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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

An air capacitor is made from two flat parallel plates 1.50 mm apart. The magnitude of charge on each plate is 0.0180 \(\mu\)C when the potential difference is 200 V. (a) What is the capacitance? (b) What is the area of each plate? (c) What maximum voltage can be applied without dielectric breakdown? (Dielectric breakdown for air occurs at an electric-field strength of 3.0 \(\times\) 10\(^6\) V/m.) (d) When the charge is 0.0180 \(\mu\)C, what total energy is stored?

A parallel-plate capacitor has capacitance \(C_0\) = 8.00 pF when there is air between the plates. The separation between the plates is 1.50 mm. (a) What is the maximum magnitude of charge Q that can be placed on each plate if the electric field in the region between the plates is not to exceed 3.00 \(\times\) 10\(^4\) V/m? (b) A dielectric with \(K = 2.70\) is inserted between the plates of the capacitor, completely filling the volume between the plates. Now what is the maximum magnitude of charge on each plate if the electric field between the plates is not to exceed 3.00 \(\times\) 10\(^4\) V/m?

A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 12.5 cm, and the outer sphere has radius 14.8 cm. A potential difference of 120 V is applied to the capacitor. (a) What is the energy density at \(r\) = 12.6 cm, just outside the inner sphere? (b) What is the energy density at \(r\) = 14.7 cm, just inside the outer sphere? (c) For a parallel-plate capacitor the energy density is uniform in the region between the plates, except near the edges of the plates. Is this also true for a spherical capacitor?

A spherical capacitor is formed from two concentric, spherical, conducting shells separated by vacuum. The inner sphere has radius 15.0 cm and the capacitance is 116 pF. (a) What is the radius of the outer sphere? (b) If the potential difference between the two spheres is 220 V, what is the magnitude of charge on each sphere?

Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is \(E = 3.20\times10^5\space V/m\). When the space is filled with dielectric, the electric field is \(E = 2.50\times10^5\space V/m\). (a) What is the charge density on each surface of the dielectric? (b) What is the dielectric constant?

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