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Two identical resistors (each with resistance \(R\) ) are connected together in series and then this combination is wired in parallel to a \(20-\Omega\) resistor. If the total equivalent resistance is \(10 \Omega,\) what is the value of \(R ?\)

Short Answer

Expert verified
The value of \(R\) is 10-\(\Omega\).

Step by step solution

01

Understand the Problem

First, understand that two identical resistors in series will have a combined resistance equal to the sum of the individual resistances. Therefore, the total resistance of the two identical resistors is \(R_1 + R_2 = 2R\). This series combination is then connected in parallel to a 20-\(\Omega\) resistor.
02

Apply the Parallel Resistance Formula

For resistors in parallel, the equivalent resistance \(R_{eq}\) can be found using: \[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \] where \(R_1 = 2R\) and \(R_2 = 20 \, \Omega\). We are given that \(R_{eq} = 10 \, \Omega\).
03

Set Up the Equation

Substitute the given and known values into the parallel resistance formula: \[ \frac{1}{10} = \frac{1}{2R} + \frac{1}{20} \] This equation can be solved to find the value of \(R\).
04

Clear the Denominators

Multiply through by the least common multiple (LCM) of the denominators to clear the fractions. Here, the LCM of \(2R\) and 20 is \(20R\), so multiply all terms by \(20R\): \[ 20R \times \frac{1}{10} = 20R \times \frac{1}{2R} + 20R \times \frac{1}{20} \] This simplifies to: \[ 2R = 10 + R \]
05

Solve for R

Rearrange the equation to isolate \(R\): \[ 2R = 10 + R \] Subtract \(R\) from both sides: \[ R = 10 \] Therefore, each identical resistor has a resistance of 10-\(\Omega\).

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Key Concepts

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

Series and Parallel Circuits
In electrical circuits, components such as resistors can be connected in various configurations to achieve the desired electrical characteristics. Two common configurations are **series circuits** and **parallel circuits**. Understanding how these configurations work is key to solving circuit-related problems.

**Series Circuits** involve connecting components end-to-end in a single path for the current to flow. For resistors in series, the total resistance is simply the sum of the individual resistances. For example, if two resistors of resistance \(R\) are connected in series, the total resistance will be \(2R\). This means all resistors contribute equally to the overall resistance.

**Parallel Circuits**, on the other hand, allow current to flow through multiple paths. In a parallel circuit, the voltage across each component is the same, but the total resistance of the circuit is reduced. The formula for calculating the equivalent resistance \(R_{eq}\) of resistors in parallel is:
  • \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots\)
Parallel circuits can decrease overall resistance, allowing more current to flow compared to a series circuit with similarly valued resistors.
Equivalent Resistance
Equivalent resistance is a way of simplifying complex resistor networks into a single resistance value that can be used in calculations as if dealing with one resistor. This concept is extremely useful in simplifying circuit analysis and making it easy to understand the behavior of an electrical circuit.

In our problem, we're dealing with a combination of series and parallel resistors. First, the two identical resistors each with resistance \(R\) are connected in series, yielding a total series resistance of \(2R\). Next, this series combination is wired in parallel to a \(20 \, \Omega\) resistor.

To find the equivalent resistance for this combination, use the parallel resistance formula which requires finding a common denominator for the resistors:
  • \(\frac{1}{R_{eq}} = \frac{1}{2R} + \frac{1}{20}\)
By understanding and applying this formula, we can determine the single equivalent resistance \(R_{eq}\) that can replace the entire network without changing the overall current or voltage in the circuit.
Ohm's Law
Ohm's Law is a fundamental principle used in electronics and electrical engineering to describe the relationship between voltage, current, and resistance. This law is expressed by the equation \(V = IR\), where \(V\) is the voltage across the resistor, \(I\) is the current flowing through it, and \(R\) is the resistance.

When analyzing circuits, Ohm's Law helps to determine one unknown quantity if the other two are known. In our given problem, although we did not explicitly use Ohm's Law to calculate the equivalent resistance, understanding this relationship is crucial in interpreting how voltage and current interact with resistors in both series and parallel circuits.

Ohm's Law assists particularly in applications where you need to ensure that components operate within their safe current and voltage ranges. It's a foundational concept for calculating power, optimizing electrical circuits, and ensuring efficiency in electronic designs.

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Most popular questions from this chapter

Two resistors \(R_{1}\) and \(R_{2}\) are in series with a \(7.0-\mathrm{V}\) battery. If \(R_{1}\) has a resistance of \(2.0 \Omega\) and \(R_{2}\) receives energy at the rate of \(6.0 \mathrm{~W}\), what is (are) the value(s) for the circuit's current(s)? (There may be more than one answer.)

Fig. 18.43 shows a schematic circuit of an instrument called a potentiometer, which is a device for determining very accurate emf values of power supplies. It consists of three batteries, an ammeter, several resistors, and a uniform wire that can be "tapped" for a specific fraction of its total resistance. \(\mathscr{E}_{\mathrm{o}}\) is the emf of a working battery, \(\mathscr{E}_{1}\) designates a battery with a precisely known emf, and \(\mathscr{E}_{2}\) designates a battery whose emf is unknown. The switch S is thrown toward battery \(1,\) and the point \(\mathrm{T}\) (for "tapped") is moved along the resistor until the ammeter reads zero. Let's call the resistance of this arrangement \(R_{1}\). This procedure is repeated with the switch thrown toward battery 2 , and the point \(\mathrm{T}\) is moved to \(\mathrm{T}^{\prime}\) until the ammeter again reads zero. Let's designate the resistance of this arrangement as \(R_{2}\). Show that the unknown emf can be determined from the following relationship: \(\mathscr{E}_{2}=\frac{R_{2}}{R_{1}} \mathscr{E}_{1}\)

Three resistors with values of \(5.0 \Omega, 10 \Omega,\) and \(15 \Omega\) are connected in series in a circuit with a \(9.0-\mathrm{V}\) battery. (a) What is the total equivalent resistance? (b) What is the current in each resistor? (c) At what rate is energy delivered to the \(15-\Omega\) resistor?

In principle, when used together, an ammeter and voltmeter allow for the measurement of the resistance of any circuit element. Let's assume that that element is a simple ohmic resistor. Suppose that the ammeter is connected in series with the resistor (which is connected to an ideal power source with voltage \(V\) ) and the voltmeter is placed across the resistor only. (a) Sketch this circuit (with instruments connected) and use it to explain why the correct resistance is not given by \(R=\frac{V}{I},\) where \(V\) is the voltmeter reading and \(I\) is the ammeter reading. (b) Show that the actual resistance of the element is larger than the result in part \((\mathrm{a})\) and is instead given by \(R=\frac{V}{I-\left(V / R_{\mathrm{V}}\right)},\) where \(R_{\mathrm{V}}\) is the resistance of the voltmeter. (c) Show that the result in part (b) reduces to \(R=\frac{V}{I}\) for an ideal voltmeter.

A battery has three cells connected in series, each with an internal resistance of \(0.020 \Omega\) and an emf of \(1.50 \mathrm{~V}\). This battery is connected to a \(10.0-\Omega\) resistor. (a) Determine the voltage across the resistor. (b) How much current is in each cell? (c) What is the rate at which heat is generated in the battery and how does it compare to the Joule heating rate in the external resistor?

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