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Chapter 26: Problem 26

26\. \(100 \mathrm{~W}\) Lightbulb A \(100 \mathrm{~W}\) lightbulb is plugged into a standard \(120 \mathrm{~V}\) outlet. (a) How much does it cost per month to leave the light turned on continuously? Assume electric energy costs \(12 \mathrm{~d} / \mathrm{kW} \cdot \mathrm{h} .\) (b) What is the resistance of the bulb? (c) What is the current in the bulb? (d) Is the resistance different when the bulb is turned off?

Short Answer

Expert verified
Monthly cost: $8.64. Bulb resistance: 144 惟. Bulb current: 0.833 A. Resistance is theoretically infinite when off but intrinsic material resistance remains the same.

Step by step solution

01

Calculate the energy consumption

First, calculate the energy consumed by the lightbulb in one month. The power rating of the lightbulb is given as 100 watts (W). Since there are 24 hours in a day and approximately 30 days in a month, total energy consumption in a month is: Energy (kWh) = Power (kW) 脳 Time (hrs) = 0.1 kW 脳 720 hrs = 72 kWh
02

Calculate the cost

The cost of electricity is given as 12 cents per kilowatt-hour (kWh). Multiply the total consumption by the cost per kWh to get the monthly cost:Cost = Energy (kWh) 脳 Cost per kWh = 72 kWh 脳 0.12 \(/kWh = \)8.64
03

Calculate the resistance of the bulb

Using the formula for power, where Power (P) = Voltage (V) 脳 Current (I), and Ohm's law, where Voltage (V) = Current (I) 脳 Resistance (R), we combine these equations to find resistance. The equation becomes:P = V^2 / R. Solving for R gives: R = V^2 / P = (120 V)^2 / 100 W = 14400 / 100 = 144 ohms (惟)
04

Calculate the current in the bulb

Using the power formula again where P = V 脳 I, solve for I to find the current:I = P / V = 100 W / 120 V = 0.833 A (amperes)
05

Determine if the resistance is different when the bulb is off

When the bulb is off, there is no current flowing through it, so theoretically the resistance can be considered infinite because there is no path for electrical flow. However, the intrinsic material resistance does not change whether the bulb is on or off.

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Key Concepts

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

Power Consumption
Power consumption refers to the amount of electrical energy used by an electrical device, such as a lightbulb, over time. This is typically measured in watts (W) and kilowatt-hours (kWh). A watt is one joule of energy per second. To find the energy consumption over a period, you multiply the power rating by the time the device is in use.
For instance, in the exercise, a 100W bulb used for 24 hours a day for 30 days consumes:
  • Energy (kWh) = Power (kW) 脳 Time (hrs) = 0.1 kW 脳 720 hrs = 72 kWh

By multiplying the total consumption by the cost per kWh, you can determine the monthly cost of running the device continuously. In this case, with an electricity cost of 12 cents per kWh, the calculation is:
  • Cost = Energy (kWh) 脳 Cost per kWh = 72 kWh 脳 0.12 = $8.64
Electric Resistance
Electric resistance is a measure of the opposition to the flow of current in a conductor. It is measured in ohms (惟). The higher the resistance, the more difficult it is for the electric current to flow through the material.
In the exercise, the resistance of the bulb is calculated using Ohm's law and the power formula:
o
  • Power (P) = Voltage (V) 脳 Current (I)

Using Ohm's law:

  • Voltage (V) = Current (I) 脳 Resistance (R)
Combining these equations, the power formula can be rearranged to solve for resistance (R):
  • P = V^2 / R
  • Solving for R:
  • R = V^2 / P
For the 100W bulb with 120V:
  • R = (120V)^2 / 100W = 144惟

This shows the resistance of the bulb when in operation.
Electric Current
Electric current is the rate at which electric charge flows past a point in a circuit. It is measured in amperes (A). One ampere is equal to one coulomb of charge passing through a point per second.
Using the power formula and the given voltage, the current (I) in the bulb can be calculated:

  • P = V 脳 I
  • Rearranging to solve for I: I = P / V
  • For a 100W bulb with 120V:
  • I = 100W / 120V = 0.833A
  • This indicates that the current flowing through the bulb is 0.833 amperes.

Understanding the current is crucial, as it affects the amount of energy the device consumes and the heat generated in the circuit.
Ohm's Law
Ohm's law is a fundamental principle in electrical engineering that relates voltage (V), current (I), and resistance (R). It states that the current through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance.
The formula for Ohm's law is:

Ul

  • V = I 脳 R

From this relationship, you can derive the other forms:
  • I = V / R
  • R = V / I
In our exercise, we used Ohm's law to find the resistance of the bulb:
  • Using the given power (P), voltage (V), and current (I)
  • P = V 脳 I, we can rearrange this as R = V^2 / P
Understanding Ohm's law is vital for analyzing and designing electrical circuits. It helps us predict how changing one element (voltage, current, or resistance) will affect the others.

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Most popular questions from this chapter

20\. Thermal Energy Thermal energy is produced in a resistor at a rate of \(100 \mathrm{~W}\) when the current is \(3.00 \mathrm{~A}\) What is the resistance?

39\. When Applied When \(115 \mathrm{~V}\) is applied across a wire that is \(10 \mathrm{~m}\) long and has a \(0.30 \mathrm{~mm}\) radius, the current density is \(1.4 \times 10^{4} \mathrm{~A} / \mathrm{m}^{2}\). Find the resistivity of the wire.

3\. Isolated Sphere An isolated conducting sphere has a \(10 \mathrm{~cm}\) radius. One wire carries a current of \(1.0000020 \mathrm{~A}\) into it. Another wire carries a current of \(1.0000000\) A out of it. How long would it take for the sphere to increase in potential by \(1000 \mathrm{~V} ?\)

11\. Longer Wire A wire with a resistance of \(6.0 \Omega\) is drawn out through a die so that its new length is three times its original length. Find the resistance of the longer wire, assuming that the resistivity and density of the material are unchanged.

31\. A Beam A beam contains \(2.0 \times 10^{8}\) doubly charged positive ions per cubic centimeter, all of which are moving north with a speed of \(1.0 \times 10^{5} \mathrm{~m} / \mathrm{s}\). (a) What are the magnitude and direction of the current density \(\vec{J}\) ? (b) Can you calculate the total current \(i\) in this ion beam? If not what additional information is needed?
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