# Unit 8 Quiz #3 Targets 7-8

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• 1.

### A total charge of 76 C passes through a cross-sectional area of a copper wire in 19 s. What is the current in the wire?

• A.

4

• B.

4 W

• C.

4 A

• D.

4 V

C. 4 A
Explanation
The current in the wire is 4 A. Current is defined as the rate at which charge flows through a conductor. In this case, a total charge of 76 C passes through the wire in 19 s. To calculate the current, we divide the total charge by the time taken: 76 C / 19 s = 4 A. Therefore, the current in the wire is 4 A.

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• 2.

### A total charge of 114 uC passes through a cross-sectional area of an aluminum wire in 0.36 s. What is the current in the wire?

• A.

41.04 A

• B.

316.67 A

• C.

4.104 x 10^-5 A

• D.

3.17 x 10^-4 A

D. 3.17 x 10^-4 A
Explanation
The current in a wire is calculated by dividing the total charge passing through the wire by the time it takes. In this case, the total charge is given as 114 uC and the time is given as 0.36 s. Dividing 114 uC by 0.36 s gives a current of 316.67 A. However, the answer is given in scientific notation as 3.17 x 10^-4 A, which is the same value.

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• 3.

### Electric eels, found in South America, can provide a potential difference of 440 V that draws a current of 0.80 A through the eel’s prey. Calculate the resistance of the circuit (the eel and prey).

• A.

550 Ohm

• B.

0.002 Ohm

• C.

352 Ohm

• D.

55 Ohm

A. 550 Ohm
Explanation
The resistance of a circuit can be calculated using Ohm's Law, which states that resistance is equal to the voltage divided by the current. In this case, the potential difference provided by the electric eel is 440 V and the current through the circuit is 0.80 A. By dividing the voltage by the current, we get 550 Ohm as the resistance of the circuit.

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• 4.

### It is claimed that a certain camcorder battery can provide a potential difference of 9.60 V and a current of 1.50 A. What is the resistance through which the battery must be discharged?

• A.

14.4 Ohm

• B.

6.4 Ohm

• C.

0.16 Ohm

• D.

11.1 Ohm

B. 6.4 Ohm
Explanation
The resistance through which the battery must be discharged 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 potential difference (V) is given as 9.60 V and the current (I) is given as 1.50 A. Therefore, the resistance (R) can be calculated as 9.60 V divided by 1.50 A, which equals 6.4 Ohm.

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• 5.

### You have probably heard that high-voltages are more dangerous than low voltages. To understand this, assume that your body has a resistance of 1.0 x 105 Ohm. What potential difference would have to be across your body to produce a current of 1.0 mA (which would cause a tingling feeling) and 15 mA (a fatal amount of current)?

• A.

1 x 10^8 V and 6.67 x 10^6 V

• B.

1 x 10^5 V and 6666.67 V

• C.

100 V and 1500 V

• D.

1 x 10^5 V and 1.5 x 10^6 V

C. 100 V and 1500 V
Explanation
The potential difference needed to produce a current of 1.0 mA is 100 V, which would cause a tingling feeling. This is because the current is directly proportional to the potential difference according to Ohm's law (V = IR), so a lower potential difference will result in a lower current. The potential difference needed to produce a fatal amount of current (15 mA) is 1500 V. This higher potential difference is required to overcome the resistance of the body and allow a larger current to flow.

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• 6.

### A generator at a central electric power plant produces electricity with a potential difference of 2.5 x 104 V across power lines which carry a current of 20.0 A. How much power does the generator produce?

• A.

1250 W

• B.

8 x 10^-4 W

• C.

500,000 W

• D.

1 x 10^7 W

C. 500,000 W
Explanation
The power produced by the generator can be calculated using the formula P = VI, where P is power, V is potential difference, and I is current. In this case, the potential difference is 2.5 x 10^4 V and the current is 20.0 A. Plugging these values into the formula, we get P = (2.5 x 10^4 V) x (20.0 A) = 5 x 10^5 W. Therefore, the generator produces 500,000 W of power.

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• 7.

### An electric sports car was developed several years ago at Texas A&M University in College Station, Texas. If the potential difference across the car’s motor is 720 V and the resistance was 0.30 Ohm, how much power was needed for the car to run?

• A.

1.73 x 10^6 W

• B.

216 W

• C.

64.8 W

• D.

2400 W

A. 1.73 x 10^6 W
Explanation
The power (P) can be calculated using the formula P = (V^2) / R, where V is the potential difference and R is the resistance. Plugging in the given values, we get P = (720^2) / 0.30 = 1.73 x 10^6 W. Therefore, the power needed for the car to run is 1.73 x 10^6 W.

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• 8.

### Suppose that you’ve just returned to work from your lunch hour. When you reach your desk, you realize that you had forgotten to turn off your computer. Fortunately, your computer was in its energy-conserving “sleep” mode. What is the power of a computer which consumed 2.7 x 108 J of energy?

• A.

2.7 x 10^8 W

• B.

9.72 x 10^11 W

• C.

75,000 W

• D.

2.62 x 10^20 W

C. 75,000 W
• 9.

### Suppose you’ve just returned to the parking lot from a 3.0 h shopping spree at the mall and realized that you had forgotten to turn off the headlights to your car. If 4.86 x 108 J of energy was spent, what is the power of the headlights?

• A.

450,000 W

• B.

1.62 x 10^8 W

• C.

1.46 x 10^9 W

• D.

4.5 x 10^4 W

D. 4.5 x 10^4 W
Explanation
The question asks for the power of the headlights, which is the rate at which energy is consumed. The energy spent is given as 4.86 x 10^8 J, and the time taken is 3.0 h. To find the power, we divide the energy by the time. Therefore, the power of the headlights is 4.86 x 10^8 J / 3.0 h = 1.62 x 10^8 W.

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• 10.

### A total charge of 76 C passes through a cross-sectional area of a copper wire in 19 s. What is the current in the wire?

• A.

4 A

• B.

4 W

• C.

1444 A

• D.

1444 W

A. 4 A
Explanation
The current in the wire 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 total charge passing through the wire is 76 C and the time is 19 s. Plugging these values into the formula, we get I = 76 C / 19 s = 4 A. Therefore, the current in the wire is 4 A.

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