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You are working on an electronics project requiring a variety of capacitors, but you have only a large supply of 100 -nF capacitors available. Show how you can connect these capacitors to produce each of the following equivalent capacitances: (a) 50 \(\mathrm{nF}\) ; (b) \(450 \mathrm{nF} ;\) (c) \(25 \mathrm{nF} ;\) (d) 75 \(\mathrm{nF}\) .

Short Answer

Expert verified
You can use series and parallel combinations of 100 nF capacitors to achieve the desired capacitances: (a) Two in series (50 nF); (b) Four in parallel with one adjusted (450 nF); (c) Four in series (25 nF); (d) Three in series parallel to one (75 nF).

Step by step solution

01

Understand Capacitor Connections

Capacitors can be connected in series or in parallel to achieve different equivalent capacitances. The formula for capacitors in series is \( \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \dots \), and for parallel, it is \( C_{eq} = C_1 + C_2 + \dots \).
02

Solve for 50 nF

To achieve 50 nF using 100 nF capacitors, we can connect two 100 nF capacitors in series. Using the series formula \( \frac{1}{C_{eq}} = \frac{1}{100} + \frac{1}{100} \), we solve for \( C_{eq} \) to get 50 nF.
03

Solve for 450 nF

To obtain 450 nF, connect four 100 nF capacitors in parallel. In parallel, capacitances add, so \( C_{eq} = 100 + 100 + 100 + 100 = 400 \) nF. Add one more 100 nF capacitor in series with one of the parallel ones, achieving \( C_{eq} = 450 \) nF via \( 400 + 50 \).
04

Solve for 25 nF

For 25 nF, four 100 nF capacitors in series will result in \( \frac{1}{C_{eq}} = \frac{1}{100} + \frac{1}{100} + \frac{1}{100} + \frac{1}{100} = \frac{1}{25} \). Therefore, \( C_{eq} = 25 \) nF.
05

Solve for 75 nF

Connect three 100 nF capacitors in series, which results in \( \frac{1}{C_{eq}} = \frac{1}{100} + \frac{1}{100} + \frac{1}{100} = \frac{1}{33.33} \). Add a fourth 100 nF capacitor in parallel to achieve \( 75 \) nF by \( C_{eq} = 33.33 + 100 = 75 \) nF.

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Key Concepts

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

Series and Parallel Circuits
In electronics, arranging components like capacitors in series or parallel circuits is vital for controlling electrical characteristics such as capacitance.

**Series Circuits:** In a series configuration, the capacitors are connected end-to-end. This means the same electrical current flows through each capacitor. The equivalent capacitance in a series of capacitors is always less than the smallest capacitor in the chain. The formula for total capacitance in series is:
  • \[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \dots \]
**Parallel Circuits:** In a parallel configuration, the capacitors are connected to the same two points. Here, the total capacitance is the sum of each capacitor's capacitance. This configuration results in a larger equivalent capacitance than any single capacitor connected:
  • \[ C_{eq} = C_1 + C_2 + \dots \]
Understanding these configurations is crucial, as they enable adjustments to fit specific electronic needs.
Capacitance Calculations
Calculating capacitance is essential when working with electronic circuits. It determines how a circuit will store and discharge electricity. Capacitance (\(C\)) is measured in farads (F), but in practice, much smaller units like nanofarads (nF) or microfarads (µF) are used.
For example, if you're tasked to achieve a 50 nF using 100 nF capacitors, you'll connect them in series. Two 100 nF capacitors in series follow:
  • \[ \frac{1}{C_{eq}} = \frac{1}{100} + \frac{1}{100} = \frac{1}{50} \Rightarrow C_{eq} = 50 \mathrm{nF} \]
To reach 450 nF, connect four 100 nF capacitors in parallel (400 nF). Then use a combination of one more series connection for precision. This calculation framework assists in configuring circuits effectively.
Electronic Components
Capacitors are fundamental electronic components. They consist of two conductive plates separated by an insulating material, known as a dielectric. They store electric charge temporarily and release it when needed, which is crucial in smoothing power supply outputs, filtering signals, or timing circuits.

Some important roles capacitors play include:
  • **Energy Storage:** Storing electrical energy and releasing it when necessary.
  • **Signal Filtering:** Allowing certain frequencies to pass while blocking others.
  • **Voltage Regulation:** Smoothing voltage fluctuations in power supplies.
Grasping how to connect multiple capacitors to achieve a desired capacitance is invaluable for designing efficient circuits with appropriate functionalities. By manipulating series and parallel connections, one can tailor circuits to specific requirements.

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Most popular questions from this chapter

A 20.0 -\muF capacitor is charged to a potential difference of 800 V. The terminals of the charged capacitor are then connected to those of an uncharged \(10.0-\mu \mathrm{F}\) capacitor. Compute (a) the original charge of the system, (b) the final potential difference across each capacitor, (c) the final energy of the system, and (d) the decrease in energy when the capacitors are connected.

Three capacitors having capacitances of \(8.4,8.4,\) and 4.2\(\mu \mathrm{F}\) are connected in series across a \(36-\mathrm{V}\) potential difference. (a) What is the charge on the \(4.2-\mu F\) capacitor? (b) What is the total energy stored in all three capacitors? (c) The capacitors are disconnected from the potential difference without allowing them to discharge. They are then reconnected in parallel with each other, with the positively charged plates connected together. What is the voltage across each capacitor in the parallel combination? (d) What is the total energy now stored in the capacitors?

A 12.5 -\muF 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 dielectric is inserted? (b) By how much did the energy change during the insertion? Did it increase or decrease?

A parallel-plate air capacitor is to store charge of magnitude 240.0 \(\mathrm{pC}\) on each plate when the potential difference between the plates is 42.0 \(\mathrm{V}\) (a) If the area of each plate is \(6.80 \mathrm{cm}^{2},\) what is the separation between the plates? (b) If the separation between the two plates is double the value calculated in part (a), what potential difference is required for the capacitor to store charge of magnitude 240.0 pC on each plate?

A parallel-plate capacitor is made from two plates 12.0 \(\mathrm{cm}\) on each side and 4.50 \(\mathrm{mm}\) apart. Half of the space between these plates contains only air, but the other half is filled with Plexiglas' of dielectric constant 3.40 (Fig. P24.72). An 18.0-V battery is connected across the plates. (a) What is the capacitance of this combination? (Hint: Can you think of this capacitor as equivalent to two capacitors in parallel? (b) How much energy is stored in the capacitor? (c) If we remove the Plexiglas' but change nothing else, how much energy will be stored in the capacitor?

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