Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 12: Problem 1720

Two resistors when connected in parallel have an equivalent of \(2 \Omega\) and when in series of \(9 \Omega\) The values of the two resistors are. (A) \(2 \Omega\) and \(9 \Omega\) (B) \(3 \Omega\) and \(6 \Omega\) (C) \(4 \Omega\) and \(5 \Omega\) (D) \(2 \Omega\) and \(7 \Omega\)

Short Answer

Expert verified
There is no valid solution for the values of the two resistors based on the given information, as we obtain a negative resistance value which is not physically possible.

Step by step solution

01

Write the formula for equivalent resistance in parallel

The formula for the equivalent resistance (Rp) when two resistors are connected in parallel is: \( \frac{1}{Rp} = \frac{1}{R1} + \frac{1}{R2} \) Since we are given that the equivalent resistance when connected in parallel is 2Ω, we can substitute the value of Rp in the equation: \( \frac{1}{2} = \frac{1}{R1} + \frac{1}{R2} \)
02

Write the formula for equivalent resistance in series

The formula for the equivalent resistance (Rs) when two resistors are connected in series is: \( Rs = R1 + R2 \) Since we are given that the equivalent resistance when connected in series is 9Ω, we can substitute the value of Rs in the equation: \( 9 = R1 + R2 \)
03

Solve the equations simultaneously

Now we have two equations with two unknowns, R1 and R2. Let's solve these equations simultaneously to find the values of R1 and R2. From the series equation, we can write R2 in terms of R1: \( R2 = 9 - R1 \) Substitute this expression for R2 in the parallel equation: \( \frac{1}{2} = \frac{1}{R1} + \frac{1}{9 - R1} \) Now, we will solve for R1, then substitute the obtained value to get R2.
04

Solve for R1

To solve for R1, first, we'll make a common denominator: \( \frac{1}{2} = \frac{R1 (9 - R1)}{R1(9 - R1)} + \frac{R1}{R1(9 - R1)} \) Multiplying both sides by the common denominator, we can eliminate the fractions: \( R1(9 - R1) = \frac{1}{2}(R1^2 - 9R1 + R1^2) \) Now, solve the quadratic equation: \( 2R1^2 - 18R1 + 2R1^2 - R1^2 = 0 \) \( R1^2 - 18R1 = 0 \) \( R1(R1 - 18) = 0 \) So, R1 can be 0 Ω or 18 Ω, but since resistors cannot have 0 Ω, we go with R1 = 18 Ω.
05

Solve for R2

Now, we can substitute the value of R1 in the expression for R2 we wrote earlier: \( R2 = 9 - 18 \) \( R2 = -9 \) Since resistance cannot be negative, there appears to be an issue in the given information. Thus, we claim that there is no valid solution for the values of the two resistors based on the given information.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Two electirc bulbs whose resistances are in the ratio of \(1: 2\) are connected in parallel to a constant voltage source the power dissipated in them have the ratio. (A) \(1: 2\) (B) \(1.1\) (C) \(2: 1\) (D) \(1: 4\)

Match the following two columns. Column I \(\quad\) Column II (a) Electrical resistance (p) \(\left[\mathrm{MLT}^{-2} \mathrm{~A}^{2}\right]\) (b) Electric potential (q) \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]\) (c) Specific resistance (r) \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right]\) (d) Specific conductance (s) None of there (A) \(a-q, b-s, c-r, d-p\) (B) \(a-q, b-r, c-s, d-s\) (C) \(a-p, b-q, c-s, d-r\) (D) \(a-p, b-r, c-q, d-s\)

A wire of resistor \(R\) is bent into a circular ring a circular ring of radius \(\mathrm{r}\) Equivalent resistance between two points \(\mathrm{X}\) and \(\mathrm{Y}\) on its circumference, when angle xoy is \(\alpha\), can be given by (A) $\left\\{(\mathrm{R} \alpha) /\left(4 \pi^{2}\right)\right\\}(2 \pi-\alpha)$ (B) \((\mathrm{R} / 2 \pi)(2 \pi-\alpha)\) (C) \(\mathrm{R}(2 \pi-\alpha)\) (D) \((4 \pi / \mathrm{R} \alpha)(2 \pi-\alpha)\)

The resistance of the series combination of two resistances is \(\mathrm{S}\), when they are joined in parallel the total resistance is \(\mathrm{P}\) If $S=n P\(, then the minimum possible value of \)n$ is.... (A) 4 (B) 3 (C) 2 (D) 1

Length of a wire of resistance \(R \Omega\) is increased to 10 times, so its resistance becomes \(1000 \Omega\), therefore \(R=\ldots .\) (The volume of the wire remains same during increase in length) (A) \(0.01 \Omega\) (B) \(0.1 \Omega\) (C) \(1 \Omega\) (D) \(10 \Omega\)
See all solutions

Recommended explanations on English Textbooks

English Grammar

Read Explanation

Email

Read Explanation

Argumentative Essay

Read Explanation

5 Paragraph Essay

Read Explanation

Synthesis Essay

Read Explanation

Language and Social Groups

Read Explanation
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.

Sign-up for free