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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 cylindrical air capacitor of length 15.0 m stores 3.20 \(\times\) 10\(^{-9}\) J of energy when the potential difference between the two conductors is 4.00 V. (a) Calculate the magnitude of the charge on each conductor. (b) Calculate the ratio of the radii of the inner and outer conductors.

You have two identical capacitors and an external potential source. (a) Compare the total energy stored in the capacitors when they are connected to the applied potential in series and in parallel. (b) Compare the maximum amount of charge stored in each case. (c) Energy storage in a capacitor can be limited by the maximum electric field between the plates. What is the ratio of the electric field for the series and parallel combinations?

A parallel-plate air capacitor is made by using two plates 12 cm square, spaced 3.7 mm apart. It is connected to a 12-V battery. (a) What is the capacitance? (b) What is the charge on each plate? (c) What is the electric field between the plates? (d) What is the energy stored in the capacitor? (e) If the battery is disconnected and then the plates are pulled apart to a separation of 7.4 mm, what are the answers to parts (a)-(d)?

Some cell walls in the human body have a layer of negative charge on the inside surface and a layer of positive charge of equal magnitude on the outside surface. Suppose that the charge density on either surface is \(\pm\) 0.50 \(\times\) 10\(^{-3}\) C/m\(^2\), the cell wall is 5.0 nm thick, and the cell- wall material is air. (a) Find the magnitude of \(\vec{E}\) S in the wall between the two layers of charge. (b) Find the potential difference between the inside and the outside of the cell. Which is at the higher potential? (c) A typical cell in the human body has a volume of 10\(^{-16}\) m\(^3\). Estimate the total electric-field energy stored in the wall of a cell of this size. (\(Hint\): Assume that the cell is spherical, and calculate the volume of the cell wall.) (d) In reality, the cell wall is made up, not of air, but of tissue with a dielectric constant of 5.4. Repeat parts (a) and (b) in this case.

A 20.0-\(\mu\)F 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-\(\mu\)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.

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