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Three identical resistors are connected in series. When a certain potential difference is applied across the combination, the total power dissipated is 45.0 W. What power would be dissipated if the three resistors were connected in parallel across the same potential difference?

Short Answer

Expert verified
405 W of power is dissipated when resistors are in parallel.

Step by step solution

01

Understand Series Circuit

In a series circuit, the same current flows through all components. The total resistance \( R_{total} \) is the sum of the individual resistances. If each resistor has resistance \( R \), then \( R_{total} = 3R \) for three resistors.
02

Calculate Resistance in Series

Using the power formula for a series circuit, \( P = \frac{V^2}{R_{total}} \). Rearrange to find \( R \): \( R_{total} = \frac{V^2}{P} \). Thus, \( 3R = \frac{V^2}{45} \).
03

Determine Individual Resistance

From \( 3R = \frac{V^2}{45} \), solving for \( R \) gives \( R = \frac{V^2}{135} \). This is the resistance of each individual resistor.
04

Analyze Parallel Circuit

In a parallel circuit, the total resistance \( R_{total} \) is given by \( \frac{1}{R_{total}} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{3}{R} \). Simplifying, \( R_{total} = \frac{R}{3} \).
05

Calculate Power in Parallel

Using the power formula for the parallel circuit \( P = \frac{V^2}{R_{total}} \), substitute for \( R_{total} \): \( P = \frac{V^2}{\frac{R}{3}} = 3 \frac{V^2}{R} \).
06

Substitute for R from Step 3

From Step 3, we know \( R = \frac{V^2}{135} \). Substitute this into the power equation: \( P = 3 \frac{V^2}{\frac{V^2}{135}} = 3 \times 135 = 405 \).

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Key Concepts

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

Series Circuit
In a series circuit, all components are connected one after another, allowing the same electric current to flow through each component sequentially. This means the overall resistance in the circuit is simply the sum of individual resistances. When resistors are connected in series, the total resistance can be calculated using the formula:
  • \( R_{total} = R_1 + R_2 + R_3 + ... + R_n \)
For example, if three identical resistors each have a resistance \( R \), the total resistance is given by \( R_{total} = 3R \). The voltage drop across the entire circuit distributes among the three resistors, but they all carry the same current. As a result, the power dissipated can be calculated by the following formula:
  • \( P = \frac{V^2}{R_{total}} \)
Parallel Circuit
A parallel circuit offers multiple different paths for the electric current to flow. If one path is obstructed or broken, the others can continue to function. In contrast to series circuits, the total resistance in a parallel circuit decreases. It can be determined using the reciprocal formula:
  • \( \frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + ... + \frac{1}{R_n} \)
For three identical resistors connected in parallel, the formula becomes
  • \( \frac{1}{R_{total}} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{3}{R} \)
Simplifying this equation results in the total resistance:
  • \( R_{total} = \frac{R}{3} \)
A key feature of a parallel circuit is that each branch receives the same voltage across it, making it different from series circuits where the voltage is divided across components.
Power Dissipation
Power dissipation refers to the process by which electrical energy is converted into thermal energy or heat within a circuit, often across resistors. It's a critical factor in both series and parallel circuits. Power dissipation is governed by the formula:
  • \( P = VI \)
  • \( P = I^2R \) (for circuits with a known current)
  • \( P = \frac{V^2}{R} \) (for circuits with a known voltage)
In a series circuit, when resistors are all identical and organized in a line, the power can be calculated as \( P = \frac{V^2}{3R} \), showing how the power dissipates over the combined resistance. In parallel circuits, power dissipation changes since each pathway supports the same voltage, and the total power dissipation increases as more parallel paths allow for more current flow, thus more energy dissipation.
Resistance Calculation
Calculating resistance is a crucial step in analyzing both series and parallel circuits. For series circuits, resistance is straightforward, as you just sum up the individual resistances. However, for parallel circuits, it becomes necessary to use the reciprocal relation for accurate calculations.
  • Series: \( R_{total} = R_1 + R_2 + R_3 + ... + R_n \)
  • Parallel: \( \frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + ... \)
Analyzing the given problem where each resistor has the same resistance \( R \), the equivalent resistance for three resistors in series is \( 3R \), while for parallel, it's \( \frac{R}{3} \). This distinction is crucial for determining how circuits will carry current and how they will respond to electrical power, especially under a fixed voltage. Computing these resistances accurately helps in predicting how much power will dissipate across such configurations.

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Most popular questions from this chapter

A 1500-W electric heater is plugged into the outlet of a 120-V circuit that has a 20-A circuit breaker. You plug an electric hair dryer into the same outlet. The hair dryer has power settings of 600 W, 900 W, 1200 W, and 1500 W. You start with the hair dryer on the 600-W setting and increase the power setting until the circuit breaker trips. What power setting caused the breaker to trip?

A \(2.00-\mu \mathrm{F}\) capacitor that is initially uncharged is connected in series with a \(6.00-\mathrm{k} \Omega\) resistor and an emf source with \(\mathcal{E}=90.0 \mathrm{~V}\) and negligible internal resistance. The circuit is completed at \(t=0 .\) (a) Just after the circuit is completed, what is the rate at which electrical energy is being dissipated in the resistor? (b) At what value of \(t\) is the rate at which electrical energy is being dissipated in the resistor equal to the rate at which electrical energy is being stored in the capacitor? (c) At the time calculated in part (b), what is the rate at which electrical energy is being dissipated in the resistor?

A circuit consists of a series combination of 6.00-k\(\Omega\) and 5.00-k\(\Omega\) resistors connected across a 50.0-V battery having negligible internal resistance. You want to measure the true potential difference (that is, the potential difference without the meter present) across the 5.00-k\(\Omega\) resistor using a voltmeter having an internal resistance of 10.0 k\(\Omega\). (a) What potential difference does the voltmeter measure across the 5.00-k\(\Omega\) resistor? (b) What is the \(true\) potential difference across this resistor when the meter is not present? (c) By what percentage is the voltmeter reading in error from the true potential difference?

Assume that a typical open ion channel spanning an axon's membrane has a resistance of 1 \(\times\) 10\(^{11}\) \(\Omega\). We can model this ion channel, with its pore, as a 12-nm-long cylinder of radius 0.3 nm. What is the resistivity of the fluid in the pore? (a) 10 \(\Omega\) \(\cdot\) m; (b) 6 \(\Omega\) \(\cdot\) m; (c) 2 \(\Omega\) \(\cdot\) m; (d) 1 \(\Omega\) \(\cdot\) m.

A resistor with \(R_1 =\) 25.0 \(\Omega\) is connected to a battery that has negligible internal resistance and electrical energy is dissipated by \(R_1\) at a rate of 36.0 W. If a second resistor with \(R_2 =\) 15.0 \(\Omega\) is connected in series with \(R_1\), what is the total rate at which electrical energy is dissipated by the two resistors?

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