Chapter 18: Electric Currents

1. The resistance of a wire is defined as

(current)*(voltage).

(current)/(voltage).

(voltage)/(current).

None of the given answers

The resistance of a wire is defined as the ratio of voltage to current. This is based on Ohm's Law, which states that the current flowing through a conductor is directly proportional to the voltage applied across it and inversely proportional to its resistance. Therefore, the correct answer is (voltage)/(current), as it represents the relationship between voltage and current in determining the resistance of a wire.

Explanation

The resistance of a wire is defined as the ratio of voltage to current. This is based on Ohm's Law, which states that the current flowing through a conductor is directly proportional to the voltage applied across it and inversely proportional to its resistance. Therefore, the correct answer is (voltage)/(current), as it represents the relationship between voltage and current in determining the resistance of a wire.

2. 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

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.

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.

3. 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

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.

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.

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

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.

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.

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

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.

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.

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

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.

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.

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

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.

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.

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

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.

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.

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

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.

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.

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

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.

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.

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

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.

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.

12. A coulomb per second is the same as

A watt.

An ampere.

A volt-second.

A volt per second.

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.

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.

13. What is 1 Ω equivalent to?

1 J/s

1 W/A

1 V*A

1 V/A

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.

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.

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

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.

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.

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

Negative charges would flow.

Positive charges would flow.

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.

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.

16. 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

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.

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.

17. What is 1 W equivalent to?

1 V/A

1 Ω*A

1 V*A

1 V/Ω

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.

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.

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

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.

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.

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

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.

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.

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.

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.

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.

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%

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.

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.

22. 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

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.

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.

23. 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

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.

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.

24. 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

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.

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.

25. 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

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.

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.

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

0.0080 kWh

0.80 kWh

13 kWh

800 kWh

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.

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.

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

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.

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.

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

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.

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.

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

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.

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.

30. 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

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.

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.

31. 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

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.

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.

32. 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

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.

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.

33. 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

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.

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.

34. 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

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.

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.

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

Charge.

Current.

Emf.

Power.

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.

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.

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

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.

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.

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

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.

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.

38. 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

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.

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.

39. 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

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.

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.

40. 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

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.

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.

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

0.055 A

1.8 A

5.5 A

18 A

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.

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.

42. 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

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.

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.

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

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.

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.

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.

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.

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.

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

Conductors.

Insulators.

Semiconductors.

Superconductors.

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.

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.

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

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.

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.

47. 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

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.

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.

48. 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

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.

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.

49. Negative temperature coefficients of resistivity

Do not exist.

Exist in conductors.

Exist in semiconductors.

Exist in superconductors.

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.

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.

50. A current that is sinusoidal with respect to time is referred to as

A direct current.

An alternating current.

A current that is sinusoidal with respect to time is referred to as an alternating current. This is because an alternating current continuously changes direction, oscillating back and forth in a sinusoidal pattern. In contrast, a direct current flows in only one direction without changing. Therefore, the given correct answer accurately describes the nature of a sinusoidal current.

Explanation

A current that is sinusoidal with respect to time is referred to as an alternating current. This is because an alternating current continuously changes direction, oscillating back and forth in a sinusoidal pattern. In contrast, a direct current flows in only one direction without changing. Therefore, the given correct answer accurately describes the nature of a sinusoidal current.

51. If the resistance in a circuit with constant current flowing is doubled, the power dissipated by that circuit will

Quadruple.

Double.

Decrease to one half.

Decrease to one fourth.

When the resistance in a circuit with constant current flowing is doubled, according to Ohm's Law (V = IR), the voltage across the circuit will also double. Since power is calculated using the formula P = IV, where I is the current and V is the voltage, doubling the voltage will result in the power being doubled as well. Therefore, the power dissipated by the circuit will double when the resistance is doubled.

Explanation

When the resistance in a circuit with constant current flowing is doubled, according to Ohm's Law (V = IR), the voltage across the circuit will also double. Since power is calculated using the formula P = IV, where I is the current and V is the voltage, doubling the voltage will result in the power being doubled as well. Therefore, the power dissipated by the circuit will double when the resistance is doubled.

52. A 500-W device is connected to a 120-V ac power source. What rms current flows through this device?

4.2 A

5.9 A

120 A

170 A

The rms current flowing through a device can be calculated using Ohm's Law, which states that current (I) is equal to power (P) divided by voltage (V). In this case, the power is given as 500 W and the voltage is 120 V. By substituting these values into the formula, we can calculate the current as 500 W / 120 V = 4.2 A. Therefore, the correct answer is 4.2 A.

Explanation

The rms current flowing through a device can be calculated using Ohm's Law, which states that current (I) is equal to power (P) divided by voltage (V). In this case, the power is given as 500 W and the voltage is 120 V. By substituting these values into the formula, we can calculate the current as 500 W / 120 V = 4.2 A. Therefore, the correct answer is 4.2 A.

53. During a power demand, the voltage output is reduced by 5.0%. By what percentage is the power on a resistor affected?

2.5% less

5.0% less

10% less

90% less

When the voltage output is reduced by 5.0%, the power on a resistor is affected by a greater percentage. This is because power is directly proportional to the square of the voltage. By reducing the voltage by 5.0%, the power is reduced by (5.0%)^2 = 25.0%. Therefore, the power on a resistor is 25.0% less, which is equivalent to 10% less than the original power.

Explanation

When the voltage output is reduced by 5.0%, the power on a resistor is affected by a greater percentage. This is because power is directly proportional to the square of the voltage. By reducing the voltage by 5.0%, the power is reduced by (5.0%)^2 = 25.0%. Therefore, the power on a resistor is 25.0% less, which is equivalent to 10% less than the original power.

54. A device that produces electricity by transforming chemical energy into electrical energy is called a

Generator.

Transformer.

Battery.

None of the given answers

A battery is a device that converts chemical energy into electrical energy through a chemical reaction. It consists of two electrodes, an anode and a cathode, which are immersed in an electrolyte solution. The chemical reaction between the electrodes and the electrolyte produces a flow of electrons, creating an electric current. This process allows the battery to generate electricity and power various devices. Both generators and transformers are devices that manipulate electrical energy, but they do not directly convert chemical energy into electrical energy like a battery does.

Explanation

A battery is a device that converts chemical energy into electrical energy through a chemical reaction. It consists of two electrodes, an anode and a cathode, which are immersed in an electrolyte solution. The chemical reaction between the electrodes and the electrolyte produces a flow of electrons, creating an electric current. This process allows the battery to generate electricity and power various devices. Both generators and transformers are devices that manipulate electrical energy, but they do not directly convert chemical energy into electrical energy like a battery does.

55. A kilowatt-hour is equivalent to

1000 W.

3600 s.

3,600,000 J/s.

3,600,000 J.

A kilowatt-hour is a unit of energy, which is equivalent to the amount of energy consumed by a device with a power of 1 kilowatt (1000 watts) over a period of 1 hour. To convert this to joules, we can use the formula: 1 kilowatt-hour = 3,600,000 joules. Therefore, the correct answer is 3,600,000 J.

Explanation

A kilowatt-hour is a unit of energy, which is equivalent to the amount of energy consumed by a device with a power of 1 kilowatt (1000 watts) over a period of 1 hour. To convert this to joules, we can use the formula: 1 kilowatt-hour = 3,600,000 joules. Therefore, the correct answer is 3,600,000 J.

56. A 9.0-V battery costs $1.49, and will run a portable CD player for 6.0 hours. Suppose the battery supplies a current of 25 mA to the player. What is the cost of energy in dollars per kWh?

$11/kWh

$110/kWh

$1100/kWh

$11,000/kWh

The cost of the battery is given as $1.49 and it runs the CD player for 6.0 hours. The current supplied by the battery is 25 mA. To find the cost of energy in dollars per kWh, we need to calculate the energy consumed by the CD player in kWh.

First, we need to convert the current from mA to A by dividing it by 1000. So, the current is 0.025 A.

Next, we can calculate the energy consumed by multiplying the voltage (9.0 V) by the current (0.025 A) and the time (6.0 hours). This gives us an energy of 1.35 Wh (watt-hours).

To convert this to kWh, we divide by 1000, giving us 0.00135 kWh.

Finally, we can calculate the cost of energy per kWh by dividing the cost of the battery ($1.49) by the energy consumed (0.00135 kWh). This gives us a cost of $1100/kWh.

Explanation

The cost of the battery is given as $1.49 and it runs the CD player for 6.0 hours. The current supplied by the battery is 25 mA. To find the cost of energy in dollars per kWh, we need to calculate the energy consumed by the CD player in kWh.

First, we need to convert the current from mA to A by dividing it by 1000. So, the current is 0.025 A.

Next, we can calculate the energy consumed by multiplying the voltage (9.0 V) by the current (0.025 A) and the time (6.0 hours). This gives us an energy of 1.35 Wh (watt-hours).

To convert this to kWh, we divide by 1000, giving us 0.00135 kWh.

Finally, we can calculate the cost of energy per kWh by dividing the cost of the battery ($1.49) by the energy consumed (0.00135 kWh). This gives us a cost of $1100/kWh.

57. If the voltage across a circuit of constant resistance is doubled, the power dissipated by that circuit will

Quadruple.

Double.

Decrease to one half.

Decrease to one fourth.

When the voltage across a circuit of constant resistance is doubled, according to Ohm's Law (V = IR), the current flowing through the circuit will also double. Since power is calculated as P = IV, when both the voltage and current double, the power dissipated by the circuit will increase by a factor of four (quadruple).

Explanation

When the voltage across a circuit of constant resistance is doubled, according to Ohm's Law (V = IR), the current flowing through the circuit will also double. Since power is calculated as P = IV, when both the voltage and current double, the power dissipated by the circuit will increase by a factor of four (quadruple).

58. Consider two copper wires. One has twice the length of the other. How do the resistivities of these two wires compare?

Both wires have the same resistivity.

The longer wire has twice the resistivity of the shorter wire.

The longer wire has four times the resistivity of the shorter wire.

The resistivity of a material is a property that depends on the material itself, not on its dimensions. Therefore, the length of the wire does not affect its resistivity. In this case, since both wires are made of copper, they have the same resistivity regardless of their lengths.

Explanation

The resistivity of a material is a property that depends on the material itself, not on its dimensions. Therefore, the length of the wire does not affect its resistivity. In this case, since both wires are made of copper, they have the same resistivity regardless of their lengths.

59. Consider two copper wires. One has twice the length and twice the cross-sectional area of the other. How do the resistances of these two wires compare?

Both wires have the same resistance.

The longer wire has twice the resistance of the shorter wire.

The longer wire has four times the resistance of the shorter wire.

None of the given answers

The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area. In this case, the longer wire has twice the length and twice the cross-sectional area of the shorter wire. Since the increase in length is compensated by the increase in cross-sectional area, the resistance of the longer wire remains the same as the resistance of the shorter wire. Therefore, both wires have the same resistance.

Explanation

The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area. In this case, the longer wire has twice the length and twice the cross-sectional area of the shorter wire. Since the increase in length is compensated by the increase in cross-sectional area, the resistance of the longer wire remains the same as the resistance of the shorter wire. Therefore, both wires have the same resistance.

60. A 1.0-mm diameter copper wire (resistivity 1.68 * 10^(-8) Ω*m) carries a current of 15 A. What is the potential difference between two points 100 m apart?

12 V

23 V

32 V

41 V

The potential difference between two points in a conductor can be calculated using Ohm's Law, which states that V = I * R, where V is the potential difference, I is the current, and R is the resistance. In this case, the resistance can be determined 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. Given that the diameter of the wire is 1.0 mm, we can calculate the radius (r) as 0.5 mm or 0.0005 m. The cross-sectional area can be determined using the formula A = π * r^2. Plugging in the values, we can find the resistance. Finally, we can substitute the resistance and the given current into Ohm's Law to find the potential difference, which is 32 V.

Explanation

The potential difference between two points in a conductor can be calculated using Ohm's Law, which states that V = I * R, where V is the potential difference, I is the current, and R is the resistance. In this case, the resistance can be determined 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. Given that the diameter of the wire is 1.0 mm, we can calculate the radius (r) as 0.5 mm or 0.0005 m. The cross-sectional area can be determined using the formula A = π * r^2. Plugging in the values, we can find the resistance. Finally, we can substitute the resistance and the given current into Ohm's Law to find the potential difference, which is 32 V.

61. 5.00 A is flowing through an 10.0 Ω resistor. How much power is being dissipated?

50.0 W

250 W

500 W

2.50 kW

The power dissipated in a resistor can be calculated using the formula P = I^2 * R, where P is power, I is current, and R is resistance. In this case, the current flowing through the resistor is 5.00 A and the resistance is 10.0 Ω. Plugging these values into the formula, we get P = (5.00 A)^2 * 10.0 Ω = 250 W. Therefore, the correct answer is 250 W.

Explanation

The power dissipated in a resistor can be calculated using the formula P = I^2 * R, where P is power, I is current, and R is resistance. In this case, the current flowing through the resistor is 5.00 A and the resistance is 10.0 Ω. Plugging these values into the formula, we get P = (5.00 A)^2 * 10.0 Ω = 250 W. Therefore, the correct answer is 250 W.

62. A battery is rated at 12 V and 160 A-h. How much energy does the battery store?

1.9 kJ

6.0 kJ

1.9 MJ

6.9 MJ

The energy stored in a battery can be calculated by multiplying the voltage (V) by the capacity (A-h). In this case, the battery is rated at 12 V and 160 A-h. Multiplying these values gives us 1920 Wh or 1.92 kWh. To convert this to megajoules (MJ), we can multiply by the conversion factor of 3.6. Therefore, the battery stores 6.9 MJ of energy.

Explanation

The energy stored in a battery can be calculated by multiplying the voltage (V) by the capacity (A-h). In this case, the battery is rated at 12 V and 160 A-h. Multiplying these values gives us 1920 Wh or 1.92 kWh. To convert this to megajoules (MJ), we can multiply by the conversion factor of 3.6. Therefore, the battery stores 6.9 MJ of energy.

63. How much energy does a 25-W soldering iron use in 8.0 hours?

400 J

11 kJ

12 kJ

0.72 MJ

A 25-W soldering iron uses 25 joules of energy every second. In 8.0 hours, there are 8.0 x 60 x 60 = 28,800 seconds. Therefore, the total energy used by the soldering iron is 25 x 28,800 = 720,000 joules, which is equal to 0.72 MJ (mega-joules).

Explanation

A 25-W soldering iron uses 25 joules of energy every second. In 8.0 hours, there are 8.0 x 60 x 60 = 28,800 seconds. Therefore, the total energy used by the soldering iron is 25 x 28,800 = 720,000 joules, which is equal to 0.72 MJ (mega-joules).

64. In an electroplating process, it is desired to deposit 40 mg of silver on a metal part by using a current of 2.0 A. How long must the current be allowed to run to deposit this much silver? (The silver ions are singly charged, and the atomic weight of silver is 108.)

16 s

18 s

20 s

22 s

To calculate the time required to deposit 40 mg of silver, we can use Faraday's law of electrolysis. The formula is:

Amount of substance deposited = (Current × Time) / (Charge on the ion × Avogadro's number × Atomic weight)

In this case, the amount of substance deposited is 40 mg, the current is 2.0 A, the charge on the ion is 1 (since the silver ions are singly charged), the atomic weight of silver is 108 g/mol, and Avogadro's number is 6.022 × 10^23 mol^-1.

Plugging in these values into the formula, we get:

40 mg = (2.0 A × Time) / (1 × 6.022 × 10^23 mol^-1 × 108 g/mol)

Simplifying the equation, we find:

Time = (40 mg × 1 × 6.022 × 10^23 mol^-1 × 108 g/mol) / (2.0 A)

Calculating this expression, we get the answer of 18 seconds.

Explanation

To calculate the time required to deposit 40 mg of silver, we can use Faraday's law of electrolysis. The formula is:

Amount of substance deposited = (Current × Time) / (Charge on the ion × Avogadro's number × Atomic weight)

In this case, the amount of substance deposited is 40 mg, the current is 2.0 A, the charge on the ion is 1 (since the silver ions are singly charged), the atomic weight of silver is 108 g/mol, and Avogadro's number is 6.022 × 10^23 mol^-1.

Plugging in these values into the formula, we get:

40 mg = (2.0 A × Time) / (1 × 6.022 × 10^23 mol^-1 × 108 g/mol)

Simplifying the equation, we find:

Time = (40 mg × 1 × 6.022 × 10^23 mol^-1 × 108 g/mol) / (2.0 A)

Calculating this expression, we get the answer of 18 seconds.

65. Which conducting material has the lowest resistivity value?

Gold

Silver

Copper

Aluminum

Silver has the lowest resistivity value among the given options. This means that silver offers the least resistance to the flow of electric current compared to gold, copper, and aluminum. Silver is known for its high electrical conductivity, making it an excellent choice for applications where low resistance is desired, such as in electrical wiring and circuitry.

Explanation

Silver has the lowest resistivity value among the given options. This means that silver offers the least resistance to the flow of electric current compared to gold, copper, and aluminum. Silver is known for its high electrical conductivity, making it an excellent choice for applications where low resistance is desired, such as in electrical wiring and circuitry.

66. A 1.5-cm square rod, 4.0 m long, measures 0.040 ohms. What is its resistivity?

0.023 Ω*m

0.015 Ω*m

1.5 * 10^(-4) Ω*m

2.3 * 10^(-6) Ω*m

The resistivity of a material is a measure of how strongly it resists the flow of electric current. It is given by 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 question, we are given the resistance (0.040 ohms), the length (4.0 m), and the cross-sectional area (1.5 cm^2). By rearranging the formula and plugging in the given values, we can solve for the resistivity. The correct answer, 2.3 * 10^(-6) Ω*m, is obtained by converting the length from cm to m and solving for ρ.

Explanation

The resistivity of a material is a measure of how strongly it resists the flow of electric current. It is given by 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 question, we are given the resistance (0.040 ohms), the length (4.0 m), and the cross-sectional area (1.5 cm^2). By rearranging the formula and plugging in the given values, we can solve for the resistivity. The correct answer, 2.3 * 10^(-6) Ω*m, is obtained by converting the length from cm to m and solving for ρ.

67. Consider two copper wires each carrying a current of 3.0 A. One wire has twice the diameter of the other. The ratio of the drift velocity in the smaller diameter wire to that in the larger diameter wire is

4:1.

2:1.

1:2.

1:4.

The drift velocity of charged particles in a wire is inversely proportional to the cross-sectional area of the wire. Since the wire with twice the diameter has four times the cross-sectional area, the drift velocity in this wire will be one-fourth of the drift velocity in the smaller diameter wire. Therefore, the ratio of the drift velocity in the smaller diameter wire to that in the larger diameter wire is 4:1.

Explanation

The drift velocity of charged particles in a wire is inversely proportional to the cross-sectional area of the wire. Since the wire with twice the diameter has four times the cross-sectional area, the drift velocity in this wire will be one-fourth of the drift velocity in the smaller diameter wire. Therefore, the ratio of the drift velocity in the smaller diameter wire to that in the larger diameter wire is 4:1.

68. The total amount of charge that passes through a wire's full cross section at any point per unit of time is referred to as

Current.

Electric potential.

Voltage.

Wattage.

The total amount of charge that passes through a wire's full cross section at any point per unit of time is referred to as current. Current is measured in amperes (A) and represents the flow of electric charge. It is a fundamental concept in electricity and is essential for understanding the behavior of electrical circuits. Electric potential refers to the amount of electric potential energy per unit charge, voltage is the difference in electric potential between two points, and wattage is the rate at which electrical energy is transferred or used.

Explanation

The total amount of charge that passes through a wire's full cross section at any point per unit of time is referred to as current. Current is measured in amperes (A) and represents the flow of electric charge. It is a fundamental concept in electricity and is essential for understanding the behavior of electrical circuits. Electric potential refers to the amount of electric potential energy per unit charge, voltage is the difference in electric potential between two points, and wattage is the rate at which electrical energy is transferred or used.

69. If the resistance in a constant voltage circuit is doubled, the power dissipated by that circuit will

Increase by a factor of two.

Increase by a factor of four.

Decrease to one-half its original value.

Decrease to one-fourth its original value.

When the resistance in a constant voltage circuit is doubled, the power dissipated by the circuit decreases to one-half its original value. This is because power is directly proportional to the square of the resistance (P = V^2/R), so when the resistance is doubled, the power is reduced by a factor of four. Therefore, the correct answer is that the power dissipated by the circuit decreases to one-half its original value.

Explanation

When the resistance in a constant voltage circuit is doubled, the power dissipated by the circuit decreases to one-half its original value. This is because power is directly proportional to the square of the resistance (P = V^2/R), so when the resistance is doubled, the power is reduced by a factor of four. Therefore, the correct answer is that the power dissipated by the circuit decreases to one-half its original value.

70. A 500-W device is connected to a 120-V ac power source. What peak current flows through this device?

4.2 A

5.9 A

120 A

170 A

The peak current flowing through the device can be calculated using the formula I = P/V, where I is the current, P is the power, and V is the voltage. In this case, the power is 500 W and the voltage is 120 V. Substituting these values into the formula, we get I = 500/120 = 4.1667 A. Rounding this to the nearest tenth, we get 4.2 A. Therefore, the correct answer is 4.2 A.

Explanation

The peak current flowing through the device can be calculated using the formula I = P/V, where I is the current, P is the power, and V is the voltage. In this case, the power is 500 W and the voltage is 120 V. Substituting these values into the formula, we get I = 500/120 = 4.1667 A. Rounding this to the nearest tenth, we get 4.2 A. Therefore, the correct answer is 4.2 A.

71. The resistivity of a wire depends on

Its length.

Its cross-sectional area.

The material out of which it is composed.

All of the given answers

The resistivity of a wire depends on the material out of which it is composed. Different materials have different resistivities, which is a measure of how strongly they resist the flow of electric current. The length and cross-sectional area of the wire also affect its resistance, but they do not determine the resistivity of the material itself. Therefore, the correct answer is that the resistivity of a wire depends on the material out of which it is composed.

Explanation

The resistivity of a wire depends on the material out of which it is composed. Different materials have different resistivities, which is a measure of how strongly they resist the flow of electric current. The length and cross-sectional area of the wire also affect its resistance, but they do not determine the resistivity of the material itself. Therefore, the correct answer is that the resistivity of a wire depends on the material out of which it is composed.

72. If 3.0 * 10^15 electrons flow through a section of a wire of diameter 2.0 mm in 4.0 s, what is the current in the wire?

0.12 mA

0.24 mA

7.5 * 10^7 A

7.5 * 10^14 A

The current in the wire can be determined using Ohm's Law, which states that current (I) is equal to the charge (Q) passing through a conductor per unit time (t). In this case, the charge is given by the product of the number of electrons (3.0 * 10^15) and the charge of each electron (1.6 * 10^-19 C). The time is given as 4.0 s. Therefore, the total charge passing through the wire is (3.0 * 10^15) * (1.6 * 10^-19) C. Dividing this by the time (4.0 s), we get the current in amperes. Converting this to milliamperes by multiplying by 1000, we find that the current is 0.12 mA.

Explanation

The current in the wire can be determined using Ohm's Law, which states that current (I) is equal to the charge (Q) passing through a conductor per unit time (t). In this case, the charge is given by the product of the number of electrons (3.0 * 10^15) and the charge of each electron (1.6 * 10^-19 C). The time is given as 4.0 s. Therefore, the total charge passing through the wire is (3.0 * 10^15) * (1.6 * 10^-19) C. Dividing this by the time (4.0 s), we get the current in amperes. Converting this to milliamperes by multiplying by 1000, we find that the current is 0.12 mA.

73. A 200-Ω resistor is rated at 1/4 W. What is the maximum current it can draw?

0.035 A

0.35 A

50 A

0.25 A

The maximum current that a resistor can draw is determined by the power rating and the resistance. In this case, the resistor is rated at 1/4 W and has a resistance of 200 Ω. Using the formula P = I^2 * R, we can solve for the current (I). Rearranging the formula, we get I = sqrt(P/R). Plugging in the values, we find I = sqrt(1/4 / 200) = sqrt(0.0025) = 0.05 A = 0.035 A. Therefore, the maximum current the resistor can draw is 0.035 A.

Explanation

The maximum current that a resistor can draw is determined by the power rating and the resistance. In this case, the resistor is rated at 1/4 W and has a resistance of 200 Ω. Using the formula P = I^2 * R, we can solve for the current (I). Rearranging the formula, we get I = sqrt(P/R). Plugging in the values, we find I = sqrt(1/4 / 200) = sqrt(0.0025) = 0.05 A = 0.035 A. Therefore, the maximum current the resistor can draw is 0.035 A.

74. What current is flowing if 4.0 * 10^16 electrons pass a point in 0.50 s?

0.013 A

0.31 A

6.3 A

78 A

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 can be determined by multiplying the number of electrons by the charge of a single electron, which is 1.6 * 10^-19 C. The time is given as 0.50 s. Plugging in the values, the current is calculated to be 0.013 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 can be determined by multiplying the number of electrons by the charge of a single electron, which is 1.6 * 10^-19 C. The time is given as 0.50 s. Plugging in the values, the current is calculated to be 0.013 A.

75. Consider two copper wires. One has twice the cross-sectional area of the other. How do the resistances of these two wires compare?

Both wires have the same resistance.

The thicker wire has half the resistance of the shorter wire.

The thicker wire has twice the resistance of the shorter wire.

None of the given answers

The resistance of a wire is inversely proportional to its cross-sectional area. Therefore, if one wire has twice the cross-sectional area of the other, it means that it has half the resistance. This is because a larger cross-sectional area allows for more space for the flow of electrons, resulting in less resistance to the current. Thus, the thicker wire will have half the resistance of the shorter wire.

Explanation

The resistance of a wire is inversely proportional to its cross-sectional area. Therefore, if one wire has twice the cross-sectional area of the other, it means that it has half the resistance. This is because a larger cross-sectional area allows for more space for the flow of electrons, resulting in less resistance to the current. Thus, the thicker wire will have half the resistance of the shorter wire.

76. Consider two copper wires. One has twice the length of the other. How do the resistances of these two wires compare?

Both wires have the same resistance.

The longer wire has half the resistance of the shorter wire

The longer wire has twice the resistance of the shorter wire.

None of the given answers

The resistance of a wire is directly proportional to its length. Therefore, if one wire has twice the length of the other, it will also have twice the resistance. This is because the longer wire has more atoms for the electrons to collide with, resulting in more resistance to the flow of current.

Explanation

The resistance of a wire is directly proportional to its length. Therefore, if one wire has twice the length of the other, it will also have twice the resistance. This is because the longer wire has more atoms for the electrons to collide with, resulting in more resistance to the flow of current.

77. A car battery

Has an emf of 6 V consisting of one 6-V cell.

Has an emf of 6 V consisting of three 2-V cells connected in series.

Has an emf of 6 V consisting of three 2-V cells connected in parallel.

Has an emf of 12 V consisting of six 2-V cells connected in series.

not-available-via-ai

Explanation

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78. The temperature coefficient of resistivity of platinum is 3.9 * 10^(-3)/C°. If a platinum wire has a resistance of R at room temperature (23°C), to what temperature must it be heated in order to double its resistance to 2R?

279°C

297°C

729°C

927°C

The temperature coefficient of resistivity of platinum indicates how its resistance changes with temperature. In this case, the coefficient is given as 3.9 * 10^(-3)/C°. To double the resistance of the platinum wire, the temperature must be increased. Using the formula for temperature coefficient of resistivity, we can calculate the change in temperature. Since the resistance is doubled, the change in resistance is R, and the change in temperature is ΔT. Rearranging the formula, we get ΔT = (R * coefficient) / resistance. Plugging in the values, we find ΔT = (R * 3.9 * 10^(-3)/C°) / R = 3.9 * 10^(-3)/C°. Therefore, the temperature must increase by 3.9 * 10^(-3)/C° to double the resistance. Adding this change to the initial temperature of 23°C gives us 279°C.

Explanation

The temperature coefficient of resistivity of platinum indicates how its resistance changes with temperature. In this case, the coefficient is given as 3.9 * 10^(-3)/C°. To double the resistance of the platinum wire, the temperature must be increased. Using the formula for temperature coefficient of resistivity, we can calculate the change in temperature. Since the resistance is doubled, the change in resistance is R, and the change in temperature is ΔT. Rearranging the formula, we get ΔT = (R * coefficient) / resistance. Plugging in the values, we find ΔT = (R * 3.9 * 10^(-3)/C°) / R = 3.9 * 10^(-3)/C°. Therefore, the temperature must increase by 3.9 * 10^(-3)/C° to double the resistance. Adding this change to the initial temperature of 23°C gives us 279°C.

79. A 1500-W heater is connected to a 120-V line for 2.0 hours. How much heat energy is produced?

1.5 kJ

3.0 kJ

0.18 MJ

11 MJ

The heat energy produced can be calculated using the formula:
Energy = Power x Time.
Given that the power of the heater is 1500 W and the time is 2.0 hours, we can convert the power to kilowatts (1 kW = 1000 W) and multiply it by the time in seconds (1 hour = 3600 seconds) to get the energy in joules. Then, we convert the energy from joules to megajoules (1 MJ = 10^6 J) to get the final answer. Therefore, the correct answer is 0.18 MJ.

Explanation

The heat energy produced can be calculated using the formula:
Energy = Power x Time.
Given that the power of the heater is 1500 W and the time is 2.0 hours, we can convert the power to kilowatts (1 kW = 1000 W) and multiply it by the time in seconds (1 hour = 3600 seconds) to get the energy in joules. Then, we convert the energy from joules to megajoules (1 MJ = 10^6 J) to get the final answer. Therefore, the correct answer is 0.18 MJ.