Problem 21 - Physics

Question

The heating element in an iron has a resistance of \(24 \Omega\). The iron is plugged into a \(120-\mathrm{V}\) outlet. What is the power delivered to the iron?

Step-by-Step Solution

Answer

The power delivered to the iron is 600 W.

Step 1: Identify the known values

We are given the resistance of the heating element and the voltage of the outlet. Specifically, the resistance \( R \) is \( 24 \Omega \) and the voltage \( V \) is \( 120 \mathrm{V} \). These values will be used in our calculations.

Step 2: Recall the formula for power

The power delivered to an electrical device can be calculated using the formula:\[ P = \frac{V^2}{R} \]where \( P \) is the power, \( V \) is the voltage, and \( R \) is the resistance.

Step 3: Substitute the known values into the formula

Now, substitute the known values \( V = 120 \mathrm{V} \) and \( R = 24 \Omega \) into the power formula:\[ P = \frac{(120)^2}{24} \]

Step 4: Solve for power

Calculate the value by performing the arithmetic:\[ P = \frac{14400}{24} = 600 \]Thus, the power delivered to the iron is \( 600 \mathrm{W} \).

Key Concepts

Ohm's Law

Ohm's Law is a fundamental principle that describes the relationship between voltage, current, and resistance in an electrical circuit. It is crucial for predicting how electricity flows through a conductor. The law is expressed by the formula:
\( V = I \times R \)
where:

• \( V \) is the voltage across the conductor in volts (V).
• \( I \) is the current flowing through the conductor in amperes (A).
• \( R \) is the resistance of the conductor in ohms (\( \Omega \)).

This formula shows that the voltage (the force pushing the electric charges) is directly proportional to the current and resistance. Therefore, if we know any two of the variables, we can calculate the third. In our exercise, Ohm’s Law helps us understand how resistance and voltage interact to determine the current passing through the heating element. This understanding is crucial when calculating the power using the formula: \( P = \frac{V^2}{R} \), where the behavior of voltage and resistance directly affects power output.

Electric Resistance

Electric resistance is a measure of how much a material opposes the flow of electric current. It's like the electrical equivalent of friction. The more resistance, the harder it is for electricity to flow through the material.

• Measured in ohms (\( \Omega \)).
• Represented by the letter \( R \).
• It depends on material, length, cross-sectional area, and temperature.

In the original exercise, the heating element in the iron has a resistance of \( 24 \Omega \). This value plays a crucial role in determining how much current flows through the iron when connected to a voltage source. High resistance means that there will be less current if the voltage is constant, impacting the power calculated using \( P = \frac{V^2}{R} \). In practical terms, materials with low resistance, like copper, are often used for efficient power distribution, whereas high-resistance materials are used to limit current flow or for heating applications, like in electric irons.

Voltage

Voltage is essentially the electrical pressure that pushes electric charges through a conductor. It is what makes current flow in a circuit in the first place. Often described as "potential difference," it represents the energy difference per charge between two points in a circuit.

• Measured in volts (V).
• Indicates how much work can be done by moving charges in an electric field.
• Can be provided by power sources like batteries or outlets.

In our problem, a voltage of \( 120 \mathrm{V} \) provides the energy needed to allow current to flow through the heating element of the iron. The higher the voltage, the more potential energy is available to do work — and in this case, to generate heat within the iron. Voltage is an essential factor in determining the power delivered to the iron using the formula \( P = \frac{V^2}{R} \). Hence, understanding voltage's role helps not only in calculating power but also in designing circuits with appropriate safety and efficiency measures.