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Chapter 26: Problem 19

Working Late! You are working late in your electronics shop and find that you need various resistors for a project. But alas, all you have is a big box of 10.0 .0 resistors. Show how you can make each of the following equivalent resistances by a combination of your \(10.0-\Omega\) resistors: (a) \(35 \Omega,(\mathrm{b}) 1.0 \Omega,(\mathrm{c}) 3.33 \Omega,\) (d) 7.5\(\Omega\)

Short Answer

Expert verified
Combine resistors in series and parallel to achieve the desired resistances: 35, 1, 3.33, and 7.5 惟.

Step by step solution

01

Understanding Resistor Combinations

There are two main ways to combine resistors: in series and in parallel. When resistors are in series, their resistances add up: \( R_{s} = R_1 + R_2 + \ldots + R_n \). When in parallel, the inverse of the total resistance is the sum of the inverses of each resistance: \( \frac{1}{R_{p}} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots + \frac{1}{R_n} \). Our aim is to use these methods to achieve the desired resistances using only 10-ohm resistors.
02

Achieving 35 惟 Resistance

To achieve a 35 惟 resistance using 10 惟 resistors in series, use 3 resistors to get 30 惟 \((10 + 10 + 10)\) and add 2 resistors in parallel to get an additional 5 惟. The two resistors in parallel have a resistance of \( \frac{1}{R_{p}} = \frac{1}{10} + \frac{1}{10} = \frac{1}{5} \implies R_{p} = 5\). Thus, \( 30 + 5 = 35 惟 \).
03

Achieving 1.0 惟 Resistance

To achieve a 1.0 惟 resistance, connect 10 resistors in parallel. Each 10 惟 resistor contributes \( \frac{1}{10} \) to the inverse of the desired total resistance. Thus, \( \frac{1}{R_{p}} = \frac{10}{10} = 1 \implies R_{p} = 1惟 \).
04

Achieving 3.33 惟 Resistance

To approximate 3.33 惟, connect 3 sets of 3 resistors in parallel. Each set, being 3 resistors in parallel, adds up to \( \frac{1}{R_{p}} = \frac{1}{10} + \frac{1}{10} + \frac{1}{10} = \frac{1}{3.33} \approx 3.33惟 \).
05

Achieving 7.5 惟 Resistance

Use 4 resistors: 2 in series and 2 in parallel. Two in series gives \(20惟\), while the other 2 in parallel result in \( \frac{1}{R_{p}} = \frac{1}{10} + \frac{1}{10} = \frac{1}{5} \implies R_{p} = 5惟 \). They are in a configuration that results in \((20脳5)/(20+5) = 7.5惟\).

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Key Concepts

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

Series Circuits
A series circuit is a type of electrical circuit in which resistors are connected end-to-end. Picture a chain, where one link follows the next in a single line. Similarly, in a series circuit, the current flows through each resistor sequentially.

One of the fundamental properties of a series circuit is that the total resistance is simply the sum of all individual resistances. If you have three resistors with resistances of 5 ohms each connected in series, the total resistance will be:
  • Resistor 1: 5惟
  • Resistor 2: 5惟
  • Resistor 3: 5惟
  • Total Resistance: 5 + 5 + 5 = 15惟
This makes series circuits simple to calculate because you just add them up.

Series circuits maintain a constant current throughout, making them useful in applications where you want current to stay the same across different components. However, keep in mind that if one component fails in a series circuit, it can cause the entire circuit to stop working.
Parallel Circuits
Parallel circuits are a bit different from series circuits. Instead of one single path for electricity to follow, a parallel circuit splits the current into multiple paths.

With parallel circuits, the total resistance is calculated differently. Here, the inverse of the total resistance ( \( \frac{1}{R_{p}} \)) is the sum of the inverses of each resistor's resistance:
  • Formula: \( \frac{1}{R_{p}} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots \)
Let's say you have two resistors, each with 10 ohms. If you connect them in parallel, the calculation would be:
  • Resistor 1: 10惟
  • Resistor 2: 10惟
  • \( \frac{1}{R_{p}} = \frac{1}{10} + \frac{1}{10} = \frac{1}{5} \Rightarrow R_{p} = 5惟 \)
This results in a lower total resistance compared to a single resistor.

Parallel circuits are very useful because if one path fails, the current can still flow through the other paths, ensuring that the circuit continues to operate. Additionally, they allow for more efficient distribution of power across devices.
Resistance Calculation
Calculating resistance in circuits can seem challenging at first, but becomes straightforward once you understand the type of circuit being analyzed.

For series circuits, resistance is a simple addition process. The overall resistance is the sum of all resistors aligned in a row. If you have resistors of 2惟, 3惟, and 5惟, the total resistance is:
  • Total Resistance = 2 + 3 + 5 = 10惟
In parallel circuits, it is a bit more complex as you'd need to use the inverse sum formula:
  • If you have two parallel resistors of 4惟 each, the total resistance is \( \frac{1}{R_{p}} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \Rightarrow R_{p} = 2惟 \)
Whenever calculating in parallel, remember to take the reciprocal of your final answer from the formula.

Through understanding how resistors work, either in series or parallel, you can manipulate and design circuits to achieve the desired resistance for any electronic projects. Practicing calculations with different configurations enhances both your proficiency and your intuition for electrical engineering tasks.

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Most popular questions from this chapter

A \(6.00-\mu \mathrm{F}\) capacitor that is initially uncharged is connected in series with a \(5.00-\Omega\) resistor and an emf source with \(\mathcal{E}=50.0 \mathrm{V}\) and negligible internal resistance. At the instant when the resistor is dissipating electrical energy at a rate of \(250 \mathrm{W},\) how much energy has been stored in the capacitor?

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?

\(120-\mathrm{V}\) circuit that has a \(20-\mathrm{A}\) circuit breaker. You plug an electric hair dryer into the same outlet. The hair dryer has power settings of \(600 \mathrm{W}, 900 \mathrm{W}, 1200 \mathrm{W},\) and 1500 \(\mathrm{W}\) . You start with the hair dryer on the \(600-\mathrm{W}\) setting and increase the power setting until the circuit breaker trips. What power setting caused the breaker to trip?

A \(150-\mathrm{V}\) voltmeter has a resistance of \(30,000 \Omega .\) When connected in series with a large resistance \(R\) across a \(110-\mathrm{V}\) line, the meter reads 74 \(\mathrm{V}\) . Find the resistance \(R\) .

A \(12.4-\mu \mathrm{F}\) capacitor is connected through a \(0.895-\mathrm{M\Omega}\) resistor to a constant potential difference of 60.0 \(\mathrm{V}\) . (a) Compute the charge on the capacitor at the following times after the connections are made: \(0,5.0 \mathrm{s}, 10.0 \mathrm{s}, 20.0 \mathrm{s},\) and 100.0 \(\mathrm{s}\) . (b) Compute the charging currents at the same instants. (c) Graph the results of parts (a) and (b) for \(t\) between 0 and 20 \(\mathrm{s}\) .
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