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Ch 26: Direct-Current Circuits
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
Chapter 26, Problem 10c

Power Rating of a Resistor. The power rating of a resistor is the maximum power the resistor can safely dissipate without too great a rise in temperature and hence damage to the resistor. A 100.0 Ω and a 150.0 Ω resistor, both rated at 2.00 W, are connected in series across a variable potential difference. What is the greatest this potential difference can be without overheating either resistor, and what is the rate of heat generated in each resistor under these conditions?

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First, understand that when resistors are connected in series, the same current flows through each resistor. The power dissipated by a resistor is given by the formula: Pi=Ii^2Ri, where Pi is the power, Ii is the current, and Ri is the resistance of the resistor.
Since both resistors are rated at 2.00 W, the maximum current through each resistor can be calculated using the formula: Ii=PiRi. Calculate the maximum current for each resistor using their respective resistances.
Once the maximum current is determined, use the smallest current value to ensure neither resistor exceeds its power rating. This is because the same current flows through both resistors in series.
Calculate the total resistance of the series circuit using the formula: R=R1+R2, where R1 and R2 are the resistances of the individual resistors.
Finally, calculate the maximum potential difference across the series circuit using Ohm's Law: V=IR, where V is the potential difference, I is the current, and R is the total resistance. This will give you the greatest potential difference that can be applied without overheating either resistor.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Power Dissipation in Resistors

Power dissipation in a resistor is the process by which electrical energy is converted into heat energy. It is calculated using the formula P = I^2R or P = V^2/R, where P is power, I is current, V is voltage, and R is resistance. Understanding this concept is crucial for determining the maximum power a resistor can handle without overheating.
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Power in Circuits

Series Circuit

In a series circuit, components are connected end-to-end, so the same current flows through each component. The total resistance is the sum of individual resistances, and the voltage across the circuit is divided among the components. This concept helps in calculating the voltage across each resistor and ensuring none exceeds its power rating.
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LRC Circuits in Series

Ohm's Law

Ohm's Law states that the current through a conductor between two points is directly proportional to the voltage across the two points, given by the formula V = IR. This fundamental principle is essential for calculating the current and voltage in the circuit, which are necessary to determine the power dissipation in each resistor.
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Related Practice
Textbook Question

A triangular array of resistors is shown in Fig. E26.5. If the battery has an internal resistance of 3.00Ω, what current will the array draw if the battery is connected across bc?

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Textbook Question

Power Rating of a Resistor. The power rating of a resistor is the maximum power the resistor can safely dissipate without too great a rise in temperature and hence damage to the resistor. If the power rating of a 15 kΩ resistor is 5.0 W, what is the maximum allowable potential difference across the terminals of the resistor?

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Textbook Question

Power Rating of a Resistor. The power rating of a resistor is the maximum power the resistor can safely dissipate without too great a rise in temperature and hence damage to the resistor. A 9.0 kΩ resistor is to be connected across a 120 V potential difference. What power rating is required?

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Textbook Question

Light Bulbs in Series and in Parallel. Two light bulbs have constant resistances of 400Ω and 800Ω. If the two light bulbs are connected in series across a 120-V line, find the current through each bulb.

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