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Compact fluorescent bulbs are much more efficient at producing light than are ordinary incandescent bulbs. They initially cost much more, but they last far longer and use much less electricity. According to one study of these bulbs, a compact bulb that produces as much light as a 100-W incandescent bulb uses only 23 W of power. The compact bulb lasts 10,000 hours, on the average, and costs \(11.00, whereas the incandescent bulb costs only \)0.75, but lasts just 750 hours. The study assumed that electricity costs $0.080 per kilowatt-hour and that the bulbs are on for 4.0 h per day. (a) What is the total cost (including the price of the bulbs) to run each bulb for 3.0 years? (b) How much do you save over 3.0 years if you use a compact fluorescent bulb instead of an incandescent bulb? (c) What is the resistance of a "100-W" fluorescent bulb? (Remember, it actually uses only 23 W of power and operates across 120 V.)

Short Answer

Expert verified
The total cost for the incandescent bulb is $39.54, and for the compact fluorescent, $19.06. You save about $20.48 with the fluorescent bulb, which has a resistance of 626.087 Ω.

Step by step solution

01

Calculate Total Operating Hours in 3 Years

First, we need to calculate how many hours the bulbs will be on over the span of 3 years. Since the bulbs are used for 4.0 hours per day, we have: \[ \text{Total hours} = 4.0 \text{ hours/day} \times 365 \text{ days/year} \times 3 \text{ years} = 4380 \text{ hours} \]
02

Determine Number of Bulbs Needed for Each Type

Compact fluorescent bulb lasts for 10,000 hours, so only 1 bulb is needed for 4380 hours. The incandescent bulb lasts 750 hours; therefore: \[ \text{Number of incandescent bulbs} = \frac{4380}{750} \approx 5.84 \] Since we cannot use a fraction of a bulb, we need 6 incandescent bulbs.
03

Calculate Total Cost for Incandescent Bulbs

For incandescent bulbs, calculate the cost of bulbs and electricity. \[ \text{Cost of bulbs} = 6 \times 0.75 = \\(4.50 \] Electricity used by one bulb: \[ 100 \text{ W} = 0.1 \text{ kW} \] \[ \text{Total electricity cost} = 4380 \text{ hours} \times 0.1 \text{ kW} \times \\)0.080/\text{kWh} = \\(35.04 \] \[ \text{Total cost} = \\)4.50 + \\(35.04 = \\)39.54 \]
04

Calculate Total Cost for Compact Fluorescent Bulbs

For compact fluorescent bulbs, calculate the cost of bulbs and electricity. \[ \text{Cost of bulb} = \\(11.00 \] Electricity used by one bulb: \[ 23 \text{ W} = 0.023 \text{ kW} \] \[ \text{Total electricity cost} = 4380 \text{ hours} \times 0.023 \text{ kW} \times \\)0.080/\text{kWh} = \\(8.0592 \] \[ \text{Total cost} = \\)11.00 + \\(8.0592 = \\)19.0592 \]
05

Calculate Savings

Calculate the savings when using a compact fluorescent bulb. \[ \text{Savings} = \\(39.54 - \\)19.0592 = \$20.4808 \]
06

Calculate Resistance of Compact Fluorescent Bulb

To find the resistance of the "100-W" compact fluorescent bulb (which uses 23 W at 120 V), use the formula \( P = \frac{V^2}{R} \) and solve for \( R \). \[ 23 = \frac{120^2}{R} \] \[ R = \frac{120^2}{23} = \frac{14400}{23} \approx 626.087 \Omega \]

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Key Concepts

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

Compact Fluorescent Bulbs
Compact fluorescent bulbs (CFLs) are a popular choice for energy efficiency today. Although they have a higher initial cost than traditional incandescent bulbs, they offer significant long-term savings in both energy use and lifespan.

A key feature of CFLs is their efficiency. They use much less electricity to produce the same amount of light. For example, a standard CFL can provide illumination equivalent to a 100-watt incandescent bulb while only using 23 watts of power. This reduced power consumption leads to lower electricity bills.

Furthermore, CFLs last significantly longer than their incandescent counterparts. On average, a CFL can last about 10,000 hours, providing consistent lighting for many years when used regularly. In contrast, incandescent bulbs typically last only around 750 hours. This means, over time, you need to purchase fewer replacement bulbs, further saving money and reducing waste.

Overall, while the upfront cost of CFLs might seem daunting, their longer lifespan and lower energy consumption result in efficient lighting solutions that are both cost-effective and environmentally friendly.
Incandescent Bulbs
Incandescent bulbs have been a staple lighting choice for many years due to their simplicity and low initial cost. These bulbs work by passing an electric current through a thin filament, heating it until it produces light.

While incandescent bulbs are inexpensive, they are not energy efficient. They typically consume more electricity to produce the same amount of light as their modern counterparts like CFLs. For instance, to achieve the same brightness as a 23-watt CFL, an incandescent bulb requires 100 watts of power.

Another downside to incandescent bulbs is their shorter lifespan. On average, these bulbs last only about 750 hours, meaning you'll need to replace them far more frequently than CFLs. This can contribute to increased costs over time despite the lower upfront price.

However, for those times when budget constraints are tight in the short term, incandescent bulbs may be a go-to option. But for long-term savings and efficiency, alternatives like CFLs offer significant benefits.
Electricity Costs
Understanding electricity costs is crucial for determining the true expense of operating various light bulbs. This involves not just the cost per bulb but also how much energy they use over time.

Electricity is typically billed by the kilowatt-hour (kWh), which is a measure of energy use. For instance, if electricity costs $0.080 per kWh, switching to energy-efficient lighting like CFLs can significantly reduce your monthly electricity bills.

To calculate total electricity costs for any bulb over a period, you multiply the wattage of the bulb by the total number of hours it's used, then divide by 1,000 to convert to kWh, and finally multiply by the cost per kWh.

For a more concrete example, running a 23-watt CFL for 4 hours per day over 3 years results in a far lower electricity cost compared to a 100-watt incandescent bulb running for the same period. This demonstrates how making informed lighting choices can lead to substantial savings over time.
Resistance Calculation
Calculating resistance is an important part of understanding how electrical devices like light bulbs work. Resistance is a measure of how much a device opposes the flow of electric current, and it's calculated using Ohm's Law.

Ohm's Law states that the power (P) used in an electrical circuit is related to the voltage (V) across the circuit and the resistance (R) of the circuit by the formula: \( P = \frac{V^2}{R} \).

To calculate the resistance of a "100-W" compact fluorescent bulb (which actually uses 23 watts at 120 volts), you rearrange the formula to solve for resistance: \( R = \frac{V^2}{P} \).

Plugging in the values, you find \( R = \frac{120^2}{23} \), which results in approximately 626.087 ohms. Understanding this concept helps to appreciate how efficiently a bulb uses electricity, which is vital for optimizing both cost and performance.

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Most popular questions from this chapter

The region between two concentric conducting spheres with radii \(a\) and \(b\) is filled with a conducting material with resistivity \(\rho\). (a) Show that the resistance between the spheres is given by $$R = {\rho\over4\pi} ({1\over a}- {1\over b})$$(b) Derive an expression for the current density as a function of radius, in terms of the potential difference \(V_{ab}\) between the spheres. (c) Show that the result in part (a) reduces to Eq. (25.10) when the separation \(L = b - a\) between the spheres is small.

In an ionic solution, a current consists of Ca\(^2+\) ions (of charge \(+2e\)) and Cl\(^-\) ions (of charge \(-e\)) traveling in opposite directions. If \(5.11 \times 1018\) Cl\(^-\) ions go from \(A\) to \(B\) every 0.50 min, while 3.24 \(\times\) 10\(^{18}\) Ca\(^2+\) ions move from \(B\) to \(A\), what is the current (in mA) through this solution, and in which direction (from \(A\) to \(B\) or from \(B\) to \(A\)) is it going?

A copper wire has a square cross section 2.3 mm on a side. The wire is 4.0 m long and carries a current of 3.6 A. The density of free electrons is 8.5 \(\times\) 10\(^{28}\)/m\({^3}\). Find the magnitudes of (a) the current density in the wire and (b) the electric field in the wire. (c) How much time is required for an electron to travel the length of the wire?

The capacity of a storage battery, such as those used in automobile electrical systems, is rated in ampere-hours (A dot h). A 50-A dot h battery can supply a current of 50 A for 1.0 h, or 25 A for 2.0 h, and so on. (a) What total energy can be supplied by a 12-V, 60-A dot h battery if its internal resistance is negligible? (b) What volume (in liters) of gasoline has a total heat of combustion equal to the energy obtained in part (a)? (See Section 17.6; the density of gasoline is 900 kg/m\(^3\).) (c) If a generator with an average electrical power output of 0.45 kW is connected to the battery, how much time will be required for it to charge the battery fully?

You apply a potential difference of 4.50 V between the ends of a wire that is 2.50 m in length and 0.654 mm in radius. The resulting current through the wire is 17.6 A. What is the resistivity of the wire?

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