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Your electronics company has several identical capacitors with capacitance \(C_1\) and several others with capacitance \(C_2\). You must determine the values of \(C_1\) and \(C_2\) but don't have access to \(C_1\) and \(C_2\) individually. Instead, you have a network with \(C_1\) and \(C_2\) connected in series and a network with \(C_1\) and \(C_2\) connected in parallel. You have a 200.0-V battery and instrumentation that measures the total energy supplied by the battery when it is connected to the network. When the parallel combination is connected to the battery, 0.180 J of energy is stored in the network. When the series combination is connected, 0.0400 J of energy is stored. You are told that \(C_1\) is greater than \(C_2\). (a) Calculate \(C_1\) and \(C_2\). (b) For the series combination, does \(C_1\) or \(C_2\) store more charge, or are the values equal? Does \(C_1\) or \(C_2\) store more energy, or are the values equal? (c) Repeat part (b) for the parallel combination.

Short Answer

Expert verified
(a) \( C_1 = 6.0 \mu F \), \( C_2 = 3.0 \mu F \). (b) Equal charge, \( C_2 \) stores more energy in series. (c) \( C_1 \) stores more charge and energy in parallel.

Step by step solution

01

Understanding Capacitor Energy in Parallel Combination

The energy stored in a parallel capacitor network is given by the formula: \( E = \frac{1}{2} C V^2 \), where \( C \) is the total capacitance of the network and \( V \) is the voltage applied. For the parallel network, this total capacitance is \( C_1 + C_2 \). Given that \( E = 0.180 \) J and \( V = 200.0 \) V, we can find \( C_1 + C_2 \).
02

Calculating Total Capacitance in Parallel Combination

Using the formula, \( 0.180 = \frac{1}{2} (C_1 + C_2) (200)^2 \). Simplifying gives \( C_1 + C_2 = \frac{0.180}{20000} = 9.0 \times 10^{-6} \) F or \( 9.0 \mu F \).
03

Understanding Capacitor Energy in Series Combination

For capacitors in series, the energy stored is also given by \( E = \frac{1}{2} C V^2 \), but the total capacitance formula is \( \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} \). Given \( E = 0.0400 \) J and \( V = 200.0 \) V, we can calculate \( C \).
04

Calculating Total Capacitance in Series Combination

Using \( 0.0400 = \frac{1}{2} C (200)^2 \), solve for \( C \): \( C = \frac{0.0400}{20000} = 4.0 \times 10^{-7} \) F or \( 0.40 \mu F \).
05

Finding Individual Capacitance Values

From the series formula \( \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} \), substitute \( C = 0.40 \mu F \) to get \( \frac{1}{0.40} = \frac{1}{C_1} + \frac{1}{C_2} \). Also, we know \( C_1 + C_2 = 9.0 \mu F \). Solving these equations simultaneously, we find \( C_1 = 6.0 \mu F \) and \( C_2 = 3.0 \mu F \).
06

Charge and Energy in Series Combination

In a series network, the charge on each capacitor is the same, due to charge conservation. Thus, both \( C_1 \) and \( C_2 \) store equal charges, but the energy stored is different: \( U = \frac{1}{2} C V^2 \). Since \( C_1 > C_2 \), \( C_2 \) will have a higher voltage across it and thus store more energy.
07

Charge and Energy in Parallel Combination

In a parallel network, each capacitor stores energy independently. The voltage across both is the same: 200 V. Since \( C_1 = 6.0 \mu F \) and \( C_2 = 3.0 \mu F \), \( C_1 \) stores more charge as \( Q = CV \), and also more energy since \( C_1 > C_2 \).

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Key Concepts

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

Series Circuit
Capacitors connected in a series circuit present a unique interaction in terms of how charge and voltage distribute across each component. When capacitors are in series, the total capacitance of the network is given by \[ \frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} \]This is different from how resistances add up. The total capacitance in series is always less than the smallest individual capacitance in the network. This interesting property is because in a series connection, each capacitor "shares" the voltage provided by the power source.
  • The charge on each capacitor remains the same: This is a key characteristic of series connections, as charge conservation ensures that all capacitors receive the same amount of charge.
  • Capacitors in series are useful when needing very specific low capacitance values not available in a single capacitor.
Understanding series circuits is crucial for developing diverse electronic functionalities, especially where voltage division is necessary.
Parallel Circuit
Parallel circuits offer a straightforward way to increase the total capacitance in a network. When capacitors are connected in parallel, the total capacitance is simply the sum of their individual capacitances:\[ C_{total} = C_1 + C_2 \]This setup ensures that the voltage across each capacitor is the same. Each capacitor in such a network fully experiences the applied voltage from the source.
  • More capacitance means more total charge can be stored in the network, which is evident in how capacitors in parallel add up.
  • Parallel configurations are particularly useful when maximal energy storage is needed.
In practical applications, parallel circuits allow flexibility in design, especially when assembling capacitors of different capacitance to achieve desired total values efficiently.
Capacitance Calculation
Calculating capacitance involves understanding how capacitors behave both individually and in networks. Techniques vary depending on their configuration, such as series or parallel. In a series network, use \[ \frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} \]while in parallel, employ \[ C_{total} = C_1 + C_2 \]
When tasked with finding the values of individual capacitors in a combination, like in our exercise, simultaneous equations derived from these principles often provide a pathway to the solution.
  • For series configurations, solving for the total capacitance then using algebraic equations can pinpoint individual values.
  • In parallel setups, direct addition simplifies finding total values, useful in calculations involving energy storage.
Capacitance calculations are fundamental in designing circuits to ensure functionality aligns with electrical needs.
Energy Storage
The energy stored in capacitors is a key aspect of their utility. It can be calculated using the expression:\[ E = \frac{1}{2} C V^2 \]This formula shows that energy depends on the capacitance and the square of the voltage across the capacitor.
  • In series circuits, energy distribution varies between capacitors, because while the charge is the same, the voltages are different, causing variations in energy stored.
  • For parallel circuits, equal voltage across capacitors translates into predictable energy calculations based entirely on capacitance.
The ability to store energy makes capacitors invaluable in various electronics where sudden energy release is needed, such as in flash photography or power conditioning scenarios.
Circuit Analysis
Circuit analysis involves applying fundamental principles to understand how complex networks function. For capacitors, this often means using known relationships such as those for series and parallel connections. By analyzing the capacitance and energy equations concurrently, one can unpack the workings of the network.
  • In our exercise, understanding both energy values and capacitance in different setups lets us calculate unknowns precisely.
  • Analyzing both charge and energy across networks enables deeper insights into capacitor functionality.
Whether for a simple or complex circuit, analysis is crucial in predicting behavior and ensuring circuits meet their design specifications, facilitating troubleshooting and optimization.

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Most popular questions from this chapter

A 0.350-m-long cylindrical capacitor consists of a solid conducting core with a radius of 1.20 mm and an outer hollow conducting tube with an inner radius of 2.00 mm. The two conductors are separated by air and charged to a potential difference of 6.00 V. Calculate (a) the charge per length for the capacitor; (b) the total charge on the capacitor; (c) the capacitance; (d) the energy stored in the capacitor when fully charged.

A budding electronics hobbyist wants to make a simple 1.0-nF capacitor for tuning her crystal radio, using two sheets of aluminum foil as plates, with a few sheets of paper between them as a dielectric. The paper has a dielectric constant of 3.0, and the thickness of one sheet of it is 0.20 mm. (a) If the sheets of paper measure 22 \(\times\)28 cm and she cuts the aluminum foil to the same dimensions, how many sheets of paper should she use between her plates to get the proper capacitance? (b) Suppose for convenience she wants to use a single sheet of posterboard, with the same dielectric constant but a thickness of 12.0 mm, instead of the paper. What area of aluminum foil will she need for her plates to get her 1.0 nF of capacitance? (c) Suppose she goes high-tech and finds a sheet of Teflon of the same thickness as the posterboard to use as a dielectric. Will she need a larger or smaller area of Teflon than of posterboard? Explain.

A parallel-plate vacuum capacitor has 8.38 J of energy stored in it. The separation between the plates is 2.30 mm. If the separation is decreased to 1.15 mm, what is the energy stored (a) if the capacitor is disconnected from the potential source so the charge on the plates remains constant, and (b) if the capacitor remains connected to the potential source so the potential difference between the plates remains constant?

The dielectric to be used in a parallel-plate capacitor has a dielectric constant of 3.60 and a dielectric strength of 1.60 \(\times\) 10\(^7\) V>m. The capacitor is to have a capacitance of 1.25 \(\times\) 10\(^{-9}\) F and must be able to withstand a maximum potential difference of 5500 V. What is the minimum area the plates of the capacitor may have?

A 10.0- \(\mu\)F parallel-plate capacitor with circular plates is connected to a 12.0-V battery. (a) What is the charge on each plate? (b) How much charge would be on the plates if their separation were doubled while the capacitor remained connected to the battery? (c) How much charge would be on the plates if the capacitor were connected to the 12.0-V battery after the radius of each plate was doubled without changing their separation?

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