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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?

Short Answer

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

Step by step solution

01

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 \((0.1\, \text{kW}) \times (31 \times 24\, \text{h}) = 74.4\, \text{kWh}\).
02

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 \(\text{Total Energy} \times \text{Cost per kWh} = 74.4\, \text{kWh} \times 0.06\, \\( / \text{kWh} = 4.464\, \\)\).
03

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, \(R = \frac{(120\, \text{V})^2}{100\, \text{W}} = 144\, \Omega\).
04

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, \(I = \frac{100\, \text{W}}{120\, \text{V}} \approx 0.8333\, \text{A}\).

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Key Concepts

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

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.

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Most popular questions from this chapter

A 1250 W radiant heater is constructed to operate at \(115 \mathrm{~V}\). (a) What is the current in the heater when the unit is operating? (b) What is the resistance of the heating coil? (c) How much thermal energy is produced in \(1.0 \mathrm{~h} ?\)

A small but measurable current of \(1.2 \times 10^{-10} \mathrm{~A}\) exists in a copper wire whose diameter is \(2.5 \mathrm{~mm} .\) The number of charge carriers per unit volume is \(8.49 \times 10^{28} \mathrm{~m}^{-3}\). Assuming the current is uniform, calculate the (a) current density and (b) electron drift speed.

A block in the shape of a rectangular solid has a cross-sectional area of \(3.50 \mathrm{~cm}^{2}\) across its width, a front-to-rear length of \(15.8 \mathrm{~cm},\) and a resistance of \(935 \Omega .\) The block's material contains \(5.33 \times 10^{22}\) conduction electrons \(/ \mathrm{m}^{3}\). A potential difference of \(35.8 \mathrm{~V}\) is maintained between its front and rear faces. (a) What is the current in the block? (b) If the current density is uniform, what is its magnitude? What are (c) the drift velocity of the conduction electrons and (d) the magnitude of the electric field in the block?

A common flashlight bulb is rated at \(0.30 \mathrm{~A}\) and \(2.9 \mathrm{~V}\) (the values of the current and voltage under operating conditions). If the resistance of the tungsten bulb filament at room temperature \(\left(20^{\circ} \mathrm{C}\right)\) is \(1.1 \Omega,\) what is the temperature of the filament when the bulb is on?

A charged belt, \(50 \mathrm{~cm}\) wide, travels at \(30 \mathrm{~m} / \mathrm{s}\) between a source of charge and a sphere. The belt carries charge into the sphere at a rate corresponding to \(100 \mu\) A. Compute the surface charge density on the belt.

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