/*! 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 97 A number of capacitors each of c... [FREE SOLUTION] | 91Ó°ÊÓ

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A number of capacitors each of capacitance \(1 \mu \mathrm{F}\) and each one of which get punctured if a potential difference just exceeding 500 volt is applied, are provided. Then an arrangement suitable for giving a capacitor of \(2 \mu \mathrm{F}\) across which 3000 volt may be applied requires at least (a) 18 component capacitors (b) 36 component capacitors (c) 72 component capacitors (d) 144 component capacitors

Short Answer

Expert verified
72 component capacitors are required.

Step by step solution

01

Determine Total Capacitance Needed

We need to design a capacitor setup with a total capacitance of \(2 \mu \mathrm{F}\).
02

Calculate Series Configuration for Voltage Requirement

Each capacitor can withstand up to 500V. To operate at 3000V, we arrange capacitors in series so that the voltage across each is 500V. Therefore, we need \( \frac{3000}{500} = 6 \) capacitors in series.
03

Determine Capacitance of Series Group

The effective capacitance of capacitors in series is given by \( \frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} + ... + \frac{1}{C_n} \). For 6 capacitors each of \(1 \mu \mathrm{F}\), the total capacitance in series is \( \frac{1}{C_{s}} = 6 \cdot \frac{1}{1 \mu \mathrm{F}} = \frac{1}{6} \mu \mathrm{F}\).
04

Calculate Number of Parallel Groups Needed

To achieve a capacitance of \(2 \mu \mathrm{F}\), we need parallel groups. If one series group has \( \frac{1}{6} \mu \mathrm{F} \), we equate \( 6 \cdot C_s = 2 \mu \mathrm{F} \) or \(n \cdot \frac{1}{6} = 2\) which gives \(n = 12\). Thus, 12 series groups are required.
05

Calculate Total Number of Capacitors

Each series group has 6 capacitors, and we need 12 such groups in parallel. The total number needed is \( 6 \times 12 = 72 \) capacitors.

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Key Concepts

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

Capacitance Calculation
Capacitance is a measure of a capacitor's ability to store electric charge for a given electric potential. It is given by the formula \( C = \frac{Q}{V} \), where \( C \) is the capacitance, \( Q \) is the charge, and \( V \) is the voltage. In practical terms, to build or modify a capacitor arrangement, we need to calculate the required capacitance accurately. In our problem, the objective is to achieve a total capacitance of \(2 \mu \text{F}\). This involves figuring out how to combine individual capacitors, each with a capacitance of \(1 \mu \text{F}\), to achieve the desired total. Key points to consider in capacitance calculation:
  • Understanding the unit of capacitance (Farads) and its subunits (such as microfarads \(\mu\text{F}\)).
  • Recognizing that larger total capacitance can be achieved by arranging capacitors in parallel.
  • Series arrangements do not increase capacitance but can distribute voltage, which is also necessary for safety and efficiency.
By properly calculating and arranging capacitors, you can meet specific electrical specifications like the voltage and total capacitance needed.
Series and Parallel Circuits
In electronic circuits, resistors, inductors, and capacitors can be arranged in series or parallel, or combinations of both. Understanding the distinction between series and parallel configurations is essential when designing circuits to meet specific voltage and capacitance needs.

Series Circuits

In a series configuration, components are connected end-to-end. For capacitors:
  • The total capacitance \(C_s\) is found using \( \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots + \frac{1}{C_n} \).
  • Voltage across the series is the sum of voltages across each capacitor; useful when higher voltage handling is required.

Parallel Circuits

For parallel configurations, each component connects across the same voltage:
  • Total capacitance \(C_p\) is simply the sum of all capacitances: \(C_p = C_1 + C_2 + \ldots + C_n\).
  • Parallel arrangements are ideal for distributing current evenly and increasing total capacitance.
In our exercise, by first arranging capacitors in series to meet the voltage requirement, and then arranging these series in parallel, we achieve both the voltage tolerance and capacitance levels required.
Voltage Rating of Capacitors
The voltage rating of a capacitor is the maximum voltage it can withstand without being damaged. Every capacitor has a specified voltage rating, and exceeding this limit risks puncturing and damaging the capacitor permanently. For our problem, each capacitor has a voltage rating of 500 volts. This means:
  • In a series arrangement, the overall applied voltage can be the sum of the ratings of each capacitor in the series. For example, with six 500V capacitors in series, the maximum allowable voltage is 3000 volts.
  • In a parallel arrangement, the voltage across each capacitor remains the same and should not exceed their individual voltage rating.
This ensures the capacitors in parallel groups each manage their voltage safely, preventing over-voltage stress. By combining the right number of capacitors in both series and parallel formations, you can safely achieve higher total rated voltages and capacitances, preventing capacitors from puncturing and keeping the device operating effectively.

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Most popular questions from this chapter

A parallel plate capacitor with air as medium between the plates has a capacitance of \(10 \mu \mathrm{F}\). The area of the capacitor is divided into two equal halves and filled with two media having dielectric constants \(K_{1}=2\) and \(K_{2}=4\). The capacitance of the system will now be (a) \(10 \mu \mathrm{F}\) (b) \(20 \mu \mathrm{F}\) (c) \(30 \mu \mathrm{F}\) (d) \(40 \mu \mathrm{F}\)

\(n\) capacitors each having capacitance \(C\) and breakdown voltage \(V\) are joined in series. The capacitance and the breakdown voltage of the combination is (a) \(C\) and \(V\) (b) \(n C\) and \(n V\) (c) \(n C\) and \(\frac{V}{n}\) (d) \(\frac{C}{n}\) and \(n V\)

The capacities of three capacitors are in the ratio \(1: 2: 3\). Their equivalent capacity when connected in parallel is \(\frac{60}{11} \mu \mathrm{F}\) more than that when they are connected in series. The individual capacitors (in \(\mu \mathrm{F}\) ) are of capacities (a) \(4,6,7\) (b) \(1,2,3\) (c) \(2,3,4\) (d) \(1,3,6\)

A capacitor charged to \(10 \mathrm{~V}\) is discharged by connecting the two plates at \(t=0 .\) If the potential difference across the plates drops to 1 volt at \(t=5 \mathrm{~ms}\) the potential difference at \(t=10 \mathrm{~ms}\) is (a) \(0.1\) volt (b) \(0.2\) volt (c) \(4.2\) volt (d) \(0.02\) volt

Two spherical conductors \(A\) and \(B\) of radii \(a\) and \(b\) \((b>a)\) are placed concentrically in air. \(A\) is given a charge \(+Q\) while \(B\) is earthed. Then the equivalent capacitance of the system is (a) \(4 \pi \varepsilon_{o}\left(\frac{a b}{b-a}\right)\) (b) \(4 \pi \varepsilon_{e}(a+b)\) (c) \(4 \pi \varepsilon_{o} b\) (d) \(4 \pi \varepsilon_{o}\left(\frac{b^{2}}{b-a}\right)\)

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