Calculate the resistance of a circuit if the voltage drop is 7.20V with a current of 50A.

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Multiple Choice

Calculate the resistance of a circuit if the voltage drop is 7.20V with a current of 50A.

Explanation:
To determine the resistance of a circuit given the voltage drop and the current, Ohm's Law can be applied, which states that resistance (R) is equal to voltage (V) divided by current (I). In this case, you have a voltage drop of 7.20 volts and a current of 50 amps. By using the formula: \[ R = \frac{V}{I} \] we can substitute in the values: \[ R = \frac{7.20 \text{ V}}{50 \text{ A}} \] Calculating this gives: \[ R = 0.144 \text{ ohms} \] When you round 0.144 to two decimal places, it corresponds to 0.14 ohms. This value falls in line with the calculated resistance, confirming that it is the correct answer. Understanding that voltage dropped across a resistance in a circuit is directly proportional to the current flowing through it is central to the application of Ohm's Law. Thus, the calculated resistance helps in analyzing the circuit's behavior and in designing components to handle specific electrical loads effectively.

To determine the resistance of a circuit given the voltage drop and the current, Ohm's Law can be applied, which states that resistance (R) is equal to voltage (V) divided by current (I).

In this case, you have a voltage drop of 7.20 volts and a current of 50 amps. By using the formula:

[ R = \frac{V}{I} ]

we can substitute in the values:

[ R = \frac{7.20 \text{ V}}{50 \text{ A}} ]

Calculating this gives:

[ R = 0.144 \text{ ohms} ]

When you round 0.144 to two decimal places, it corresponds to 0.14 ohms. This value falls in line with the calculated resistance, confirming that it is the correct answer.

Understanding that voltage dropped across a resistance in a circuit is directly proportional to the current flowing through it is central to the application of Ohm's Law. Thus, the calculated resistance helps in analyzing the circuit's behavior and in designing components to handle specific electrical loads effectively.

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