Problem 34 - University Physics with Modern Physics

Question

When a resistor with resistance \(R\) is connected to a 1.50-V flashlight battery, the resistor consumes 0.0625 W of electrical power. (Throughout, assume that each battery has negligible internal resistance.) (a) What power does the resistor consume if it is connected to a 12.6-V car battery? Assume that \(R\) remains constant when the power consumption changes. (b) The resistor is connected to a battery and consumes 5.00 W. What is the voltage of this battery?

Step-by-Step Solution

Answer

(a) 4.41 W with the 12.6-V battery. (b) Voltage is 13.42 V when the resistor consumes 5.00 W.

Step 1: Understanding Power Formula

The power consumed by a resistor is given by the formula \[ P = \frac{V^2}{R} \]where \(P\) is power, \(V\) is voltage, and \(R\) is the resistance.

Step 2: Determine Resistance using 1.50-V Battery

Given that with a 1.50-V battery, the resistor consumes 0.0625 W of power, we can use the formula for power:\[ 0.0625 = \frac{1.50^2}{R} \]Solving for \(R\),\[ R = \frac{1.50^2}{0.0625} = \frac{2.25}{0.0625} = 36 \ \Omega \]

Step 3: Calculate Power with 12.6-V Battery

Now, knowing \(R = 36 \ \Omega\), calculate the power consumed with a 12.6-V battery.\[ P = \frac{12.6^2}{36} \]\[ P = \frac{158.76}{36} = 4.41 \ W \]

Step 4: Solve for Voltage when Power is 5.00 W

Given the same \(R = 36 \ \Omega\) and a power consumption of 5.00 W, use the power formula:\[ 5.00 = \frac{V^2}{36} \]Solving for \(V^2\),\[ V^2 = 5.00 \times 36 = 180 \]Therefore,\[ V = \sqrt{180} = 13.42 \ V \]

Key Concepts

Resistance Calculation

Resistance is a fundamental concept in electrical circuits. It's a measure of how much a resistor resists the flow of electric current. In this exercise, we started with a given power consumption and a known voltage to determine the resistance of a resistor. Using the power formula:\[ P = \frac{V^2}{R} \]The resistance \( R \) can be calculated if you rearrange the formula to:\[ R = \frac{V^2}{P} \]By substituting the given values from the exercise, with the voltage \( V = 1.50 \, V \) and power \( P = 0.0625 \, W \), we found that the resistance \( R = 36 \, \Omega \). This means the resistor restricts the flow of current such that it maintains this power consumption at the given voltage. Whenever you are achieving a resistance calculation, ensure all the units are consistent and check for accuracy by substituting back into the original power formula.

Power Formula

The power formula is an essential tool in electrical engineering and physics. It helps us calculate the power consumed by a circuit component using its voltage and resistance. The formula used in the task is:\[ P = \frac{V^2}{R} \]Here, \( P \) denotes the power in watts (W), \( V \) is the voltage in volts (V), and \( R \) is the resistance in ohms (\( \Omega \)).Power: Reflects how fast energy is being used or consumed by the resistor. It's a direct outcome of both voltage and resistance. The higher the power, the more energy is consumed quickly.

• Voltage (\( V \)): How much potential energy is provided to push current through the resistor.
• Resistance (\( R \)): Determines how much the resistor will oppose this flow of current.

Understanding this formula allows you to predict how a circuit will behave as different voltages are applied. It is a crucial stepping stone for many electrical calculations.

Voltage Calculation

Voltage represents the potential difference across a resistor. In this problem, we needed to calculate voltage when the power and resistance are known. To find voltage, the formula derived from the power equation is:\[ V = \sqrt{P \times R} \]For instance, to find the voltage needed to drive a 5.00 W power consumption through a resistor of 36 \( \Omega \), we used:\[ V^2 = 5.00 \times 36 = 180 \]And solved for \( V \):\[ V = \sqrt{180} \approx 13.42 \, V \]This voltage calculation helps us understand the potential energy necessary for a given power consumption. It also shows the tight relationship between power, resistance, and voltage — changing one inevitably affects the others, a key consideration when designing and troubleshooting circuits.

Ohm's Law

Though Ohm's Law was not directly used in the steps, it's a core principle underlying the calculations for resistance, power, and voltage. It states the relationship between voltage (V), current (I), and resistance (R) in an electric circuit:\[ V = I \times R \]What Ohm's Law tells us is:

• Voltage (V): The potential difference that pushes the current through a resistor.
• Current (I): The flow of electric charge through the resistor affected by the voltage and resistance.
• Resistance (R): Property of the resistor that limits current flow.

This foundational law helps explain how various components interact within an electrical system. It serves as the background for understanding why calculations like those done in the exercise can predict circuit behavior. Knowing Ohm's Law equips you with the ability to navigate complex circuits and solve for unknown quantities.