/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 Two resistances having values of... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Two resistances having values of \(2 R\) and \(3 R\) are in parallel. \(R\) and the equivalent resistance are both integers. What are the possible values for \(R\) ?

Short Answer

Expert verified
Possible values for \( R \) are multiples of 5.

Step by step solution

01

Understanding Parallel Resistors Formula

When resistors are in parallel, their equivalent resistance \( R_{eq} \) can be found using the formula: \[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \] where \( R_1 = 2R \) and \( R_2 = 3R \).
02

Substitute Given Values into Formula

Substitute the values \( R_1 = 2R \) and \( R_2 = 3R \) into the parallel resistance formula: \[ \frac{1}{R_{eq}} = \frac{1}{2R} + \frac{1}{3R} \]
03

Find a Common Denominator

Find a common denominator for the fractions on the right-hand side. The least common denominator of \( \frac{1}{2R} \) and \( \frac{1}{3R} \) is \( 6R \). Rewrite each fraction: \[ \frac{1}{2R} = \frac{3}{6R}, \quad \frac{1}{3R} = \frac{2}{6R} \]
04

Combine Fractions on Right-Hand Side

Combine the fractions: \[ \frac{1}{R_{eq}} = \frac{3}{6R} + \frac{2}{6R} = \frac{5}{6R} \]
05

Solve for Equivalent Resistance

Now, flip the equation to solve for \( R_{eq} \): \[ R_{eq} = \frac{6R}{5} \] Since \( R_{eq} \) is an integer, \( \frac{6R}{5} \) must also be an integer.
06

Find Possible Values of \( R \)

Since \( \frac{6R}{5} \) is an integer, \( R \) must be a multiple of 5. Let \( R = 5k \) where \( k \) is an integer. Substitute this back to check: \[ R_{eq} = \frac{6(5k)}{5} = 6k \] Since \( R_{eq} \) is an integer for any integer \( k \), \( R \) can be any multiple of 5.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Equivalent Resistance
When resistors are combined in a circuit, they impact how easily current can flow. If you have multiple resistors, you can combine their values to find what is known as the "equivalent resistance". This tells you the resistance that would produce the same effect in the circuit as the combination of all resistors together.

For resistors in parallel, things are a bit different than when they are in series. In such setups, the formula for equivalent resistance, denoted as \( R_{eq} \), is:
  • \( \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots \)
This formula acknowledges that in a parallel configuration, there are multiple paths for the current. Lower equivalent resistance values typically result, compared to series circuits. In the equality, each component provides a path for the current, effectively reducing the overall resistance.

Thus, figuring out the equivalent resistance in parallel is crucial for understanding the entire circuit's behavior.
Resistors in Parallel
In electrical circuits, resistors can be arranged in different ways, with one common configuration being in parallel. When resistors are in parallel, they connect across the same two points. Consequently, the voltage across each resistor is the same, but the current through each can vary.

The behavior of resistors in parallel is defined by their configuration, allowing the total or equivalent resistance to be smaller than any individual resistor. This is because each resistor creates an additional path for the current flow. Using the formula:
  • \( \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \)
allows you to understand how these pathways contribute to the circuit's overall resistance. In the exercise, substituting given resistors \( R_1 = 2R \) and \( R_2 = 3R \), maintaining calculations precise with common denominators like \( 6R \), helps to accurately determine how the circuit behaviors integrate.

The ability to calculate resistors in parallel effectively aids in designing circuits that meet specific resistance and current needs.
Integer Solutions in Electrical Circuits
In our exercise, both the resistance \( R \) and the equivalent resistance \( R_{eq} \) were required to be integers. This requirement imposes particular conditions on the circuit's design. For the mathematical expression \( \frac{6R}{5} \), where \( 6R/5 \) represents \( R_{eq} \), division by 5 indicates that \( R \) must be a multiple of 5 for the result to remain an integer.

This integer constraint is common in circuit problems to simplify real-world configurations using discrete components. To find all possible values of \( R \) that meet these conditions, we understand \( R = 5k \), where \( k \) is an integer, ensuring that the expression simplifies correctly.

Understanding these integer solutions assists not just in theoretical calculations but also in practical applications, allowing for efficient design and stability in circuits. It highlights the importance of careful consideration of resistance values during circuit design and serves as a fundamental aspect of electrical engineering.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Starting from the Norton equivalent circuit with a resistive load \(R_{L}\) attached, find an expression for the power delivered to the load in terms of \(I_{n}, R_{t}\), and \(R_{L}\). Assuming that \(I_{n}\) and \(R_{t}\) are fixed values and that \(R_{L}\) is a variable, show that maximum power is delivered for \(R_{L}=R_{t} .\) Find an expression for maximum power delivered to the load in terms of \(I_{n}\) and \(R_{t}\).

What conductance in series with a 7025 conductance results in an equivalent conductance of \(2025 ?\)

The heating element of an electric cook top has two resistive elements, \(R_{1}=40 \Omega\) and \(R_{2}=100 \Omega\), which can be operated separately, in series, or in parallel from voltages of either \(120 \mathrm{~V}\) or \(240 \mathrm{~V}\). For the lowest power, \(R_{1}\) is in series with \(R_{2}\), and the combination is operated from \(120 \mathrm{~V}\). What is the lowest power? For the highest power, how should the elements be operated? What power results? List three more modes of operation and the resulting power for each.

A somewhat discharged automotive battery has an open-circuit voltage of \(12.5 \mathrm{~V}\) and supplies \(50 \mathrm{~A}\) when a \(0.1-\Omega\) resistance is connected across the battery terminals. Draw the Thévenin and Norton equivalent circuits, including values for the circuit parameters. What current can this battery deliver to a short circuit? Considering that the energy stored in the battery remains constant under open-circuit conditions, which of these equivalent circuits seems more realistic? Explain.

We can model a certain battery as a voltage source in series with a resistance. The opencircuit voltage of the battery is \(9 \mathrm{~V}\). When a \(100-\Omega\) resistor is placed across the terminals of the battery, the voltage drops to \(6 \mathrm{~V}\). Determine the internal resistance (Thévenin resistance) of the battery.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.