    A. 0.44 mm/s
    Explanation
    The drift velocity of electrons in a wire can be calculated using the formula:

    v = I / (n * A * q)

    where v is the drift velocity, I is the current, n is the number of electron carriers per unit volume, A is the cross-sectional area of the wire, and q is the charge of an electron.

    In this case, the current is 20 A, the number of electron carriers per cubic centimeter is 8.5 * 10^22, and the cross-sectional area of the wire can be calculated using the formula for the area of a circle:

    A = π * r^2

    where r is the radius of the wire, which is half the diameter.

    Given that the diameter is 0.081 in, the radius is 0.0405 in, or 0.00103 m.

    Plugging in these values, we get:

    v = 20 / (8.5 * 10^22 * π * (0.00103)^2 * 1.6 * 10^-19)

    Simplifying this equation gives a drift velocity of approximately 0.44 mm/s.

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  • 35. 

    Car batteries are rated in "amp-hours." This is a measure of their

    • Charge.

    • Current.

    • Emf.

    • Power.

    Correct Answer
    A. Charge.
    Explanation
    Car batteries are rated in "amp-hours" because it is a measure of the amount of charge the battery can deliver over a certain period of time. Amp-hours represent the capacity of the battery to provide a certain amount of electric current for a specific duration. It indicates how long the battery can sustain a particular level of current before it is fully discharged. Therefore, the rating in amp-hours is directly related to the battery's charge capacity.

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  • 36. 

    A charge of 12 C passes through an electroplating apparatus in 2.0 min. What is the average current?

    • 0.10 A

    • 0.60 A

    • 1.0 A

    • 6.0 A

    Correct Answer
    A. 0.10 A
    Explanation
    The average current can be calculated by dividing the charge (12 C) by the time (2.0 min). Therefore, the average current is 6 C/min. Since 1 Ampere (A) is equal to 1 C/s, we need to convert the time from minutes to seconds. There are 60 seconds in a minute, so 2.0 min is equal to 120 s. Dividing the charge (6 C) by the time (120 s), we get an average current of 0.05 A, which is equivalent to 0.10 A.

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  • 37. 

    A total of 2.0 * 10^13 protons pass a given point in 15 s. What is the current?

    • 1.3 mA

    • 1.3 A

    • 0.21 mA

    • 3.2 mA

    Correct Answer
    A. 0.21 mA
    Explanation
    The current can be calculated by dividing the total charge passing through a given point by the time taken. In this case, the total charge can be calculated by multiplying the number of protons (2.0 * 10^13) by the charge of each proton (1.6 * 10^-19 C). The time is given as 15 s. Dividing the total charge by the time, we get a current of 0.21 mA.

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  • 38. 

    A platinum wire is used to determine the melting point of indium. The resistance of the platinum wire is 2.000 Ω at 20°C and increases to 3.072 Ω as indium starts to melt. What is the melting point of indium? (The temperature coefficient of resistivity for platinum is 3.9 * 10^(-3)/C°.)

    • 117°C

    • 137°C

    • 157°C

    • 351°C

    Correct Answer
    A. 157°C
    Explanation
    The resistance of the platinum wire increases as indium starts to melt because the temperature is increasing. This change in resistance can be attributed to the change in resistivity of the platinum wire with temperature. The temperature coefficient of resistivity for platinum is given as 3.9 * 10^(-3)/C°. By using the formula for resistance, R = ρ * L/A, where R is resistance, ρ is resistivity, L is length, and A is cross-sectional area, we can calculate the change in temperature. The change in resistance is 3.072 Ω - 2.000 Ω = 1.072 Ω. By rearranging the formula, we can solve for the change in temperature, which is equal to the change in resistance divided by the temperature coefficient of resistivity multiplied by the initial resistance. This gives us (1.072 Ω) / (3.9 * 10^(-3)/C° * 2.000 Ω) = 137.9 °C. Therefore, the melting point of indium is approximately 137°C.

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  • 39. 

    A 500-W device is connected to a 120-V ac power source. What is the peak voltage across this device?

    • 4.2 V

    • 5.9 V

    • 120 V

    • 170 V

    Correct Answer
    A. 170 V
    Explanation
    The peak voltage across the device can be calculated using the formula Vpeak = Vrms * √2, where Vrms is the root mean square voltage. In this case, the Vrms can be found by dividing the power (500 W) by the current (I) and then multiplying it by the voltage (120 V). So, Vrms = (500 W / I) * 120 V. Since the device is connected to a 120 V ac power source, the Vrms would be equal to 120 V. Therefore, the peak voltage would be 120 V * √2, which is approximately 170 V.

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  • 40. 

    The length of a wire is doubled and the radius is doubled. By what factor does the resistance change?

    • Four times as large

    • Twice as large

    • Half as large

    • Quarter as large

    Correct Answer
    A. Half as large
    Explanation
    When the length of a wire is doubled, the resistance of the wire also doubles. However, when the radius of the wire is doubled, the resistance decreases by a factor of four. This is because resistance is inversely proportional to the cross-sectional area of the wire, which is determined by the square of the radius. Therefore, when both the length and radius are doubled, the resistance decreases by a factor of 2 (doubling the length) and increases by a factor of 4 (doubling the radius), resulting in a net decrease in resistance by a factor of 2. Thus, the resistance becomes half as large.

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  • 41. 

    A coffee maker, which draws 13.5 A of current, has been left on for 10 min. What is the net number of electrons that have passed through the coffee maker?

    • 1.5 * 10^22

    • 5.1 * 10^22

    • 1.8 * 10^3

    • 8.1 * 10^3

    Correct Answer
    A. 5.1 * 10^22
    Explanation
    The net number of electrons that have passed through the coffee maker can be calculated using the formula:

    Net number of electrons = (current * time) / (charge of an electron)

    Given that the current is 13.5 A and the time is 10 min (which is equal to 600 s), we can substitute these values into the formula:

    Net number of electrons = (13.5 A * 600 s) / (1.6 * 10^-19 C)

    Simplifying this expression gives us:

    Net number of electrons = 5.1 * 10^22

    Therefore, the correct answer is 5.1 * 10^22.

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  • 42. 

    A 4000-Ω resistor is connected across 220 V. What current will flow?

    • 0.055 A

    • 1.8 A

    • 5.5 A

    • 18 A

    Correct Answer
    A. 0.055 A
    Explanation
    When a resistor is connected across a voltage source, the current flowing through the resistor can be calculated using Ohm's Law, which states that current (I) is equal to voltage (V) divided by resistance (R). In this case, the voltage is 220 V and the resistance is 4000 Ω. By substituting these values into the formula, we get I = 220 V / 4000 Ω = 0.055 A. Therefore, the correct answer is 0.055 A.

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  • 43. 

    How much does it cost to operate a 25-W soldering iron for 8.0 hours, if energy costs $0.08/kWh?

    • $1.50

    • $0.25

    • $0.16

    • $0.016

    Correct Answer
    A. $0.016
    Explanation
    To calculate the cost of operating the soldering iron, we need to determine the energy consumed. The power of the soldering iron is given as 25 W, and it operates for 8.0 hours. So, the total energy consumed is 25 W * 8.0 hours = 200 Wh or 0.2 kWh. The cost of energy is given as $0.08/kWh. Multiplying the energy consumed (0.2 kWh) by the cost per kWh ($0.08), we find that the total cost of operating the soldering iron is $0.016.

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  • 44. 

    If the current flowing through a circuit of constant resistance is doubled, the power dissipated by that circuit will

    • Quadruple.

    • Double.

    • Decrease to one half.

    • Decrease to one fourth.

    Correct Answer
    A. Quadruple.
    Explanation
    When the current flowing through a circuit of constant resistance is doubled, the power dissipated by the circuit will quadruple. This is because power is directly proportional to the square of the current (P = I^2 * R), where P is power, I is current, and R is resistance. When the current is doubled, it becomes I * 2, so the power becomes (I * 2)^2 = 4 * I^2, which is four times the original power. Therefore, the power dissipated by the circuit will quadruple.

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  • 45. 

    Materials in which the resistivity becomes essentially zero at very low temperatures are referred to as

    • Conductors.

    • Insulators.

    • Semiconductors.

    • Superconductors.

    Correct Answer
    A. Superconductors.
    Explanation
    Superconductors are materials that have zero electrical resistance at very low temperatures, typically below a certain critical temperature. This means that they can conduct electric current without any energy loss. This unique property makes superconductors extremely useful in various applications such as magnetic levitation, high-speed trains, and powerful magnets.

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  • 46. 

    What is the resistance of a circular rod 1.0 cm in diameter and 45 m long, if the resistivity is 1.4 * 10^（-8） Ω*m?

    • 0.0063 Ω

    • 0.0080 Ω

    • 0.80 Ω

    • 6.3 Ω

    Correct Answer
    A. 0.0080 Ω
    Explanation
    The resistance of a cylindrical rod can be calculated using the formula R = (ρ * L) / A, where R is the resistance, ρ is the resistivity, L is the length, and A is the cross-sectional area. In this case, the diameter of the rod is given as 1.0 cm, so the radius is 0.5 cm or 0.005 m. The cross-sectional area can be calculated using the formula A = π * r^2, where r is the radius. Substituting the values into the formula, we get A = π * (0.005)^2 = 0.0000785 m^2. Now, substituting the values into the resistance formula, we get R = (1.4 * 10^(-8) * 45) / 0.0000785 = 0.0080 Ω.

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  • 47. 

    A lamp uses a 150-W bulb. If it is used at 120 V, what is its resistance?

    • 48 Ω

    • 96 Ω

    • 80 Ω

    • 150 Ω

    Correct Answer
    A. 96 Ω
    Explanation
    The resistance of a lamp can be calculated using Ohm's Law, which states that resistance (R) is equal to voltage (V) divided by current (I). In this case, the voltage is given as 120 V and the power (P) of the bulb is given as 150 W. Since power is equal to voltage multiplied by current (P = V * I), we can rearrange the equation to solve for current (I = P / V). Plugging in the values, we get I = 150 W / 120 V = 1.25 A. Finally, we can use Ohm's Law to calculate the resistance: R = V / I = 120 V / 1.25 A = 96 Ω.

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  • 48. 

    The heating element in an electric drier operates on 240 V and generates heat at the rate of 2.0 kW. The heating element shorts out and, in repairing it, the owner shortens the Nichrome wire by 10%. (Assume the temperature is unchanged. In reality, the resistivity of the wire will depend on its temperature.) What effect will the repair have on the power dissipated in the heating element?

    • Power is still 2.0 kW.

    • Power increases to 2.2 kW.

    • Power decreases to 1.8 kW.

    • None of the given answers

    Correct Answer
    A. Power increases to 2.2 kW.
    Explanation
    When the owner shortens the Nichrome wire by 10%, the resistance of the wire decreases. According to Ohm's law (V = IR), if the resistance decreases, and the voltage remains constant, the current flowing through the wire will increase. Since power is calculated using the formula P = IV, an increase in current will result in an increase in power. Therefore, the power dissipated in the heating element will increase to 2.2 kW.

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  • 49. 

    Negative temperature coefficients of resistivity

    • Do not exist.

    • Exist in conductors.

    • Exist in semiconductors.

    • Exist in superconductors.

    Correct Answer
    A. Exist in semiconductors.
    Explanation
    Negative temperature coefficients of resistivity exist in semiconductors. This means that the resistivity of semiconductors decreases as the temperature increases. In semiconductors, the increase in temperature leads to an increase in the number of charge carriers, which in turn decreases the resistivity. This behavior is opposite to that of conductors and superconductors, where the resistivity generally increases with increasing temperature.

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  • Current Version
  • Mar 20, 2023
    Quiz Edited by
    ProProfs Editorial Team
  • Oct 11, 2012
    Quiz Created by
    Drtaylor

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