Problem 33 - College Physics

Question

A resistor with a 15.0 \(\mathrm{V}\) potential difference across its ends develops thermal energy at a rate of 327 \(\mathrm{W}\) . (a) What is the current in the resistor? (b) What is its resistance?

Step-by-Step Solution

Answer

The current is 21.8 A and the resistance is approximately 0.688 Ω.

Step 1: Understand Power Formula

The power (\(P\)) dissipated by a resistor in an electrical circuit is given by the formula: \( P = V \cdot I \), where \( V \) is the potential difference across the resistor and \( I \) is the current flowing through it. Rearrange this formula to solve for the current \(I\).

Step 2: Calculate Current

Given: \( V = 15.0 \mathrm{V} \) and \( P = 327 \mathrm{W} \). Use the formula \( P = V \cdot I \) and substitute the given values: \( 327 = 15.0 \times I \). Rearrange to solve for \( I \):\[ I = \frac{327}{15.0} \]Calculate \( I \):\[ I = 21.8 \mathrm{A} \] (to one decimal place).

Step 3: Understand Ohm's Law

Ohm's Law states \( V = I \cdot R \), where \( R \) is the resistance. Rearrange this formula to solve for the resistance \( R \):\[ R = \frac{V}{I} \].

Step 4: Calculate Resistance

Use the Ohm's Law formula for resistance. Substitute the known values of \( V = 15.0 \mathrm{V} \) and \( I = 21.8 \mathrm{A} \):\[ R = \frac{15.0}{21.8} \]Calculate \( R \):\[ R \approx 0.688 \Omega \] (to three decimal places).

Key Concepts

Electric Current

Electric current refers to the flow of electric charge through a conductor. It is analogous to the flow of water in a pipeline. Just as water flows from a region of high pressure to low pressure, electric current flows from a region of higher potential (voltage) to a region of lower potential.

Electric current is measured in amperes (A), named after André-Marie Ampère, a pioneer in electromagnetism. One ampere corresponds to a charge flow of one coulomb per second.

• The formula that relates electric current with power and voltage is: \[ I = \frac{P}{V} \]
• Where, \( I \) is the current in amperes, \( P \) is the power in watts, and \( V \) is the voltage in volts.

In the context of our exercise, the current through the resistor is calculated using the given power and voltage, resulting in a current of 21.8 A.

Power Dissipation

Power dissipation in a resistor refers to the process by which an electrical component converts electric energy into heat. This occurs due to the resistance offered to the flow of current.

According to the formula \( P = V \cdot I \), power is the product of the potential difference (voltage) across the resistor and the current flowing through it. This power loss manifests as heat, a phenomenon harnessed in devices such as electric heaters and incandescent bulbs.

• To calculate power dissipation: \[ P = V \cdot I \]Where \( P \) is in watts (W), \( V \) is in volts (V), and \( I \) is in amperes (A).
• Another useful formula involves resistance: \[ P = I^2 \cdot R \]Where \( R \) is resistance in ohms.

In the given problem, 327 watts of power is dissipated across the resistor, causing it to convert this energy into heat.

Resistor

A resistor is a fundamental component in electrical circuits, designed to limit the flow of electric current. Think of it like a narrow section of a pipe that slows the flow of water. The resistor controls the flow of current and divides voltage within the circuit.

Ohm's Law, \( V = I \cdot R \), is the primary relationship used to understand the behavior of resistors in circuits.

• Here, \( R \) denotes the resistance, measured in ohms (Ω). The unit honors George Ohm, who explored the relationship between voltage, current, and resistance.
• Resistance is calculated using the rearranged Ohm's Law: \[ R = \frac{V}{I} \]

In the provided exercise, the resistance is calculated to be approximately 0.688 ohms, illustrating the resistor's opposition to the current flow. Resistors are key in both managing current levels, and safely distributing voltage in an electrical circuit.