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24.58. Several \(0.25-\mu \mathrm{F}\) capacitors are available. The voltage across each is not to exceed 600 \(\mathrm{V}\) . You need to make a capacitor with capacitance 0.25\(\mu \mathrm{F}\) to be connected across a potential difference of 960 \(\mathrm{V}\) . (a) Show in a diagram how an cquivalent capacitor with the desired properties can be obtained. (b) No dielectric is a perfect insulator that would not permit the flow of any charge through its volume. Suppose that the dielectric in one of the capacitors in your diagram is a moderately good conductor. What will happen in this case when your combination of capacitors is connected across the \(960-\mathbf{V}\) potential difference?

Short Answer

Expert verified
Use two 0.25 µF capacitors in series (to handle 960 V), then double the setup in parallel for effective 0.25 µF.

Step by step solution

01

Understanding the Problem

We're given capacitors each having 0.25 µF capacitance with a maximum voltage of 600 V. We need to create an equivalent 0.25 µF capacitor that can handle 960 V.
02

Analyzing Voltage Requirement

The required 960 V is more than 600 V, requiring us to connect capacitors in series to divide the voltage. For series connection, total voltage is divided equally, so to handle 960 V, use two capacitors in series (600 V each can handle the combined 960 V).
03

Calculating Series Capacitance

In series connection, the total capacitance is given by \( \frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots \). Using two 0.25 µF capacitors in series: \( \frac{1}{C_{total}} = \frac{1}{0.25} + \frac{1}{0.25} \). Therefore, \( C_{total} = 0.125 \) µF.
04

Adjust Total Capacitance

To achieve 0.25 µF with two series capacitors giving 0.125 µF, we need to duplicate this series arrangement in parallel. This provides the effective capacitance \( 0.125 \text{ µF} + 0.125 \text{ µF} = 0.25 \text{ µF} \).
05

Draw the Diagram

Draw two sets of two capacitors in series. Connect these two sets in parallel. Each vertical pair handles 960 V, and the horizontal parallel sets sum to 0.25 µF.
06

Exploring Impact of Good Conductor

If one capacitor becomes a good conductor, it effectively shorts, causing more voltage across the remaining capacitors. This increases risk of exceeding 600 V in individual capacitors, potentially damaging them.

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Key Concepts

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

Capacitance in Series and Parallel
When dealing with capacitor networks, understanding how capacitance works in series and parallel arrangements is crucial. A series connection will divide the voltage across all capacitors evenly, but it decreases the total capacitance. This is because the total capacitance in a series, denoted as \( C_{total} \), is found using the formula: \[\frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots + \frac{1}{C_n}\]This results in a smaller overall capacitance compared to any individual capacitor in the series.
  • For a single capacitor of 0.25 µF enduring 600 V, connecting two in series divides the voltage to 600 V per capacitor.
  • However, this also halves the overall capacitance to 0.125 µF in this case.To restore the desired capacitance of 0.25 µF, the arrangement is duplicated and placed in parallel.
In a parallel connection, total capacitance is simply the sum of all capacitors involved:\[C_{total} = C_1 + C_2 + \ldots + C_n\]This configuration allows the capacitors to retain their total capacitance at 0.25 µF while sustaining the required voltage.
Voltage Tolerance in Capacitors
Capacitors have a specific voltage tolerance, indicating the maximum potential difference they can safely handle. This is crucial to ensure their longevity and prevent damage. If a capacitor is subjected to a voltage beyond its tolerance level, it might break down, possibly leading to a short circuit.
To manage a high voltage requirement, like 960 V in our case, placing capacitors in series is necessary. This divides the voltage among them. By connecting capacitors with a 600 V voltage rating in series, each capacitor can manage its portion of the load without exceeding its capacity.
  • In a scenario demanding higher voltage than a single capacitor can handle, such as our example, a series arrangement ensures each capacitor remains within safe operating conditions.
  • It’s essential to match the total voltage the network must withstand with the sum of individual voltage tolerances.
Another aspect in maintaining the correct voltage tolerance is to ensure that no capacitor inadvertently becomes a conductor, which can severely disrupt the balance.
Effects of Dielectric Conductivity
Dielectric materials within capacitors ideally act as insulators. However, real-world materials possess some degree of conductivity, which can lead to issues over time. If the dielectric becomes too conductive, it allows additional current to flow through, weakening the capacitor's effectiveness.
  • In our described capacitor network, if one becomes a good conductor due to a fault in the dielectric, it effectively creates a short circuit.
  • This would result in the remaining capacitors bearing a greater voltage strain, potentially breaching their maximum tolerance.
Such a condition dramatically increases the risk of the remaining capacitors failing, as their voltage tolerances are exceeded.
A conductive dielectric can lead to overheating or even catastrophic failure. It's vital to use high-quality dielectric materials and regularly check the health of capacitors employed in critical networks, especially those dealing with high voltages. Regular maintenance and inspections can preempt such issues, ensuring safe operation over the long term.

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Most popular questions from this chapter

24.77. Three square metal plates \(A, B,\) and \(C,\) each 12.0 \(\mathrm{cm}\) on a side and 1.50 \(\mathrm{mm}\) thick, are arranged as in Fig. 24.43 . The plates are separated by sheets of paper 0.45 \(\mathrm{mm}\) thick and with dielectric constant \(4.2 .\) The outer plates are connected together and connected to point \(b\) . The inner plate is connected to point \(a\) (a) Copy the diagram and show by plus and minus signs the charge distribution on the plates when point \(a\) is maintained at a positive potential relative to point \(b\) . (b) What is the capacitance between points \(a\) and \(b ?\)

24.48 . A parallel-plate capacitor has the volume between its plates filled with plastic with dielectric constant \(K\) . The magnitude of the charge on each plate is \(Q\) . Each plate has area \(A\) , and the distance between the plates is \(d\) . (a) Use Gauss's law as stated in Eq. \((24.23)\) to calculate the magnitude of the electric field in the

24.3. A parallel-plate air capacitor of capacitance 245 \(\mathrm{pF}\) has a charge of magnitude 0.148\(\mu \mathrm{C}\) on each plate. The plates are 0.328 \(\mathrm{mm}\) apart. (a) What is the potential difference between the plates? (b) What is the area of each plate? (c) What is the electric-field magnitude between the plates? (d) What is the surface charge density on each plate?

24.47. A \(12.5-\mu F\) capacitor is connected to a power supply that keeps a constant potential difference of 24.0 \(\mathrm{V}\) across the plates. A piece of material having a dielectric constant of 3.75 is placed between the plates, completely filling the space between them. (a) How much energy is stored in the capacitor before and after the diclectric is inserted? (b) By how much did the energy change dur- ing the insertion? Did it increase or decrease?

24.11. A cylindrical capacitor has an inner conductor of radius 1.5 \(\mathrm{mm}\) and an outer conductor of radius 3.5 \(\mathrm{mm}\) . The two conductors are separated by vacuum, and the entire capacitor is 2.8 \(\mathrm{m}\) long. (a) What is the capacitance per unit length? (b) The potential of the inner conductor is 350 \(\mathrm{mV}\) higher than that of the outer conductor. Find the charge (magnitude and sign) on both conductors.

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