# Chapter 18: Electric Currents

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• 1.

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

• A.

Generator.

• B.

Transformer.

• C.

Battery.

• D.

C. Battery.
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.

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• 2.

### A car battery

• A.

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

• B.

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

• C.

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

• D.

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

D. Has an emf of 12 V consisting of six 2-V cells connected in series.
• 3.

### 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

• A.

Current.

• B.

Electric potential.

• C.

Voltage.

• D.

Wattage.

A. Current.
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.

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• 4.

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

• A.

Negative charges would flow.

• B.

Positive charges would flow.

B. 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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• 5.

### A coulomb per second is the same as

• A.

A watt.

• B.

An ampere.

• C.

A volt-second.

• D.

A volt per second.

B. 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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• 6.

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

• A.

Charge.

• B.

Current.

• C.

Emf.

• D.

Power.

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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• 7.

### The resistance of a wire is defined as

• A.

(current)*(voltage).

• B.

(current)/(voltage).

• C.

(voltage)/(current).

• D.

C. (voltage)/(current).
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.

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• 8.

### What is 1 Ω equivalent to?

• A.

1 J/s

• B.

1 W/A

• C.

1 V*A

• D.

1 V/A

D. 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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• 9.

### The resistance of a wire is

• A.

Proportional to its length and its cross-sectional area.

• B.

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

• C.

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

• D.

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

B. 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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• 10.

### The resistivity of a wire depends on

• A.

Its length.

• B.

Its cross-sectional area.

• C.

The material out of which it is composed.

• D.

C. 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.

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• 11.

### Which conducting material has the lowest resistivity value?

• A.

Gold

• B.

Silver

• C.

Copper

• D.

Aluminum

B. Silver
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.

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• 12.

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

• A.

Both wires have the same resistivity.

• B.

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

• C.

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

• D.

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

A. Both wires have the same resistivity.
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.

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• 13.

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

• A.

Both wires have the same resistance.

• B.

The longer wire has half the resistance of the shorter wire

• C.

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

• D.

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

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• 14.

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

• A.

Both wires have the same resistance.

• B.

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

• C.

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

• D.

B. The thicker wire has 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.

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• 15.

### 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?

• A.

Both wires have the same resistance.

• B.

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

• C.

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

• D.

A. 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.

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• 16.

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

• A.

Four times as large

• B.

Twice as large

• C.

Half as large

• D.

Quarter as large

C. 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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• 17.

### 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?

• A.

75%

• B.

100%

• C.

300%

• D.

400%

C. 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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• 18.

### The resistivity of most common metals

• A.

Remains constant over wide temperature ranges.

• B.

Increases as the temperature increases.

• C.

Decreases as the temperature increases.

• D.

Varies randomly as the temperature increases

B. 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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• 19.

### Negative temperature coefficients of resistivity

• A.

Do not exist.

• B.

Exist in conductors.

• C.

Exist in semiconductors.

• D.

Exist in superconductors.

C. 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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• 20.

### What is 1 W equivalent to?

• A.

1 V/A

• B.

1 Ω*A

• C.

1 V*A

• D.

1 V/Ω

C. 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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• 21.

### A kilowatt-hour is equivalent to

• A.

1000 W.

• B.

3600 s.

• C.

3,600,000 J/s.

• D.

3,600,000 J.

D. 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.

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• 22.

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

• A.

Increase by a factor of two.

• B.

Increase by a factor of four.

• C.

Decrease to one-half its original value.

• D.

Decrease to one-fourth its original value.

C. Decrease 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.

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• 23.

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

• A.

• B.

Double.

• C.

Decrease to one half.

• D.

Decrease to one fourth.

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).

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• 24.

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

• A.

• B.

Double.

• C.

Decrease to one half.

• D.

Decrease to one fourth.

B. Double.
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.

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• 25.

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

• A.

• B.

Double.

• C.

Decrease to one half.

• D.

Decrease to one fourth.

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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• 26.

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

• A.

2.5% less

• B.

5.0% less

• C.

10% less

• D.

90% less

C. 10% less
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.

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• 27.

### A current that is sinusoidal with respect to time is referred to as

• A.

A direct current.

• B.

An alternating current.

B. An alternating 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.

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• 28.

### 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

• A.

4:1.

• B.

2:1.

• C.

1:2.

• D.

1:4.

A. 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.

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• 29.

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

• A.

Conductors.

• B.

Insulators.

• C.

Semiconductors.

• D.

Superconductors.

D. 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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• 30.

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

• A.

2.2 A

• B.

0.67 A

• C.

0.30 A

• D.

0.20 A

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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• 31.

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

• A.

0.10 A

• B.

0.60 A

• C.

1.0 A

• D.

6.0 A

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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• 32.

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

• A.

20 C

• B.

2.0 C

• C.

5.0 C

• D.

0.050 C

C. 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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• 33.

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

• A.

1.3 mA

• B.

1.3 A

• C.

0.21 mA

• D.

3.2 mA

C. 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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• 34.

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

• A.

0.013 A

• B.

0.31 A

• C.

6.3 A

• D.

78 A

A. 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.

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• 35.

### 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?

• A.

0.12 mA

• B.

0.24 mA

• C.

7.5 * 10^7 A

• D.

7.5 * 10^14 A

A. 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.

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• 36.

### 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?

• A.

1.5 * 10^22

• B.

5.1 * 10^22

• C.

1.8 * 10^3

• D.

8.1 * 10^3

B. 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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• 37.

### 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.)

• A.

16 s

• B.

18 s

• C.

20 s

• D.

22 s

B. 18 s
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.

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• 38.

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

• A.

12.1 V

• B.

82.5 V

• C.

334 V

• D.

1320 V

D. 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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• 39.

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

• A.

100 V

• B.

25 V

• C.

4.0 V

• D.

1.0 V

B. 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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• 40.

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

• A.

0.055 A

• B.

1.8 A

• C.

5.5 A

• D.

18 A

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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• 41.

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

• A.

12.7 Ω

• B.

78.6 Ω

• C.

109 Ω

• D.

154 Ω

B. 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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• 42.

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

• A.

1.5 * 10^19

• B.

2.5 * 10^19

• C.

3.5 * 10^19

• D.

4.5 * 10^19

D. 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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• 43.

### 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.)

• A.

0.00012 Ω

• B.

0.00021 Ω

• C.

0.0012 Ω

• D.

0.0021 Ω

B. 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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• 44.

### 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?

• A.

0.0063 Ω

• B.

0.0080 Ω

• C.

0.80 Ω

• D.

6.3 Ω

B. 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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• 45.

### 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?

• A.

16 mm

• B.

16 cm

• C.

1.6 m

• D.

16 m

D. 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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• 46.

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

• A.

0.023 Ω*m

• B.

0.015 Ω*m

• C.

1.5 * 10^(-4) Ω*m

• D.

2.3 * 10^(-6) Ω*m

D. 2.3 * 10^(-6) Ω*m
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.

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• 47.

### 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.)

• A.

30.9 A

• B.

61.8 A

• C.

0.388 A

• D.

0.776 A

C. 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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• 48.

### 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.)

• A.

0.67 V

• B.

0.34 V

• C.

0.067 V

• D.

0.034 V

C. 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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• 49.

### 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?

• A.

0.0030 V

• B.

0.32 V

• C.

0.64 V

• D.

1.6 V

C. 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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• 50.

### 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?

• A.

12 V

• B.

23 V

• C.

32 V

• D.

41 V

C. 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.

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• Mar 20, 2023
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• Oct 11, 2012
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