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Two identical \(3.00-\Omega\) 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
The equivalent resistance is \(3.75 \, \Omega\.\).

Step by step solution

01

Identify the Problem

We have two identical wires with resistance of \( R = 3.00 \, \Omega \) each laid side by side, soldered to touch each other for half their length. We need to find the equivalent resistance of this setup.
02

Model the Circuit Path

Visualize that for half the length, the wires are soldered together, forming a parallel configuration, and for the other half they act in series.
03

Calculate Resistance for Parallel Section

For the half-length where the wires are in parallel, the equivalent resistance \( R_p \) is given by the parallel resistance formula: \[ \frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} = \frac{2}{3} \]. Thus, \( R_p = \frac{3}{2} \, \Omega \).
04

Calculate Resistance for Series Section

For the half-length where the wires are in series, the resistance is simply the sum of both resistances: \( R_s = R + R = 3 + 3 = 6 \, \Omega \).
05

Average the Resistance

Since the wires have equal lengths, we average the resistance of the parallel and series sections. Thus, the equivalent resistance \( R_{eq} \) is: \[ R_{eq} = \frac{R_p + R_s}{2} = \frac{1.5 + 6}{2} = \frac{7.5}{2} = 3.75 \, \Omega \]

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Key Concepts

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

Understanding the Parallel Resistance Formula
In circuits, resistors can be connected in parallel and this setup influences the overall resistance within the circuit. When resistors are in parallel, they have their ends connected to the same two points in the circuit, essentially providing multiple paths for current to travel.
This parallel arrangement reduces the total resistance compared to any single resistor in the group.The formula to calculate total or equivalent resistance, \( R_p \), for resistors in parallel is:
  • \( \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots + \frac{1}{R_n} \)
For our specific problem, both wires are only \( 3.00 \, \Omega \) each and merged halfway as a parallel setup. Thus it simplified to:
  • \( \frac{1}{R_p} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \)
  • Simplifying gives \( R_p = 1.5 \, \Omega \)
This shows how parallel configurations can lessen the resistance, helping more current to pass through multiple paths.
Grasping Series Resistance
In a series circuit, components are connected end-to-end in a single path, meaning the current flows through one component to get to the next. The total resistance in a series is simply the sum of all resistors along that path.
This setup effectively increases the resistance because the current has to pass through each resistor one after another.The formula for the total resistance, \( R_s \), for resistors in a series is straightforward:
  • \( R_s = R_1 + R_2 + \ldots + R_n \)
In the example problem, two \( 3.00 \, \Omega \) wires are part of a series for half their length. This equates to:
  • \( R_s = 3 + 3 = 6 \, \Omega \)
Understanding this allows us to predict how series arrangements increase the total resistance, offering a higher resistance path for electric current.
Basics of Circuit Path Modeling
Circuit path modeling involves depicting and understanding how different components in a circuit are laid out and how current flows through them. It is key to grasp how resistors combine either in series or parallel.
This understanding assists in precisely calculating equivalent resistance for complex circuits.In the given problem, we model the circuit path by visualizing that for one half of its length the wires are laid parallel and for the other half in series:
  • First, model the parallel section (half-length) using the parallel resistance formula.
  • Second, model the series section (complementary half-length) with the series resistance summation.
  • Finally, average the two calculated resistances given both sections are equal in length.
Circuit path modeling helps visualize existing connections between components, like how the problem's wire connection transforms from parallel to series, leading to the final combined resistance of \( 3.75 \, \Omega \). It's a crucial method for understanding electrical circuits deeply.

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Most popular questions from this chapter

A \(4.60-\mu F\) capacitor that is initially uncharged is connected in series with a \(7.50-\mathrm{k} \Omega\) resistor and an emf source with \(\mathcal{E}=245 \mathrm{V}\) and negligible internal resistance. Just after the circuit is completed, what are (a) the voltage drop across the capacitor: (b) the voltage drop across the resistor: (c) the charge on the capacitor: (d) the current through the resistor'? (e) A long time after the circuit is completed (after many time constants) what are the values of the quantities in parts \((a)-(d) ?\)

A circuit consists of a series combination of \(6.00-\mathrm{k} \Omega\) and \(5.00-\mathrm{k} \Omega\) resistors connected across a \(50.0-\mathrm{V}\) battery having negligible internal resistance. You want to measure the true potential difference (that is, the potential difference without the meter present) across the \(5.00-\mathrm{k} \Omega\) resistor using a voltmeter having an internal resistance of 10.0 \(\mathrm{k} \Omega\) . (a) What potential difference does the voltmeter measure across the \(5.00-\mathrm{k} \Omega\) resistor? (b) What is the true potential difference across this resistor when the meter is not present? (c) By what percentage is the voltmeter reading in error from the true potential difference?

CP A 1.00 -km cable having a cross-sectional area of 0.500 \(\mathrm{cm}^{2}\) is to be constructed out of equal lengths of copper and aluminum. This could be accomplished either by making a \(0.50-\mathrm{km}\) cable of each one and welding them together end to end or by making two parallel 1.00 -km cables, one of each metal (Fig. \(\mathrm{P26.69)..~Calculate~the~}\) resistance of the \(1.00-\mathrm{km}\) cable for both designs to see which one provides the least resistance.

CP A 20.0 -m-long cable consists of a solid-inner, cylindrical, nickel core 10.0 \(\mathrm{cm}\) in diameter surrounded by a solid-outer cylindrical shell of copper 10.0 \(\mathrm{cm}\) in inside diameter and 20.0 \(\mathrm{cm}\) in outside diameter. The resistivity of nickel is \(7.8 \times 10^{-8} \Omega \cdot \mathrm{cm} .\) (a) What is the resistance of this cable? (b) If we think of this cable as a single material, what is its equivalent resistivity?

A resistor and a capacitor are connected in series to an emf source. The time constant for the circuit is 0.870 s. (a) A second capacitor, identical to the first, is added in series. What is the time constant for this new circuit? (b) In the original circuit a second capacitor, identical to the first, is connected in parallel with the first capacitor. What is the time constant for this new circuit?

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