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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?

Short Answer

Expert verified
(a) Parallel: \( P_1 + P_2 \); (b) Series: \( \frac{\varepsilon^2}{R_1 + R_2} \)

Step by step solution

01

Understanding Power Dissipation

The power dissipated by a resistor is given by the formula \( P = \frac{\varepsilon^2}{R} \). For resistor \( R_1 \), this can be written as \( P_1 = \frac{\varepsilon^2}{R_1} \), and for \( R_2 \), \( P_2 = \frac{\varepsilon^2}{R_2} \).
02

Calculating Power in Parallel Connection

When resistors \( R_1 \) and \( R_2 \) are connected in parallel, the total power consumed \( P_{\text{parallel}} \) is the sum of the power dissipated by each resistor. Therefore, \( P_{\text{parallel}} = P_1 + P_2 \).
03

Calculating Equivalent Resistance in Series

For a series connection, the total resistance \( R_{\text{series}} \) is the sum of the resistances of the individual resistors: \( R_{\text{series}} = R_1 + R_2 \).
04

Calculating Power in Series Connection

Using the equivalent resistance from Step 3, the total power consumed in series \( P_{\text{series}} \) is calculated using the power formula: \( P_{\text{series}} = \frac{\varepsilon^2}{R_{\text{series}}} \). Substitute \( R_{\text{series}} = R_1 + R_2 \) into the formula to get \( P_{\text{series}} = \frac{\varepsilon^2}{R_1 + R_2} \).

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Key Concepts

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

Resistors in Series
When resistors are arranged in a series, they are connected end-to-end, forming a single path for the current to flow. This configuration is straightforward and has a crucial property: the total resistance of the series is simply the sum of the individual resistances.

For two resistors, this is calculated as:
  • \( R_{\text{series}} = R_1 + R_2 \)
This means all the resistors in the series have the same current passing through them because there is only one path for the current.

However, the voltage across each resistor can be different unless the resistors are identical. The total voltage across the series is the sum of the voltages across each resistor. This property leads to interesting implications when calculating power dissipation in series resistors.
Resistors in Parallel
In parallel circuits, resistors are connected so that each one has the same voltage across it. Each resistor provides a separate path for current to flow.

To find the total resistance in a parallel circuit, you use the reciprocal formula:
  • \( \frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2} \)
After finding the reciprocal of the sum, you get the equivalent resistance for the parallel configuration.

The fantastic part of parallel circuits is that they can lead to a total resistance value that's smaller than the smallest resistor in the network. This allows each path to consume its share of the power, providing a more balanced load across different components.
Power Dissipation in Resistors
When resistors consume electrical energy, they convert it into heat. This process is referred to as power dissipation.

The basic formula to determine the power dissipated by a resistor is:
  • \( P = \frac{\varepsilon^2}{R} \)
For a series connection, calculating the power dissipation requires using the total series resistance:
  • \( P_{\text{series}} = \frac{\varepsilon^2}{R_1 + R_2} \)
In parallel circuits, the power dissipation is simply the sum of the power dissipated by each resistor:
  • \( P_{\text{parallel}} = P_1 + P_2 \)
This approach shows the clear difference in power consumption between series and parallel circuits, emphasizing the importance of choosing the right configuration for specific electrical needs.

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Most popular questions from this chapter

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?

A 224-\(\Omega\) resistor and a 589-\(\Omega\) resistor are connected in series across a 90.0-V line. (a) What is the voltage across each resistor? (b) A voltmeter connected across the 224-\(\Omega\) resistor reads 23.8 V. Find the voltmeter resistance. (c) Find the reading of the same voltmeter if it is connected across the 589-\(\Omega\) resistor. (d) The readings on this voltmeter are lower than the "true" voltages (that is, without the voltmeter present). Would it be possible to design a voltmeter that gave readings \(higher\) than the "true" voltages? Explain.

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?

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?

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?

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