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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

An emf source with \(\varepsilon =\) 120 V, a resistor with \(R =\) 80.0 \(\Omega\), and a capacitor with \(C =\) 4.00 \(\mu\)F are connected in series. As the capacitor charges, when the current in the resistor is 0.900 A, what is the magnitude of the charge on each plate of the capacitor?

A 12.0-mF capacitor is charged to a potential of 50.0 V and then discharged through a 225-\(\Omega\) resistor. How long does it take the capacitor to lose (a) half of its charge and (b) half of its stored energy?

A resistor \(R_1\) consumes electrical power \(P_1\) when connected to an emf \(\varepsilon\). When resistor \(R_2\) is connected to the same emf, it consumes electrical power \(P_2\). In terms of \(P_1\) and \(P_2\), what is the total electrical power consumed when they are both connected to this emf source (a) in parallel and (b) in series?

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-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?

Two light bulbs have constant resistances of 400 \(\Omega\) and 800 \(\Omega\). If the two light bulbs are connected in series across a 120-V line, find (a) the current through each bulb; (b) the power dissipated in each bulb; (c) the total power dissipated in both bulbs. The two light bulbs are now connected in parallel across the 120-V line. Find (d) the current through each bulb; (e) the power dissipated in each bulb; (f) the total power dissipated in both bulbs. (g) In each situation, which of the two bulbs glows the brightest? (h) In which situation is there a greater total light output from both bulbs combined?

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