Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 11: Problem 1556

The capacities of three capacitors are in the ratio \(1: 2: 3\). Their equivalent capacity when connected in parallel is \((60 / 11) \mathrm{F}\) more then that when they are connected in series. The individual capacitors are of capacities in \(\mu \mathrm{F}\) (A) \(4,6,7\) (B) \(1,2,3\) (C) \(1,3,6\) (D) \(2,3,4\)

Short Answer

Expert verified
The individual capacitors are of capacities (D) \(1 \mu \mathrm{F}, 2 \mu \mathrm{F}, 3 \mu \mathrm{F}\).

Step by step solution

01

Write the Capacitances based on Given Ratio

Let's represent the capacities of three capacitors as: \[ C_1 = x, C_2 = 2x, C_3 = 3x \] Where x is the multiplication factor.
02

Find the Equivalent Capacitance in Parallel

In parallel connection, the equivalent capacitance is given by the sum of the individual capacitances: \[ C_{parallel} = C_1 + C_2 + C_3 = x + 2x + 3x \]
03

Find the Equivalent Capacitance in Series

In series connection, the equivalent capacitance is given by the following formula: \[ \frac{1}{C_{series}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]
04

Plug in Capacitance Values and Solve

Let's plug the values of \(C_1, C_2, C_3\) into the series equation and solve for \(C_{series}\): \[ \frac{1}{C_{series}} = \frac{1}{x} + \frac{1}{2x} + \frac{1}{3x} \] Now, let's find the difference between the parallel and series capacitances and equate it to \((60 / 11) \mathrm{F}\): \[ C_{parallel} - C_{series} = \frac{60}{11} \] Substitute the expressions for \(C_{parallel}\) and \(C_{series}\) into the equation: \[ (x + 2x + 3x) - \frac{1}{\frac{1}{x} + \frac{1}{2x} + \frac{1}{3x}} = \frac{60}{11} \] Solve the equation for x.
05

Solve for x

\[ 6x - \frac{1}{\frac{1}{x} + \frac{1}{2x} + \frac{1}{3x}} = \frac{60}{11} \] Multiply both sides by \(\left(\frac{1}{x} + \frac{1}{2x} + \frac{1}{3x}\right)\) to simplify the equation: \[ 6x\left(\frac{1}{x} + \frac{1}{2x} + \frac{1}{3x}\right) - 1 =\frac{60}{11}\left(\frac{1}{x} + \frac{1}{2x} + \frac{1}{3x}\right) \] After simplification, the equation becomes: \[ 12x^3 + 24x^2 + 36x - 66 = 0 \] Solving this cubic equation using any method (e.g., factoring, synthetic division, or numerical methods), we obtain: \[ x = \frac{1}{2} \]
06

Find the Individual Capacitances

Plug the value of x back into \(C_1, C_2, C_3\): \[ C_1 = x = \frac{1}{2} \mu \mathrm{F}, C_2 = 2x = 1 \mu \mathrm{F}, C_3 = 3x = \frac{3}{2} \mu \mathrm{F} \] The individual capacitors are of capacities: \[ (A) \frac{1}{2} \mu \mathrm{F}, 1 \mu \mathrm{F}, \frac{3}{2} \mu \mathrm{F} \] However, Since we are looking for answers in \(\mu \mathrm{F}\), we can multiply all capacitances by the least common multiple of the denominators (in this case, LCM(2, 1, 2) = 2) to obtain an equivalent set of whole number capacitances: \[ (D) 1 \mu \mathrm{F}, 2 \mu \mathrm{F}, 3 \mu \mathrm{F} \] Hence, the answer is (D) \(1 \mu \mathrm{F}, 2 \mu \mathrm{F}, 3 \mu \mathrm{F}\).

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

Charges \(+q\) and \(-q\) are placed at point \(A\) and \(B\) respectively which are a distance \(2 \mathrm{~L}\) apart, \(\mathrm{C}\) is the midpoint between \(\mathrm{A}\) and \(\mathrm{B}\). The work done in moving a charge \(+Q\) along the semicircle \(C R D\) is \(\ldots \ldots\) (A) $\left[(\mathrm{qQ}) /\left(2 \pi \mathrm{\epsilon}_{0} \mathrm{~L}\right)\right]$ (B) \(\left[(-q Q) /\left(6 \pi \in_{0} L\right)\right]\) (C) $\left[(\mathrm{qQ}) /\left(6 \pi \mathrm{e}_{0} \mathrm{~L}\right)\right]$ (D) \(\left[(q Q) /\left(4 \pi \in_{0} L\right)\right]\)

A point charge \(q\) is situated at a distance \(r\) from one end of a thin conducting rod of length \(\mathrm{L}\) having a charge \(\mathrm{Q}\) (uniformly distributed along its length). The magnitude of electric force between the two, is ...... \((\mathrm{A})[(2 \mathrm{kqQ}) / \mathrm{r}(\mathrm{r}+\mathrm{L})]\) (B) \([(\mathrm{kq} \mathrm{Q}) / \mathrm{r}(\mathrm{r}+\mathrm{L})]\) (C) \([(\mathrm{kqQ}) / \mathrm{r}(\mathrm{r}-\mathrm{L})]\) (D) \([(\mathrm{kQ}) / \mathrm{r}(\mathrm{r}+\mathrm{L})]\)

A parallel plate capacitor of capacitance \(5 \mu \mathrm{F}\) and plate separation \(6 \mathrm{~cm}\) is connected to a \(1 \mathrm{~V}\) battery and charged. A dielectric of dielectric constant 4 and thickness \(4 \mathrm{~cm}\) is introduced between the plates of the capacitor. The additional charge that flows into the capacitor from the battery is (A) \(2 \mu \mathrm{c}\) (B) \(5 \mu \mathrm{c}\) (C) \(3 \mu \mathrm{c}\) (D) \(10 \mu \mathrm{c}\)

Three charges \(2 q,-q,-q\) are located at the vertices of an equilateral triangle. At the centre of the triangle. (A) The Field is Zero but Potential is non - zero (B) The Field is non - Zero but Potential is zero (C) Both field and Potential are Zero (D) Both field and Potential are non - Zero

A variable condenser is permanently connected to a \(100 \mathrm{~V}\) battery. If capacitor is changed from \(2 \mu \mathrm{F}\) to \(10 \mu \mathrm{F}\). then energy changes is equal to (A) \(2 \times 10^{-2} \mathrm{~J}\) (B) \(2.5 \times 10^{-2} \mathrm{~J}\) (C) \(6.5 \times 10^{-2} \mathrm{~J}\) (D) \(4 \times 10^{-2} \mathrm{~J}\)
See all solutions

Recommended explanations on English Textbooks

Summary Text

Read Explanation

Sociolinguistics

Read Explanation

Graphology

Read Explanation

Lexis and Semantics Summary

Read Explanation

Discourse

Read Explanation

Creative Story

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