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To repair a power supply for a stereo amplifier, an electronics technician needs a \(100-\mu \mathrm{F}\) capacitor capable of withstanding a potential difference of \(90 \mathrm{V}\) between the plates. The only available supply is a box of five \(100-\mu \mathrm{F}\) capacitors, each having a maximum voltage capability of \(50 \mathrm{V} .\) Can the technician substitute a combination of these capacitors that has the proper electrical characteristics? If so, what will be the maximum voltage across any of the capacitors used? (Suggestion: The technician may not have to use all the capacitors in the box.)

Short Answer

Expert verified
Yes, the technician can use a combination of two capacitors in series from the available supply. The maximum voltage across any of the capacitors used in the series configuration would be 50 V.

Step by step solution

01

- Understanding of Capacitors in Series

When in series, capacitors combine in such a way that the total capacitance decreases but their total voltage capability increases. The total voltage \(V_T\) is the sum of voltages across each capacitor. The total capacitance \(C_T\) of capacitors in series is calculated using the formula, \(\frac{1}{C_T} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots + \frac{1}{C_n}\)
02

- Calculation total voltage

The total voltage capability is the sum of the individual voltages. Therefore, if the technician connects two capacitors in series, the total voltage capability would be \(50V + 50V = 100V\), which is greater than the required 90V. So, the technician can use two capacitors in series.
03

- Verifying the Capacitance Requirement

Now proceed with checking whether the capacitance requirement of the resulting series configuration is met. For the two capacitors of equal capacitance \(C_1 = C_2 = 100 \, \mu F\), the total capacitance \(C_T\) can be calculated as follows: \(\frac{1}{C_T} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{100\, \mu F} + \frac{1}{100\,\mu F} = \frac{2}{100\, \mu F}\), which means that \(C_T = \frac{100}{2} = 50\, \mu F\). Although the capacitance requirement is not met, it is not a problem since it is not a requirement for the repair. The critical requirement was the voltage capability, which is satisfied.
04

- Determining the Maximum Voltage

The maximum voltage that can be applied across any of the capacitors in this series configuration is 50V as per the individual capacitor's specification.

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Key Concepts

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

Total Capacitance
When connecting capacitors in series, it's essential to understand how their total capacitance behaves. The formula to calculate the total capacitance \( C_T \) for capacitors in series is given by:
  • \( \frac{1}{C_T} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots + \frac{1}{C_n} \)
This equation shows that, unlike resistors, when capacitors are in series, the total capacitance \( C_T \) is always less than that of any individual capacitor in the series. This occurs because the series connection increases the separation between the charges, effectively reducing the capability to store charge when viewed as a whole system.
In simpler terms, while each additional capacitor in series contributes to an increased voltage rating, it actually divides the total capacitance. Thus, even though it's necessary to use series connections to achieve desired voltage capability, it will result in reduced capacitance.
Voltage Capability
Voltage capability is one of the critical factors when configuring capacitors in series. Each capacitor has a specific maximum voltage it can tolerate, often referred to as the breakdown voltage. In a series connection, capacitors effectively split the total applied voltage along the series.
  • The total voltage capability across the series is the sum of the individual voltages the capacitors can handle.
  • For example, two capacitors rated at 50V each in series can support a combined total voltage of 100V.
This arrangement allows electronics technicians to meet higher voltage requirements without exceeding the limitations of individual capacitors. However, the maximum voltage allowed across any single capacitor will still be limited to its breakdown voltage, which in this problem is 50V.
Stereo Amplifier Repair
In the context of repairing a stereo amplifier, ensuring that the capacitors can sustain both the necessary voltage and capacitance is crucial. Here, the challenge is not just about replacing a faulty component but ensuring the repair effectively restores functionality with available components. For example, say you need a single 100-\(\mu F\) capacitor capable of 90V, but only have capacitors with 50V capability. By connecting two of these in series, the voltage requirement is comfortably met at 100V, although the total capacitance drops to 50-\(\mu F\). While this lower capacitance may not affect the device's ability to function properly, it might influence the performance in ways that the technician must consider, perhaps in the amplifier's sound fidelity or power efficiency.
Electrical Characteristics of Capacitors
Capacitors are fundamental components in electronic circuits, owing to their ability to store and release electrical energy. Their electrical characteristics include factors like capacitance value, voltage rating, tolerance, and equivalent series resistance (ESR). Understanding these characteristics can aid in selecting the proper capacitor for a specific application.
  • Capacitance Value: Determines how much charge a capacitor can hold. Larger capacitance means more energy storage.
  • Voltage Rating: The maximum voltage a capacitor can withstand without risk of failure—crucial when working with high-voltage circuits.
  • Tolerance: Indicates how much the capacitance can vary from its rated value, affecting precision circuits.
  • Equivalent Series Resistance (ESR): Represents real-world inefficiencies within a capacitor, influencing performance in high-frequency applications.
Balancing these characteristics with a circuit's requirements ensures reliable and efficient operation, especially in specialized equipment like stereo amplifiers. Understanding these factors helps in making informed decisions during repairs and optimizations.

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Most popular questions from this chapter

The general form of Gauss's law describes how a charge creates an electric field in a material, as well as in vacuum. It is $$\oint \mathbf{E} \cdot d \mathbf{A}=\frac{q}{\epsilon}$$ where \(\epsilon=\kappa \epsilon_{0}\) is the permittivity of the material. (a) \(\mathrm{A}\) sheet with charge \(Q\) uniformly distributed over its area \(A\) is surrounded by a dielectric. Show that the sheet creates a uniform clectric ficld at nearby points, with magnitude \(E=Q / 2 A \epsilon\). (b) Two large sheets of area \(A\), carrying opposite charges of equal magnitude \(Q,\) are a small distance \(d\) apart. Show that they create uniform electric field in the space between them, with magnitude \(E=Q / A \epsilon\) (c) Assume that the negative plate is at zero potential. Show that the positive plate is at potential Qd/Ae. (d) Show that the capacitance of the pair of plates is \(A \epsilon / d=\kappa A \epsilon_{0} / d\).

A \(20.0-\mu \mathrm{F}\) spherical capacitor is composed of two concentric metal spheres, one having a radius twice as large as the other. The region between the spheres is a vacuum. Determine the volume of this region.

A parallel-plate capacitor of plate separation \(d\) is charged to a potential difference \(\Delta V_{0}\). A dielectric slab of thickness \(d\) and dielectric constant \(\kappa\) is introduced between the plates while the battery remains connected to the plates. (a) Show that the ratio of energy stored after the diclectric is introduced to the energy stored in the empty capacitor is \(U / U_{0}=\kappa .\) Give a physical explanation for this increase in stored energy. (b) What happens to the charge on the capacitor? (Note that this situation is not the same as in Example 26.7 , in which the battery was removed from the circuit before the dielectric was introduced.)

A small object of mass \(m\) carries a charge \(q\) and is suspended by a thread between the vertical plates of a parallel-plate capacitor. The plate separation is \(d\). If the thread makes an angle \(\theta\) with the vertical, what is the potential difference between the plates?

A \(10.0-\mu \mathrm{F}\) capacitor has plates with vacuum between them. Each plate carries a charge of magnitude \(1000 \mu \mathrm{C}\). A particle with charge \(-3.00 \mu \mathrm{C}\) and mass \(2.00 \times 10^{-16} \mathrm{kg}\) is fired from the positive plate toward the negative plate with an initial speed of \(2.00 \times 10^{6} \mathrm{m} / \mathrm{s} .\) Does it reach the negative plate? If so, find its impact speed. If not, what fraction of the way across the capacitor does it travel?

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