/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 44 Multiple-Concept Example 9 discu... [FREE SOLUTION] | 91Ó°ÊÓ

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Multiple-Concept Example 9 discusses the physics principles used in this problem. Three resistors, \(2.0,4.0,\) and \(6.0 \Omega,\) are connected in series across a \(24-\mathrm{V}\) battery. Find the power delivered to each resistor

Short Answer

Expert verified
Resistances receive 8.0W, 16.0W, and 24.0W respectively.

Step by step solution

01

Determine Total Resistance

In a series circuit, the total resistance \( R_{\text{total}} \) is the sum of the individual resistances. Given the resistors are \( 2.0\, \Omega \), \( 4.0\, \Omega \), and \( 6.0\, \Omega \), we calculate:\[ R_{\text{total}} = 2.0 + 4.0 + 6.0 = 12.0\, \Omega \]
02

Calculate the Current in the Circuit

Using Ohm's Law, the current \( I \) in the circuit can be found. Ohm's Law states \( V = I \times R \), so: \[ I = \frac{V}{R_{\text{total}}} = \frac{24\, \text{V}}{12.0\, \Omega} = 2.0\, \text{A} \]
03

Calculate Power Delivered to Each Resistor

The power \( P \) delivered to each resistor is given by \( P = I^2 \times R \). For each resistor, calculate:- For the \(2.0\, \Omega\) resistor: \[ P = (2.0\, \text{A})^2 \times 2.0\, \Omega = 8.0\, \text{W} \]- For the \(4.0\, \Omega\) resistor: \[ P = (2.0\, \text{A})^2 \times 4.0\, \Omega = 16.0\, \text{W} \]- For the \(6.0\, \Omega\) resistor: \[ P = (2.0\, \text{A})^2 \times 6.0\, \Omega = 24.0\, \text{W} \]

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Key Concepts

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

Ohm's Law
Ohm's Law is one of the fundamental principles used to calculate electrical quantities in electric circuits. Named after Georg Simon Ohm, this law establishes a direct relationship between voltage, current, and resistance in an electrical circuit. The formula is given by:\[ V = I \times R \]Where:
  • \( V \) is the voltage across the circuit (in volts).
  • \( I \) is the current flowing through the circuit (in amperes).
  • \( R \) is the resistance in the circuit (in ohms).
In essence, Ohm's Law tells us that the amount of current passing through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance. This means that for a constant resistance, if the voltage increases, the current increases proportionally. Conversely, if the resistance increases, the current decreases, provided the voltage remains constant. Understanding Ohm's Law is crucial for analyzing any circuit and is especially useful when dealing with series and parallel resistor networks.
Power Calculation
Power in electric circuits is the rate at which electrical energy is converted into another form, such as heat, light, or motion. The power consumed by a resistor in a circuit can be calculated using the formula:\[ P = I^2 \times R \]Alternatively, using Ohm's Law, we can express power in terms of voltage and current as:\[ P = V \times I \]Where:
  • \( P \) is the power (in watts).
  • \( I \) is the current through the resistor (in amperes).
  • \( R \) is the resistance (in ohms).
  • \( V \) is the voltage across the resistor (in volts).
The power calculation allows us to understand how much electrical energy is being used or dissipated by each resistor in the circuit. By computing power delivered to each resistor, as shown in the step-by-step solution, we can ensure that components in the circuit operate efficiently and safely, preventing overheating and potential damage.
Resistors in Series
Resistors in series are a common configuration in electric circuits, where the resistors are connected in a single path for the current to flow. In such setups, the same current flows through each resistor, but the voltage across each resistor may vary depending on its resistance.In a series circuit, the total resistance \( R_{\text{total}} \) is the sum of all individual resistances. Thus, for resistors \( R_1 \), \( R_2 \), and \( R_3 \) connected in series:\[ R_{\text{total}} = R_1 + R_2 + R_3 \]This series combination affects how the total current in the circuit is determined based on the total resistance. One key advantage of resistors in series is that it allows for easy division of voltage across each resistor, providing a simple way to distribute electrical energy.Understanding series circuits is essential as they are widely used in electronic devices and systems, where specific voltage levels are needed across different components for optimal operation.

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Most popular questions from this chapter

The average power used by a stereo speaker is 55 W. Assuming that the speaker can be treated as a 4.0-\Omega resistance, find the peak value of the ac voltage applied to the speaker.

Two capacitors are connected in parallel across the terminals of a battery. One has a capacitance of \(2.0 \mu \mathrm{F}\) and the other a capacitance of \(4.0 \mu \mathrm{F} .\) These two capacitors together store \(5.4 \times 10^{-5} \mathrm{C}\) of charge. What is the voltage of the battery?

In Section 12.3 it was mentioned that temperatures are often measured with electrical resistance thermometers made of platinum wire. Suppose that the resistance of a platinum resistance thermometer is \(125 \Omega\) when its temperature is \(20.0^{\circ} \mathrm{C}\). The wire is then immersed in boiling chlorine, and the resistance drops to \(99.6 \Omega\). The temperature coefficient of resistivity of platinum is \(\alpha=3.72 \times 10^{-3}\left(\mathrm{C}^{\circ}\right)^{-1} .\) What is the temperature of the boiling chlorine?

A blow-dryer and a vacuum cleaner each operate with a voltage of \(120 \mathrm{V}\). The current rating of the blow-dryer is \(11 \mathrm{A}\), and that of the vacuum cleaner is \(4.0 \mathrm{A}\). Determine the power consumed by (a) the blow-dryer and (b) the vacuum cleaner. (c) Determine the ratio of the energy used by the blow-dryer in 15 minutes to the energy used by the vacuum cleaner in onehalf hour.

A coffee cup heater and a lamp are connected in parallel to the same \(120-\mathrm{V}\) outlet. Together, they use a total of \(111 \mathrm{W}\) of power. The resistance of the heater is \(4.0 \times 10^{2} \Omega .\) Find the resistance of the lamp.

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