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Chapter 27: Problem 86

Two resistors \(R_{1}\) and \(R_{2}\) may be connected either in series or in parallel across an ideal battery with emf \(\mathscr{E}\). We desire the rate of energy dissipation of the parallel combination to be five times that of the series combination. If \(R_{1}=100 \Omega\), what are the (a) smaller and (b) larger of the two values of \(R_{2}\) that result in that dissipation rate?

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Step by step solution

01

Understanding the Formulas

To solve this problem, we will use the formulas for energy dissipation (power) in both series and parallel resistor configurations. The power in a resistor is given by the formula \(P = \frac{V^2}{R}\), where \(V\) is the voltage across the resistor. In a series circuit, resistors are added \((R_{series} = R_{1} + R_{2})\). In a parallel circuit, the equivalent resistance is found through \(\frac{1}{R_{parallel}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}\). The relationship given is that the power in the parallel configuration is 5 times that of the series configuration.
02

Express Power in Both Configurations

For both configurations, the battery provides the same emf \(\mathscr{E}\). Therefore, for the series configuration, the power is \(P_{series} = \frac{\mathscr{E}^2}{R_{1} + R_{2}}\). For the parallel configuration, the power is \(P_{parallel} = \frac{\mathscr{E}^2}{\frac{1}{R_{1}} + \frac{1}{R_{2}}}\), which simplifies to \(P_{parallel} = \frac{\mathscr{E}^2 R_{2}}{R_{1} R_{2}}\).

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Key Concepts

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

Series Circuit
In a series circuit, elements are connected end-to-end, forming a single path for the flow of electric current. The total resistance in a series circuit is the sum of the individual resistances of each component. This can be expressed as:
  • \( R_{\text{series}} = R_{1} + R_{2} + R_{3} + \ldots \)
Since the current flows through each resistor equally, the total voltage provided by any power source is divided among them. This leads to:
  • The same current running through each part of the circuit.
  • A different voltage drop across each resistor, proportional to its resistance.
Understanding how resistors influence the voltage drop is crucial, as it helps determine how much energy is dissipated or used by each resistor.
Parallel Circuit
Parallel circuits allow multiple paths for current to flow. In this arrangement, each component is connected across the same potential difference. To find the equivalent resistance of resistors in parallel, we use:
  • \( \frac{1}{R_{\text{parallel}}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \dots \)
Here are key characteristics of parallel circuits:
  • The voltage across each resistor is identical to the voltage from the power source.
  • The total current flowing into the circuit from the power source is the sum of the currents through each individual resistor.
This division of current often makes parallel circuits preferable in household circuits, as they ensure a consistent voltage supply to all components.
Energy Dissipation
Energy dissipation in an electric circuit typically occurs in the form of heat, which results from resistors impeding current flow. The rate at which a resistor dissipates energy can be quantified using the power formula:
  • \( P = \frac{V^2}{R} \)
This formula is derived from Ohm's law and shows that power depends on both the voltage across and the resistance of the circuit component.In practical scenarios, like in the given exercise, comparing energy dissipation in series and parallel configurations helps understand circuit efficiency. In general:
  • A lower resistance results in higher energy dissipation, assuming constant voltage.
  • In the provided problem, the parallel circuit dissipates five times the energy as the series one, aiding in specific circuit designs, particularly where higher power output is necessary.
Resistor Calculations
Resistor calculations are integral to designing and analyzing electric circuits. For designing circuits, we use specific formulas to calculate total resistance, which varies depending on whether components are in series or parallel.**In Series:**
  • Add the resistances directly: \( R_{\text{total}} = R_{1} + R_{2} \)
**In Parallel:**
  • Combine using the reciprocal formula: \( \frac{1}{R_{\text{total}}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} \)
This concept was pivotal in solving the original exercise, providing the simplifications needed to derive the relationship between series and parallel configurations. These calculations not only influence resistance but also determine the power, or energy dissipation, of a circuit. Clear understanding allows us to use components efficiently and to design circuits tailored for specific applications.

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Most popular questions from this chapter

A wire of resistance \(5.0 \Omega\) is connected to a battery whose emf \(\mathscr{E}\) is \(2.0 \mathrm{~V}\) and whose internal resistance is \(1.0 \Omega\). In \(2.0 \mathrm{~min}\), how much energy is (a) transferred from chemical form in the battery, (b) dissipated as thermal energy in the wire, and (c) dissipated as thermal energy in the battery?

A \(15.0 \mathrm{k} \Omega\) resistor and a capacitor are connected in series, and then a \(12.0 \mathrm{~V}\) potential difference is suddenly applied across them. The potential difference across the capacitor rises to \(5.00 \mathrm{~V}\) in \(1.30 \mu \mathrm{s}\). (a) Calculate the time constant of the circuit. (b) Find the capacitance of the capacitor.

Thermal energy is to be generated in a \(0.10 \Omega\) resistor at the rate of \(10 \mathrm{~W}\) by connecting the resistor to a battery whose emf is \(1.5 \mathrm{~V} .\) (a) What potential difference must exist across the resistor? (b) What must be the internal resistance of the battery?

A certain car battery with a \(12.0 \mathrm{~V}\) emf has an initial charge of \(120 \mathrm{~A} \cdot \mathrm{h}\). Assuming that the potential across the terminals stays constant until the battery is completely discharged, for how many hours can it deliver energy at the rate of \(100 \mathrm{~W}\) ?

A 10 -km-long underground cable extends east to west and consists of two parallel wires, each of which has resistance 13 \(\Omega / \mathrm{km} .\) An electrical short develops at distance \(x\) from the west end when a conducting path of resistance \(R\) connects the wires (Fig. 27 31). The resistance of the wires and the short is then \(100 \Omega\) when measured from the east end and \(200 \Omega\) when measured from the west end. What are (a) \(x\) and (b) \(R\) ?
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