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Chapter 26: Problem 56

Two identical \(1.00-\Omega\) wires are laid side by side and soldered together so they touch each other for half of their lengths. What is the equivalent resistance of this combination?

Short Answer

Expert verified
0.75 \(\Omega\)

Step by step solution

01

Understanding the Problem

We have two wires, each with a resistance of \(1.00\,\Omega\). They are touching each other for half their length, which implies they are in parallel for that section and in series for the other half. We need to find the equivalent resistance of this configuration.
02

Calculating Resistance of Wires in Parallel

When two resistors are in parallel, the formula for the equivalent resistance \(R_{parallel}\) is \[\frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2}\]. In this case, \(R_1 = R_2 = 1.00\,\Omega\). Thus, for the half-length in parallel, \[\frac{1}{R_{parallel}} = \frac{1}{1.00} + \frac{1}{1.00} = 2\]. Using the formula, \(R_{parallel} = \frac{1}{2} = 0.50\,\Omega\).
03

Calculating Resistance of Wires in Series

For the portion not touching, the wires are in series. The equivalent resistance \(R_{series}\) for two resistances in series is just the sum of both resistances: \(R_{series} = 1.00 + 1.00 = 2.00\,\Omega\). However, here we are considering the whole length equivalent, so during the non-touching half sections, \(R = 1.00\,\Omega + 1.00\,\Omega = 2.00\,\Omega\) divided by 2 (since it's half) is \(1.00\,\Omega\) to contrast to the parallel part.
04

Final Equivalent Resistance Calculation

For computing the entire equivalent resistance of half parallel and half series configuration, we take the sum of each calculated resistances for half: \(R_{equivalent} = R_{parallel} + R_{parallel}\) where \(R_{parallel} = \frac{0.50+1.00}{2} = 0.75\,\Omega\), ensuring when split the left and right are same.

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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 placed in series, it means they are connected end-to-end, so the electricity flows through one resistor directly into the next. This configuration leads to the total or equivalent resistance being the sum of all individual resistances in the series.

Imagine you have a string of Christmas lights, where each bulb represents a resistor. In such an arrangement, if one bulb goes out, the entire string goes out!
  • The formula for resistors in series is: \( R_{total} = R_1 + R_2 + R_3 + \ldots \)
  • This is because the current must pass through each resistor in turn, facing each resistance along the way.
In our exercise's context, the wires not touching each other act like resistors in series for that part of their length. Thus, total resistance is calculated by simply adding their resistance values.
Resistors in Parallel
When resistors are in parallel, they are connected side by side, and the electric current has various paths to take. In this setup, the total or equivalent resistance is lower than any of the individual resistances due to the additional pathways.

Think of it like multiple lanes on a highway, allowing more cars to pass through without queueing up. If one lane is blocked, others can still let the cars through.
  • The formula for calculating resistors in parallel: \( \frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \ldots \)
  • This formula takes the reciprocal of the sum of reciprocals of individual resistances.
In our original problem, the portion of the wire that is touching acts as resistors in parallel, which considerably reduces the resistance for that section.
Electrical Circuits
An electrical circuit is essentially a path allowing electricity to flow from one point to another typically involving resistors, a power source, and conductive paths. Understanding how resistors function within a circuit is crucial for predicting how electrons will behave.
  • Series circuits have a single pathway for current, while parallel circuits have multiple paths.
  • Combining both series and parallel connections allows for more complex and functional designs.
In practical applications, circuits can manage different loads and functionalities effectively by using combinations of series and parallel resistors. The exercise you've encountered is an example of such a combination, where analyzing each segment separately can help deduce the overall behavior of the circuit.
Resistance Calculation
Resistance calculation aids in determining just how much an electrical circuit restricts the flow of electrons. Various configurations will alter the behavior. Having a clear understanding of calculating the equivalent resistance when combining resistors either in series or parallel is essential.

In our exercise, you saw two half-length wires behaving partly in series and partly in parallel, necessitating calculations using both formulas:
  • For series: \( R_{series} = R_1 + R_2 \)
  • For parallel: \( \frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2} \)
By thoughtfully applying these formulas step by step, you can determine the total resistance in any mixed-configuration circuit. This skill is pivotal in designing real-world electrical systems and optimizing circuit efficiencies.

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Most popular questions from this chapter

A resistor with \(R=850 \Omega\) is connected to the plates of a charged capacitor with capacitance \(C=4.62 \mu F\) . Just before the connection is made. the charge on the capacitor is 8.10 \(\mathrm{mC}\) .connection is made, the charge on the capacitor is 8.10 \(\mathrm{mC}\) . (a) What is the energy initially stored in the capacitor? (b) What is the electrical power dissipated in the resistor just after the connec- tion is made? (c) What is the electrical power dissipated in the resistor at the instant when the energy stored in the capacitor has decreased to half the value calculated in part (a)?

The heating element of an electric stove consists of a heater wire embedded within an electrically insulating material, which in turn is inside a metal casing. The heater wire has a resistance of 20\(\Omega\) at room temperature \(\left(23.0^{\circ} \mathrm{C}\right)\) and a temperature coefficient of resistivity \(\alpha=2.8 \times 10^{-3}\left(\mathrm{C}^{\circ}\right)^{-1}\) . The heating element oper- ates from a 120 \(\mathrm{V}\) line. (a) When the heating element is first turned on, what current does it draw and what electrical power does it dis- sipate? (b) When the heating element has reached an operating temperature of \(280^{\circ} \mathrm{C}\left(536^{\circ} \mathrm{F}\right),\) what current does it draw and what electrical power does it dissipate?

The resistance of a galvanometer coil is \(25.0 \Omega,\) and the current required for full-scale defiection is 500\(\mu A\) . (a) Show in a diagram how to convert the galvanometer to an ammeter reading 20.0 \(\mathrm{mA}\) full scale, and compute the shumt resistance. (b) Show how to convert the galvanometer to a voltmeter reading 500 \(\mathrm{mV}\)

A \(1.50-\mu F\) capacitor is charging through a \(12.0-\Omega\) resistor using a \(10.0-\mathrm{V}\) battery. What will be the current when the capacitor has acquired \(\frac{1}{4}\) of its maximum charge? Will it be \(\frac{1}{4}\) of the maximum current?

Consider the potentiometer circuit of Fig. 26.19 \(\mathrm{a}\) . The resistor between \(a\) and \(b\) is a uniform wire with length \(l,\) with a sliding contact \(c\) at a distance \(x\) from \(b\) . An unknown \(\mathrm{emf} \mathcal{E}_{2}\) is measured by \(\mathcal{E}_{2}=(x / l) \mathcal{E}_{1} .\) (b) Why is the intermal resistance of the galvanometer not important? (c) Suppose \(\mathcal{E}_{1}=9.15 \mathrm{V}\) and \(l=1.000 \mathrm{m} .\) The galvanometer \(G\) reads zero when \(x=0.365 \mathrm{m} .\) What is the emf \(\mathcal{E}_{2} ?\)
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