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Chapter 26: Problem 72

Three identical resistors are connected in series. When a certain potential difference is applied across the combination, the total power dissipated is 36 \(\mathrm{W}\) . What power would be dissipated if the three resistors were connected in parallel across the same potential difference?

Short Answer

Expert verified
The power dissipated in parallel is 324 W.

Step by step solution

01

Understand Series and Parallel Connections

When resistors are connected in series, the total resistance \( R_s \) is the sum of the individual resistances. When resistors are connected in parallel, the total resistance \( R_p \) is calculated using the formula \( R_p = \frac{R}{3} \) for three identical resistors (assuming each resistor has resistance \( R \)).
02

Calculate Individual Resistance in Series

We know the power dissipated \( P_s \) for resistors in series is given by \( P_s = \frac{V^2}{R_s} = 36 \text{ W} \). With \( R_s = 3R \), rearranging gives \( V^2 = 36 \times 3R = 108R \).
03

Calculate Total Resistance in Parallel

For parallel connection, the total resistance \( R_p \) is \( \frac{R}{3} \), where \( R \) is the resistance of a single resistor.
04

Derive Power Dissipation in Parallel

Using \( P_p = \frac{V^2}{R_p} \) for the parallel circuit, substitute \( V^2 = 108R \) from Step 2 and \( R_p = \frac{R}{3} \). Thus, \( P_p = \frac{108R}{\frac{R}{3}} = 324 \text{ W} \).

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Key Concepts

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

Resistor Networks
Understanding how resistors are connected in a circuit is crucial to grasping how electrical systems work. In electrical circuits, resistor networks can be configured in different ways, most commonly in series or parallel arrangements. Each of these configurations impacts the total resistance and the current flowing through the circuit.
  • Series Circuit: When resistors are connected in series, their resistances simply add up. This means the total resistance is the sum of the individual resistors. Mathematically, this can be expressed as \( R_s = R_1 + R_2 + R_3 + ... + R_n \). When all resistors are identical, like in our example, the total resistance is \( R_s = nR \), where \( n \) is the number of resistors.
  • Parallel Circuit: In parallel circuits, the reciprocal of the total resistance is the sum of the reciprocals of each individual resistance. For three identical resistors, this simplifies to \( R_p = \frac{R}{3} \).
Recognizing these patterns helps in calculating both the total resistance and the power dissipated in complex circuits.
Power Dissipation
Power dissipation in resistor networks is the process of electrical energy being converted into heat energy. This is a key concept required to understand energy management in circuits. The power dissipated in a circuit, symbolized as \( P \), can be calculated using the formula:
  • \( P = \frac{V^2}{R} \), where \( V \) is the voltage across the resistor and \( R \) is the resistance.
  • For series resistors, the voltage is distributed among them, and the power can be calculated using the combined resistance \( R_s \).
  • For parallel resistors, each shares the same voltage, allowing the formula \( P = \frac{V^2}{R_p} \) to compute the total power dissipation.
In the exercise, when the resistors are connected in series, a total power of 36W is dissipated. However, in a parallel setup, the arrangement results in a much lower net resistance, leading to a significantly higher power dissipation of 324W when powered by the same voltage.
Series and Parallel Circuits
Series and parallel configurations have distinct characteristics that influence how current flows and power is distributed in a circuit.
  • Series Circuits: These circuits have a single path for the current to flow through. As a result, the same current flows through each resistor. Voltage across the circuit is divided among the resistors, explaining why total resistance increases.
  • Parallel Circuits: Multiple paths exist for the current to travel. Each resistor receives the same voltage from the source, but different paths mean current can be variable through each branch, depending on its resistance. This results in a lower total resistance than any individual path resistance.
Knowing the effect of these configurations helps when analyzing the circuit's behavior. In series circuits, the increased resistance lowers the current for a given voltage, decreasing power dissipation. In contrast, the decreased resistance in parallel circuits increases the current, resulting in higher power dissipation when connected to the same voltage source. Thus, understanding these factors is essential for designing and troubleshooting practical electrical networks.

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Most popular questions from this chapter

Working Late! You are working late in your electronics shop and find that you need various resistors for a project. But alas, all you have is a big box of 10.0 .0 resistors. Show how you can make each of the following equivalent resistances by a combination of your \(10.0-\Omega\) resistors: (a) \(35 \Omega,(\mathrm{b}) 1.0 \Omega,(\mathrm{c}) 3.33 \Omega,\) (d) 7.5\(\Omega\)

A resistor and a capacitor are connected in series to an emf source. The time constant for the circuit is 0.870 s. (a) A second capacitor, identical to the first, is added in series. What is the time constant for this new circuit? (b) In the original circuit a second capacitor, identical to the first, is connected in parallel with the first capacitor. What is the time constant for this new circuit?

A galvanometer having a resistance of 25.0\(\Omega\) has a \(1.00-\Omega\) shunt resistance installed to convert it to an ammeter. It is then used to measure the current in a circuit consisting of a \(15.0-\Omega\) resistor connected across the terminals of a \(25.0-\mathrm{V}\) battery having no appreciable internal resistance. (a) What current does the ammeter measure? (b) What should be the true current in the circuit (that is, the current without the ammeter present)? (c) By what percentage is the ammeter reading in error from the true current?

CP A 1.00 -km cable having a cross-sectional area of 0.500 \(\mathrm{cm}^{2}\) is to be constructed out of equal lengths of copper and aluminum. This could be accomplished either by making a \(0.50-\mathrm{km}\) cable of each one and welding them together end to end or by making two parallel 1.00 -km cables, one of each metal (Fig. \(\mathrm{P26.69)..~Calculate~the~}\) resistance of the \(1.00-\mathrm{km}\) cable for both designs to see which one provides the least resistance.

The heating element of an electric dryer is rated at 4.1 \(\mathrm{kW}\) when connected to a \(240-\mathrm{V}\) line. (a) What is the current in the heating element? Is 12 -gauge wire large enough to supply this current? (b) What is the resistance of the dryer's heating element at its operating temperature? (c) At 11 cents per kWh, how much does it cost per hour to operate the dryer?
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