Problem 43 - Fundamentals of Physics

Question

An unknown resistor is connected between the terminals of a \(3.00 \mathrm{~V}\) battery. Energy is dissipated in the resistor at the rate of \(0.540 \mathrm{~W}\). The same resistor is then connected between the terminals of a \(1.50 \mathrm{~V}\) battery. At what rate is energy now dissipated?

Step-by-Step Solution

Answer

The energy is dissipated at a rate of approximately 0.135 W.

Step 1: Use Ohm’s Law and Power Formula

The formulas to use here are Ohm’s Law, \( V = IR \), and the power formula, \( P = IV \). From the given data, let \( R \) be the resistance of the resistor, \( V = 3.00 \text{ V} \), and \( P = 0.540 \text{ W} \). We need to find \( R \) first using these formulas.

Step 2: Calculate Resistance

Using the power formula \( P = IV \) and Ohm’s Law \( V = IR \), we can express power as \( P = \frac{V^2}{R} \). Solving for \( R \) gives \( R = \frac{V^2}{P} = \frac{(3.00)^2}{0.540} = \frac{9.00}{0.540} \approx 16.67 \text{ ohms} \). Thus, the resistance \( R \) is approximately 16.67 ohms.

Step 3: Determine New Power Dissipation

Now, use the same formula \( P = \frac{V^2}{R} \) to find the power dissipated when the resistor is connected to the \( 1.50 \text{ V} \) battery. Substitute \( R = 16.67 \text{ ohms} \) and \( V = 1.50 \text{ V} \). So \( P = \frac{(1.50)^2}{16.67} = \frac{2.25}{16.67} \approx 0.135 \text{ W} \). Thus, the rate of energy dissipation is approximately 0.135 W.

Key Concepts

Power Formula

The power formula is an essential expression in understanding how electrical energy is converted into different forms, like heat, in a resistor. In a circuit, the power (P) dissipated by a resistor is directly related to the voltage (V) across it and the current (I) flowing through it. The formula can be written as:\[ P = IV \]This equation tells us that power is the product of voltage and current. However, using Ohm’s Law (V = IR), we can transform this formula for practical applications where resistance (R) is known. By substituting Ohm’s Law into the power formula, we get:\[ P = \frac{V^2}{R} \]This is especially useful when the voltage across the resistor and its resistance are known. It allows us to calculate the power dissipation without needing to know the current. Understanding the interplay between these variables helps in designing and analyzing electrical circuits efficiently.

Resistance Calculation

Resistance calculation is key to understanding how different materials and components will react in an electrical circuit. Resistance (R) refers to how much a device opposes the flow of electric current. We can calculate it using the formulas derived from Ohm’s Law and the power formula.From the power equation \( P = \frac{V^2}{R} \), we can rearrange it to solve for resistance:\[ R = \frac{V^2}{P} \]This formula shows that resistance can be computed if the voltage across the component and the power it consumes are known. In our exercise, when connected to a 3.00 V battery, the unknown resistor dissipates 0.540 W of power. Plugging these values into the formula gives:\[ R = \frac{(3.00)^2}{0.540} \approx 16.67 \text{ ohms} \]Hence, the resistance is approximately 16.67 ohms. Knowing the resistance aids in predicting how the electrical component will behave under different voltages.

Electrical Circuits

Electrical circuits are pathways through which electric current flows, consisting of various connected components. An understanding of circuits involves analyzing how voltage, current, and resistance affect each other. Circuits follow the principles of Ohm’s Law and the concept of energy conservation.In our example, an unknown resistor behaves differently when connected to different voltage sources. Initially attached to a 3.00 V battery, it dissipates energy at 0.540 W. When reconnected to a 1.50 V battery, the power dissipation changes, emphasizing how voltage affects power across a fixed resistance.The new power dissipation can be calculated with the power formula:\[ P = \frac{(1.50)^2}{16.67} \approx 0.135 \text{ W} \]This showcases an essential aspect of circuits: lowering the voltage across a resistor reduces its power dissipation, which is crucial for energy efficient designs. Understanding electrical circuits helps in engineering solutions and optimizing energy use.