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Chapter 19: Problem 33

The wattage rating of a lightbulb is the power it consumes when it is connected across a \(120 \mathrm{~V}\) potential difference. For example, a \(60 \mathrm{~W}\) lightbulb consumes \(60.0 \mathrm{~W}\) of electric power only when it is connected across a \(120 \mathrm{~V}\) potential difference. (a) What is the resistance of a \(60 \mathrm{~W}\) lightbulb? (b) Without doing any calculations, would you expect a \(100 \mathrm{~W}\) bulb to have more or less resistance than a \(60 \mathrm{~W}\) bulb? Calculate and find out.

Short Answer

Expert verified
The 60W bulb's resistance is 240 惟, and the 100W bulb's resistance is 144 惟. A 100W bulb has less resistance than a 60W bulb.

Step by step solution

01

Understanding Ohm's Law and Power Formula

To solve this problem, we need to use Ohm's Law and the power formula. Ohm's Law is given by \( V = IR \), where \( V \) is voltage, \( I \) is current, and \( R \) is resistance. The power formula is \( P = VI \), where \( P \) is power.
02

Rearrange and Use Power Formula

We can express the current \( I \) in terms of power and voltage from the formula \( P = VI \). So, \( I = \frac{P}{V} \). Substitute this into Ohm's Law to find resistance: \[ R = \frac{V}{I} = \frac{V}{\frac{P}{V}} = \frac{V^2}{P} \].
03

Calculate Resistance of a 60W Lightbulb

Now, substitute \( P = 60 \, \text{W} \) and \( V = 120 \, \text{V} \) into the equation:\[ R = \frac{120^2}{60} = \frac{14400}{60} = 240 \, \Omega \]. Thus, the resistance of a 60W lightbulb is \( 240 \, \Omega \).
04

Compare Resistance Expectations for Other Wattage

Generally, as the power \( P \) increases, resistance \( R \) decreases since \( R = \frac{V^2}{P} \). Therefore, a 100W bulb should have less resistance than a 60W bulb because the denominator in the resistance formula is larger, decreasing the overall value.
05

Calculate Resistance of a 100W Lightbulb

For a 100W lightbulb, use \( P = 100 \, \text{W} \) and \( V = 120 \, \text{V} \):\[ R = \frac{120^2}{100} = \frac{14400}{100} = 144 \, \Omega \]. Therefore, the resistance of the 100W bulb is \( 144 \, \Omega \), which is indeed less than the 60W bulb's resistance.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Electrical Power
Electrical power measures how much electrical energy is consumed or generated by a device over a certain period.
Power is denoted by the symbol \( P \) and is measured in watts (W). It defines how fast electrical work is done.
For example, a 60W lightbulb uses 60 watts of power to produce light.
  • This means it requires 60 joules of energy each second while operating.
  • The formula to calculate electrical power is \( P = VI \), where \( V \) is the potential difference (voltage), and \( I \) is the current in amperes (A).
Key Point: Devices with higher wattage ratings typically use more electrical energy.
Resistance Calculation Basics
Resistance is a property of a material or component that resists the flow of electric current.
The symbol for resistance is \( R \) and it is measured in ohms (\( \Omega \)). Resistance can affect how much current flows for a given voltage.
  • Ohm's Law links the concepts of voltage, current, and resistance: \( V = IR \), where \( V \) is voltage, \( I \) is current, and \( R \) is resistance.
  • The formula for resistance when power and voltage are known is \( R = \frac{V^2}{P} \).
Using this formula, we find the resistance of a 60W bulb is 240\( \Omega \) and a 100W bulb's resistance is 144\( \Omega \). Insight: As a bulb's power rating increases, its resistance generally decreases, assuming the same voltage is applied.
Potential Difference Clarification
Potential difference, or voltage, is the measure of electric potential energy per charge between two points in a circuit.
It is denoted by \( V \) and measured in volts (V).
  • Voltage is what "pushes" electric charges through a circuit.
  • A 120V potential difference means that each coulomb of charge experiences 120 joules of energy.
Relation to Power and Resistance: The potential difference, in conjunction with current and resistance, determines how much power a device uses and how much resistance it has. For a lightbulb, power usage is specified at a certain voltage, like the 120V in this problem.
Engaging in Physics Education
Physics education empowers students to understand and apply foundational principles like Ohm's Law and energy concepts.
This problem not only involves arithmetic calculations but encourages students to think critically about how these principles interconnect.
  • It demonstrates real-world applications of physics.
  • It reinforces the concept that increasing resistance on a fixed voltage reduces power consumption.
  • It also emphasizes problem-solving skills through formula manipulation and understanding concepts behind the equations.
Takeaway: Understanding these basic laws lays the groundwork for more complex physics and engineering topics. This exercise shows the significance of electrical concepts in everyday objects, such as lightbulbs.

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Most popular questions from this chapter

A 5.0 A current flows through the leads of an electrical appliance. (a) Calculate how much charge passes through the appliance in \(1.0 \mathrm{~min} .\) (b) How many electrons are in that amount of charge?

A toaster using a Nichrome heating element operates on \(120 \mathrm{~V}\). When it is switched on at \(20^{\circ} \mathrm{C}\), the heating element carries an initial current of 1.35 A. A few seconds later, the current reaches the steady value of 1.23 A. (a) What is the final temperature of the element? The average value of the temperature coefficient of resistivity for Nichrome over the temperature range from \(20^{\circ} \mathrm{C}\) to the final temperature of the element is \(4.5 \times 10^{-4}\left(\mathrm{C}^{\circ}\right)^{-1} .\) (b) What is the power dissipated in the heating element (i) initially; (ii) when the current reaches a steady value?

A fully charged \(6.0 \mu \mathrm{F}\) capacitor is connected in series with a \(1.5 \times 10^{5} \Omega\) resistor. What percentage of the original charge is left on the capacitor after 1.8 s of discharging?

An idealized voltmeter is connected across the terminals of a \(15.0 \mathrm{~V}\) battery, and a \(75.0 \Omega\) appliance is also connected across its terminals. If the voltmeter reads \(11.3 \mathrm{~V}\) : (a) how much power is being dissipated by the appliance, and (b) what is the internal resistance of the battery?

A capacitor is charged to a potential of \(12.0 \mathrm{~V}\) and is then connected to a voltmeter having an internal resistance of \(3.40 \mathrm{M} \Omega\). After a time of \(4.00 \mathrm{~s}\), the voltmeter reads \(3.0 \mathrm{~V}\). What are (a) the capacitance and (b) the time constant of the circuit?
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