/*! 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 69 An unknown capacitor \(C\) is co... [FREE SOLUTION] | 91Ó°ÊÓ

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An unknown capacitor \(C\) is connected in series with a \(3.0-\mu \mathrm{F}\) capacitor; this pair is placed in parallel with a 1.0 - \(\mu\) F capacitor, and the entire combination is put in series with a 2.0 - \(\mu\) F capacitor. (a) Make a circuit diagram of this network. (b) When a potential difference of \(100 \mathrm{V}\) is applied across the open ends of the network, the total energy stored in all the capacitors is \(5.8 \mathrm{mJ}\). Find \(C\).

Short Answer

Expert verified
To find the unknown capacitor, draw the circuit and apply the rules of series and parallel circuit analysis to find the total capacitance of the network. Then, use the energy storage equation to solve for the unknown capacitor value.

Step by step solution

01

Drawing the Circuit Diagram

Draw the circuit diagram as described. The 3.0 µF and unknown capacitor C are in series. They are in parallel with the 1.0 µF capacitor. Then, this whole combination is in series with a 2.0 µF capacitor.
02

Calculate Series and Parallel Capacitances

Apply the formula for capacitors in series for the 3.0 µF and \(C\) capacitor to find their equivalent capacitance. For capacitors in series, \(\frac{1}{C_eq} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{3.0}+\frac{1}{C}\) . This equivalent capacitor is in parallel with a 1.0 µF capacitor. So, the total capacitance of that combination is simply their sum, \(C_t = C_eq + 1.0\). Finally, this combination of capacitors is in series with a 2.0 µF capacitor, so the total capacitance \(C_n\) of the network is given by: \(\frac{1}{C_n} = \frac{1}{C_t} + \frac{1}{2.0}\)
03

Use Energy Formula to Find the Unknown Capacitor \(C\)

The energy stored in a capacitor is given by \(U = \frac{1}{2}C_nV^2\). Substituting \(C_n\) and \(V = 100V\) into this equation gives \(U = \frac{1}{2}(C_n)(100)^2 = 5.8mJ\). Solving this equation gives \(C_n\). Substituting \(C_n\) into the equation in step 2 and solving for \(C\) will yield the value for the unknown capacitor \(C\).

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Key Concepts

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

Series and Parallel Capacitances
Understanding how capacitors behave when connected in different configurations is fundamental in circuit analysis. When capacitors are placed in series, they have a combined effect that reduces the overall capacitance. Think of it like a narrow pipe in a water system: the narrowest point restricts the flow. Similarly, the total capacitance of capacitors in series is less than the smallest individual capacitor in that series.

To calculate the series capacitance, the reciprocal of the total capacitance (\(C_{eq}\text{ for series}\)) is the sum of the reciprocals of each capacitance: \[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots + \frac{1}{C_n} \]Now, in contrast, when capacitors are connected in parallel, they essentially provide more 'space' for charge storage, much like adding more lanes to a highway to increase traffic flow. The total capacitance of parallel capacitors is simply the sum of all individual capacitances: \[ C_{eq}\text{ for parallel} = C_1 + C_2 + \cdots + C_n \]
Given these basics, you can start combining capacitors step by step, always swapping a series or parallel combination with its equivalent, until you compute the total equivalent capacitance for a complex network.
Energy Stored in Capacitors
Capacitors don't just store charge; they store energy. This is crucial for many applications, like providing power bursts in flash photography or stabilizing power supply in electronic devices. The amount of energy (\(U\text{ for energy}\)) stored in a capacitor is related to the voltage across it and its capacitance.

The formula for the energy stored is: \[ U = \frac{1}{2}CV^2 \]where \(C\) is the capacitance and \(V\) is the voltage across the capacitor. Why the squared voltage term? It's because energy is proportional to the charge the capacitor holds and to the voltage that moves that charge into place.

From this relationship, you can infer that doubling the voltage will quadruple the stored energy, reflecting the importance of voltage in energy storage capacity. Also, the greater the capacitance, the more energy can be stored at a given voltage.
Equivalent Capacitance Calculation
Calculating the equivalent capacitance in a complex network requires attention to how capacitors are interconnected. Start by identifying simple series and parallel combinations, and replace them with their equivalent capacitances. Iteratively apply these simplifications until you have computed the overall capacitance from one end of the circuit to the other.

Let's take the written exercise example:
  • Determine equivalent capacitance for the capacitors in series.
  • Combine that result with any parallel capacitances.
  • And then, if required, tackle the next series or parallel combination.
Once the equivalent capacitance (\(C_n\text{ for network}\)) of the entire network is known, you can relate it to the energy stored using the voltage across the network. This helps in solving for any unknowns in the circuit as the exercise demonstrated, effectively turning a complex network into a solvable equation.

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Most popular questions from this chapter

A capacitor consists of square conducting plates \(25 \mathrm{cm}\) on a side and \(5.0 \mathrm{mm}\) apart, carrying charges \(\pm 1.1 \mu \mathrm{C}\). Find (a) the electric field, (b) the potential difference between the plates, and (c) the stored energy.

Your company's purchasing department bought lots of cheap 2.0-\muF, 50-V capacitors. Your budget is maxed out and they won't let you buy additional capacitors for a circuit you're designing. You need \(2.0-\mu \mathrm{F}, 100-\mathrm{V}\) capacitors and \(0.5-\mu \mathrm{F}, 50-\mathrm{V}\) capacitors. How will you combine the available capacitors to make these?

Show that the units of \(\epsilon_{0}\) may be written as \(\mathrm{F} / \mathrm{m}\)

Your company is still stuck with those 2 - \(\mu\) F capacitors from Problem 44. They turn out to be so cheap that their capacitances are all too low, ranging from \(1.7 \mu \mathrm{F}\) to \(1.9 \mu \mathrm{F}\). A colleague suggests you put variable "trimmer" capacitors in parallel with the cheap capacitors and adjust the combination to precisely \(2.00 \mu \mathrm{F}\) The available trimmers have variable capacitance from \(25 \mathrm{nF}\) to \(350 \mathrm{nF} .\) Will they work?

A cubical region \(1.0 \mathrm{m}\) on a side is located between \(x=0\) and \(x=1 \mathrm{m} .\) The region contains an electric field whose magnitude varies with \(x\) but is independent of \(y\) and \(z: E=E_{0}\left(x / x_{0}\right),\) where \(E_{0}=24 \mathrm{kV} / \mathrm{m}\) and \(x_{0}=6.0 \mathrm{m} .\) Find the total energy in the region.

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