Problem 49 - Fundamentals of Physics

Question

A \(100 \mathrm{~W}\) lightbulb is plugged into a standard \(120 \mathrm{~V}\) outlet. (a) How much does it cost per 31 -day month to leave the light turned on continuously? Assume electrical energy costs US \(\$ 0.06 / \mathrm{kW} \cdot \mathrm{h}\). (b) What is the resistance of the bulb? (c) What is the current in the bulb?

Step-by-Step Solution

Answer

(a) $4.464; (b) 144 \Omega; (c) 0.8333 A.

Step 1: Calculate Total Energy Consumption

To find the total energy consumed by the lightbulb over 31 days, we first convert the power rating from watts to kilowatts. The power consumption is given by \(100\, \text{W} = 0.1\, \text{kW}\). The lightbulb is used continuously for \(31 \times 24\) hours. Thus, the total energy consumed is \((0.1\, \text{kW}) \times (31 \times 24\, \text{h}) = 74.4\, \text{kWh}\).

Step 2: Calculate Cost of Electricity

The cost of using the lightbulb is calculated using the total energy consumption and the cost per kilowatt-hour. The cost is given by \(\text{Total Energy} \times \text{Cost per kWh} = 74.4\, \text{kWh} \times 0.06\, \\( / \text{kWh} = 4.464\, \\)\).

Step 3: Calculate Resistance of the Lightbulb

To find the resistance, we use Ohm's Law which states \(V = I \cdot R\), and the power formula \(P = V \cdot I\). Rearranging for resistance, we get \(R = \frac{V^2}{P}\). Substituting the given values, \(R = \frac{(120\, \text{V})^2}{100\, \text{W}} = 144\, \Omega\).

Step 4: Calculate Current Through the Lightbulb

We use the power formula \(P = V \cdot I\), and rearrange it to find the current: \(I = \frac{P}{V}\). Substituting the values, \(I = \frac{100\, \text{W}}{120\, \text{V}} \approx 0.8333\, \text{A}\).

Key Concepts

Electrical Power Calculation

Electrical power is a fundamental concept in understanding how electrical devices consume energy. Power can be thought of as the rate at which energy is used or produced. For electrical devices, power use is commonly measured in watts (W) or kilowatts (kW). One kilowatt equals 1,000 watts. To calculate electrical power consumption, you use the formula:

• Power (\(P\)) = Voltage (\(V\)) \(\times\) Current (\(I\))

However, when talking about energy consumption, we often convert watts to kilowatts and consider time to find out how much energy was used over a specific period.

For example, if a device runs at 100 watts continuously, to find the energy consumption, you must convert this power into kilowatts by dividing by 1,000:

• \(100\, \text{W} = 0.1\, \text{kW}\)

Next, multiply this by the time the device is in use. For a lightbulb used continuously for 31 days:

• Energy (\(E\)) = Power (\(\text{kW}\)) \(\times\) Time (\(\text{hours}\))
• \(E = 0.1\, \text{kW} \times 744\, \text{h} = 74.4\, \text{kWh}\)

Understanding this helps calculate how much you will pay for electricity:

• Cost = Energy \(\times\) price per kWh
• Cost = 74.4 kWh \(\times\) \(0.06 \text{ USD/kWh} = \text{USD 4.464}\)

Resistance of a Lightbulb

The resistance of a lightbulb is crucial in determining how much current it will draw at a given voltage. Resistance, measured in ohms (\(\Omega\)), is calculated using Ohm's Law:

• \(V = I \cdot R\)

It can also be expressed via the power formula. By rearranging the power formula, we derive the resistance calculation:

• Power (\(P\)) is also equal to Voltage (\(V\)) squared, divided by Resistance (\(R\))
• \(R = \frac{V^2}{P}\)

In our case with a bulb (plugged into a 120 V outlet and consuming 100 watts of power):

• \(R = \frac{120^2}{100} = 144 \Omega\)

This means the bulb has a resistance of 144 ohms. Knowing the resistance helps in determining how much current the bulb will need to function properly.

Electric Current Calculation

Current is the flow of electric charge through a conductor, usually measured in amperes (A). The amount of current flowing through a lightbulb depends on the power it uses and the voltage of the supply. To calculate the current, you rearrange the basic power formula:

• Power (\(P\)) = Voltage (\(V\)) \(\times\) Current (\(I\))

This becomes:

• \(I = \frac{P}{V}\)

Substituting the values for a 100 watt bulb connected to a 120 volt supply:

• \(I = \frac{100}{120} \approx 0.8333\, \text{A}\)

Thus, the bulb draws a current of about 0.8333 amperes. This calculation ensures the bulb operates safely within the designed electrical systems and helps in circuit design and troubleshooting.