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Chapter 25: Problem 78

You have many \(2.0 \mu \mathrm{F}\) capacitors, each capable of withstanding \(200 \mathrm{~V}\) without undergoing electrical breakdown (in which they conduct charge instead of storing it). How would you assemble a combination having an equivalent capacitance of (a) \(0.40 \mu \mathrm{F}\) and (b) \(1.2 \mu \mathrm{F}\), each combination capable of withstanding \(1000 \mathrm{~V} ?\)

Short Answer

Expert verified
Use 5 in series for (a) 0.40 渭F. For (b) 1.2 渭F, use three such series sets in parallel.

Step by step solution

01

Understanding Series and Parallel Capacitance

In a series connection, the total capacitance \( C_s \) is given by \( \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots \). In parallel, the total capacitance \( C_p \) is \( C_p = C_1 + C_2 + \cdots \). Series connections increase the voltage tolerance, while parallel connections increase capacitance.
02

Voltage Requirements and Strategy

To withstand \(1000\, \text{V}\), connect several capacitors in series to distribute the voltage. Given each capacitor can withstand \(200\, \text{V}\), use \(5\) in series since \(5 \times 200 = 1000\). This connection will share the voltage across the capacitors.
03

Calculating Capacitance for 5 Series Capacitors

Each capacitor is \(2.0 \mu \text{F}\). For \(5\) capacitors in series, the equivalent capacitance \( C_s \) is \( \frac{1}{C_s} = \frac{1}{2.0} + \frac{1}{2.0} + \cdots + \frac{1}{2.0} \) repeated 5 times. This simplifies to \( C_s = \frac{2.0}{5} = 0.4 \mu \text{F} \).
04

Constructing 1.2 渭F with 1000 V tolerance

Since \(0.40 \mu \text{F}\) is achieved by \(5\) capacitors in series, to reach \(1.2 \mu \text{F}\), use three \(0.40 \mu \text{F}\) stacks in parallel. So we need \(5 \times 3 = 15\) capacitors in total, organized as \(3\) parallel groups of \(5\) series capacitors.

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Key Concepts

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

Equivalent Capacitance
When working with capacitor networks, equivalent capacitance is a measure that helps understand the overall storage capacity of a particular configuration of capacitors. If you want to replace multiple capacitors with a single equivalent capacitor that has the same function, you鈥檒l need to calculate this equivalent capacitance properly. Here's how different arrangements affect capacitance:
  • **In Series:** When capacitors are connected end-to-end, the total or equivalent capacitance, denoted as \( C_s \), decreases. The formula to find \( C_s \) is:\[ \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots \]Here, adding more capacitors in series results in a capacitance that is smaller than that of any individual capacitor in the sequence.
  • **In Parallel:** When capacitors are connected side-by-side, the total capacitance, \( C_p \), increases. Only by summing up the individual capacitances do you get:\[ C_p = C_1 + C_2 + \cdots \]Parallel arrangements allow the capability to store more charge at the same voltage.
Series and Parallel Connections
Understanding the behavior of capacitors in series and parallel connections is vital for designing circuits that meet specific requirements. Capacitors in series and parallel provide different qualities to the network:
  • **Series Connections:**
    • Are often used to withstand higher voltages.
    • Each capacitor shares the total voltage, which means if you need the network to survive high voltages, placing several capacitors in series is a smart choice.
    • Lower the total capacitance, as calculated by the formula for series capacitance.
  • **Parallel Connections:**
    • Are typically used to increase the overall capacitance of a circuit without changing the voltage tolerance of individual capacitors.
    • Help in storing more charge, making it useful in applications where higher energy storage is necessary.
Voltage Tolerance
Voltage tolerance is a critical parameter in designing capacitor networks, especially when you need the setup to withstand specific operational voltages. When capacitors are combined, their configuration impacts how they handle voltage:
  • **Series Capacitors:**
    • Divides the total applied voltage across each capacitor in the series. For example, if each capacitor withstands \(200\, \mathrm{V}\) and you need them to handle up to \(1000\, \mathrm{V}\), you would connect five in series. This way, each capacitor ensures the network can handle the full \(1000\, \mathrm{V}\).
  • **Parallel Capacitors:**
    • Share the same voltage across all capacitors in the group. Thus, for each to withstand the applied voltage, the voltage rating of the individual capacitors must meet or exceed that voltage.
Ensuring the correct voltage tolerance prevents electrical breakdown, where capacitors start conducting rather than storing charge, thereby compromising the circuit's safety and functionality.

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Most popular questions from this chapter

The parallel plates in a capacitor, with a plate area of \(8.50 \mathrm{~cm}^{2}\) and an air-filled separation of \(3.00 \mathrm{~mm}\), are charged by a \(6.00 \mathrm{~V}\) battery. They are then disconnected from the battery and pulled apart (without discharge) to a separation of \(8.00\) \(\mathrm{mm}\). Neglecting fringing, find (a) the potential difference between the plates, (b) the initial stored energy, (c) the final stored energy, and (d) the work required to separate the plates.

Two parallel-plate capacitors, \(6.0 \mu \mathrm{F}\) each, are connected in parallel to a \(10 \mathrm{~V}\) battery. One of Fig. 25-29 Problems 11, the capacitors is then squeezed so 17, and 38 . that its plate separation is \(50.0 \%\) of its initial value. Because of the squeezing, (a) how much additional charge is transferred to the capacitors by the battery and (b) what is the increase in the total charge stored on the capacitors?

Assume that a stationary electron is a point of charge. What is the energy density \(u\) of its electric field at radial distances (a) \(r=\) \(1.00 \mathrm{~mm}\), (b) \(r=1.00 \mu \mathrm{m}\), (c) \(r=1.00 \mathrm{~nm}\), and \((\mathrm{d}) r=1.00 \mathrm{pm} ?\) (e) What is \(u\) in the limit as \(r \rightarrow 0 ?\)

An air-filled parallel-plate capacitor has a capacitance of \(1.3\) pF. The separation of the plates is doubled, and wax is inserted between them. The new capacitance is \(2.6 \mathrm{pF}\). Find the dielectric constant of the wax.

A certain capacitor is charged to a potential difference \(V\). If you wish to increase its stored energy by \(10 \%\), by what percentage should you increase \(V ?\)
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