Problem 100 - Physics

Question

A cigarette lighter in a car is a resistor that, when activated, is connected across the 12\(V\) battery. Suppose a lighter uses \(33 \mathrm{~W}\) of power. Find (a) the resistance of the lighter and (b) the current that the battery delivers to the lighter.

Step-by-Step Solution

Answer

The resistance of the lighter is approximately 4.36 Ω, and the current is approximately 2.75 A.

Step 1: Determining the Resistance - Step 1

Use the power formula for electrical systems, which is given by \( P = V^2 / R \) where \( P \) represents power, \( V \) represents voltage, and \( R \) represents resistance. Rearrange it to solve for the resistance \( R = V^2 / P \).

Step 2: Calculate the Resistance - Step 2

Plug in the known values into the rearranged power formula: \( P = 33 \, W \) and \( V = 12 \, V \). Calculate \( R \) using \( R = (12)^2 / 33 \). This results in \( R = 144 / 33 \approx 4.36 \, \Omega \).

Step 3: Determining the Current - Step 3

Use Ohm's Law, which states that \( I = V / R \), where \( I \) is the current. We will use \( V = 12 \, V \) and the previously calculated resistance \( R \approx 4.36 \, \Omega \).

Step 4: Calculate the Current - Final Step

Substitute the given values into Ohm’s Law: \( I = 12 / 4.36 \). This results in \( I \approx 2.75 \, A \).

Key Concepts

Resistance

Resistance is an essential concept in understanding electrical circuits. It opposes or resists the flow of electric current through a conductor, much like how friction resists motion. Its unit of measurement is ohms, symbolized by \( \Omega \). When calculating resistance, one of the frequently used formulas involves power and voltage, expressed as \( R = V^2 / P \), where \( R \) is the resistance, \( V \) is the voltage, and \( P \) is the power.

In the scenario with the car's cigarette lighter, we have a power \( P = 33 \, W \) and a voltage \( V = 12 \, V \). By rearranging and applying these values to the formula, we find:
- \( R = (12)^2 / 33 = 144 / 33 \)
- This calculation results in a resistance of approximately \( 4.36 \, \Omega \).

This value indicates how much the lighter opposes the flow of electric current. A higher resistance would mean less current flows through the lighter for the same voltage. Understanding resistance aids in diagnosing how components will behave within a circuit.

Electric Current

Electric current is all about the flow of electric charge. It's like the traffic of electrons in the circuit highway and is measured in amperes (\( A \)). To better grasp current, use Ohm's Law, which states \( I = V / R \). Here, \( I \) represents the current, \( V \) stands for the voltage across the component, and \( R \) is the resistance.

For the cigarette lighter example, after determining the resistance of \( 4.36 \, \Omega \), we can solve for the current using the car's 12\( V \) supply. Using Ohm's Law:
- Plug the values into the formula: \( I = 12 / 4.36 \).
- Calculate to find \( I \approx 2.75 \, A \).

This result means the battery delivers approximately \( 2.75 \, A \) of current to the lighter. Understanding electric current is crucial because it's the rate at which electric charge passes through a point in the circuit. A higher current often implies more energy is being transferred.

Power Formula

The power formula in electrical engineering connects the dots between power, voltage, and resistance. It helps determine how much work an electrical device performs over time. Power is measured in watts (\( W \)) and can be calculated using \( P = V^2 / R \). This version of the formula is particularly helpful when you know the voltage and resistance, but need to find the power consumption of a device. However, it can be rearranged to solve for any variable when the others are known.

In our cigarette lighter problem, the power \( P = 33 \, W \) is given, and this formula was rearranged to find the resistance shown by \( R = V^2 / P \). Calculating further with given inputs brings us to \( R \) equating to \( 4.36 \, \Omega \), showing how the formula efficiently ties these electric quantities together.

Additionally, knowing the power allows us to foresee how much energy is being used, crucial for applications from conserving energy to predicting battery life. Understanding the power formula is key in the design and analysis of electrical systems.