Problem 35 - University Physics with Modern Physics

Question

The power rating of a light bulb (such as a 100-W bulb) is the power it dissipates when connected across a 120-V potential difference. What is the resistance of (a) a 100-W bulb and (b) a 60-W bulb? (c) How much current does each bulb draw in normal use?

Step-by-Step Solution

Answer

(a) 144 Ω, (b) 240 Ω; (c) 0.833 A (100-W), 0.5 A (60-W)

Step 1: Understanding Ohm's Law and Power

Ohm's Law states that the potential difference (V) across a resistor is equal to the current (I) through it times its resistance (R), or \( V = I \times R \). The power (P) dissipated by a resistor can be expressed as \( P = V \times I \). These equations are crucial for determining the resistance and current.

Step 2: Calculating Resistance for a 100-W Bulb

Using the power formula \( P = V^2 / R \), we can rearrange to solve for resistance: \( R = V^2 / P \). For a 100-W bulb with 120-V:\[ R = \frac{120^2}{100} = \frac{14400}{100} = 144 \, \Omega \].

Step 3: Calculating Resistance for a 60-W Bulb

Apply the same formula for the 60-W bulb: \( R = V^2 / P \). Inserting the values:\[ R = \frac{120^2}{60} = \frac{14400}{60} = 240 \, \Omega \].

Step 4: Finding Current for a 100-W Bulb

Using the power formula \( P = V \times I \), solve for current: \( I = P / V \). For the 100-W bulb:\[ I = \frac{100}{120} \approx 0.833 \, \text{A} \].

Step 5: Finding Current for a 60-W Bulb

For the 60-W bulb, use the same method: \( I = P / V \). Therefore:\[ I = \frac{60}{120} = 0.5 \, \text{A} \].

Key Concepts

Electric Resistance

Electric resistance is a fundamental concept in understanding how electrical circuits work. It measures how much a material opposes the flow of electric current. Think of it as a type of friction for electricity moving through a wire. Resistance is measured in ohms, represented by the symbol \( \Omega \). High resistance means less current flows at a given voltage, while low resistance allows more current to pass through.

To determine the resistance of a device like a light bulb, you can use the formula derived from Ohm’s Law:

• \( R = \frac{V^2}{P} \)
• \( R \) is the resistance in ohms
• \( V \) is the voltage across the bulb in volts
• \( P \) is the power in watts

The calculated resistance helps determine how efficient a bulb is under a specific voltage. Knowing the resistance of a bulb helps us predict its behavior in a circuit, maintain safety, and ensure proper functionality.

Power Dissipation

Power dissipation in the context of a light bulb refers to the process where electrical energy is converted into heat and light. The efficiency of this conversion partly determines how bright a bulb will be.

When electricity passes through a bulb, its filament resists the flow, causing the filament to heat up. This heat is dissipated, with a portion of it being emitted as visible light. The power rating of a bulb, such as 100-W or 60-W, represents the rate at which it can convert electrical energy into light and heat. This is why higher wattage bulbs are generally brighter.

• The basic formula for power is \( P = V \times I \), where \( P \) is power in watts, \( V \) is voltage, and \( I \) is current.
• Alternatively, \( P = \frac{V^2}{R} \) shows the relationship between power, voltage, and resistance.

Understanding power dissipation is crucial for choosing the right components for circuits to ensure they do not overheat or fail prematurely. It also helps predict energy consumption.

Current Calculation

Calculating the electric current flowing through a component is an essential part of circuit analysis. The current, measured in amperes (A), indicates the flow of electric charge through a conductor.

Using the formula based on Ohm's Law and power relations, current can be calculated as:

• \( I = \frac{P}{V} \)
• \( I \) represents current in amperes
• \( P \) is the power in watts
• \( V \) is the voltage in volts

By knowing the power rating and voltage of a device, we can easily find the current it draws. For example, a 100-W bulb drawing from a 120-V supply uses approximately \( 0.833 \) amperes. Calculating current helps ensure that wires and components are properly rated to handle the electrical load, preventing overheating or electrical hazards.