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Two capacitors are connected to a battery. The battery voltage is \(V=60.0 \mathrm{V}\) , and the capacitances are \(C_{1}=2.00 \mu \mathrm{F}\) and \(C_{2}=4.00 \mu \mathrm{F}\) . Determine the total energy stored by the two capacitors when they are wired \((\mathrm{a})\) in parallel and \((\mathrm{b})\) in series.

Short Answer

Expert verified
Parallel: 10.8 mJ, Series: 2.394 mJ.

Step by step solution

01

Understanding the Parallel Configuration

When two capacitors are connected in parallel, their equivalent capacitance \( C_{eq} \) is the sum of their individual capacitances. Thus, for the capacitors \( C_1 \) and \( C_2 \): \[ C_{eq} = C_1 + C_2 = 2.00 \, \mu F + 4.00 \, \mu F = 6.00 \, \mu F \]
02

Calculating Energy in Parallel Configuration

The energy \( E \) stored in capacitors is given by the formula: \[ E = \frac{1}{2} C_{eq} V^2 \] Substituting the values for parallel configuration: \[ E = \frac{1}{2} \times 6.00 \, \mu F \times (60.0 \, V)^2 \]Convert microfarads to farads: \[ E = \frac{1}{2} \times 6.00 \times 10^{-6} \, F \times 3600 \, V^2 \] Calculate the energy: \[ E = \frac{1}{2} \times 6.00 \times 10^{-6} \times 3600 = 10.8 \times 10^{-3} \, J \] Thus, \( E = 10.8 \, mJ \).
03

Understanding the Series Configuration

For capacitors in series, the equivalent capacitance \( C_{eq} \) is calculated using:\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \] \[ \frac{1}{C_{eq}} = \frac{1}{2.00 \, \mu F} + \frac{1}{4.00 \, \mu F} \] \[ \frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{4} = \frac{3}{4} \] \[ C_{eq} = \frac{4}{3} \, \mu F \approx 1.33 \, \mu F \]
04

Calculating Energy in Series Configuration

Using the energy formula, substitute \( C_{eq} \) for series: \[ E = \frac{1}{2} \times 1.33 \, \mu F \times (60.0 \, V)^2 \] Convert microfarads to farads: \[ E = \frac{1}{2} \times 1.33 \times 10^{-6} \, F \times 3600 \, V^2 \] Calculate the energy: \[ E = \frac{1}{2} \times 1.33 \times 10^{-6} \times 3600 = 2.394 \times 10^{-3} \, J \] Thus, \( E = 2.394 \, mJ \).

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Key Concepts

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

Capacitors
Capacitors are essential components in electrical circuits. They store electrical energy in an electric field internally. A capacitor consists of two conductive plates separated by an insulator or dielectric. Here’s how capacitors operate:
  • When voltage is applied across the plates, electric charges accumulate on the plates.
  • This creates an electric field that stores energy.
  • The quantity of charge a capacitor can hold is directly proportional to its capacitance, which is measured in farads (F).
Whether used in electronic devices, power systems, or even camera flashes, capacitors help stabilize voltages, filter signals, and more. Understanding their role and behavior is vital for anyone working with electrical circuits.
Energy Storage
The energy stored in a capacitor is part of its core functionality. Energy storage enhances the capacity of circuits to perform tasks like maintaining voltage levels and supplying power when transitions happen abruptly. The energy \[ E \]stored in a capacitor is calculated using the formula:\[ E = \frac{1}{2} C V^2 \]where \( C \) is the capacitance in farads and \( V \) is the voltage in volts. Some key points are:
  • Energy increases with higher capacitance and greater voltage.
  • The energy is directly proportional to the square of the voltage.
  • For the same capacitance, doubling the voltage increases energy four-fold.
Effectively leveraging energy storage in capacitors allows for more efficient and effective circuit designs.
Series and Parallel Circuits
Understanding the configuration of capacitors is critical because it affects how they store energy. **Series and parallel** configurations are the two primary ways to connect capacitors.In a **parallel configuration**, capacitors are connected end-to-end, making them function as a single larger capacitor. The total equivalent capacitance is simply the sum of the individual capacitances:\[ C_{eq} = C_1 + C_2 \]In a **series configuration**, the charge on each capacitor is the same, which affects the equivalent capacitance in an inversely proportional way:\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \]
  • Series connections generally result in a smaller equivalent capacitance.
  • Parallel connections increase overall capacitance and energy storage capacity.
Depending on the need for voltage, capacitance, or specific applications, choosing the right configuration is crucial.
Equivalent Capacitance
When dealing with multiple capacitors, determining the **equivalent capacitance** simplifies calculations for energy storage, charge, and current flow. It essentially combines multiple capacitors into a single equivalent capacitor.
  • For parallel capacitors: Join up their capacitances directly since they combine their electric charge storage capacity, leading to a greater equivalent capacitance.
  • For series capacitors: Calculate the reciprocal of the sum of the reciprocals of individual capacitances, resulting in a smaller equivalent capacitance.
Equivalent capacitance lets engineers and designers tailor circuits to have specific energy, voltage, current, and space requirements. Simplified models with equivalent capacitance lead to better optimization and improved circuit performance.

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Most popular questions from this chapter

Two identical resistors are connected in parallel across a 25-V battery, which supplies them with a total power of 9.6 W. While the battery is still connected, one of the resistors is heated so that its resistance doubles. The resistance of the other resistor remains unchanged. Find (a) the initial resistance of each resistor and (b) the total power delivered to the resistors after one resistor has been heated.

A \(1.40-\Omega\) resistor is connected across a \(9.00-\mathrm{V}\) battery. The voltage between the terminals of the battery is observed to be only 8.30 \(\mathrm{V}\) . Find the internal resistance of the battery.

You have three capacitors: \(C_{1}=67 \mu \mathrm{F}, C_{2}=45 \mu \mathrm{F},\) and \(C_{3}=33 \mu \mathrm{F}\) . Determine the maximum equivalent capacitance you can obtain by connecting two of the capacitors in parallel and then connecting the parallel combination in series with the remaining capacitor.

The rear window defogger of a car consists of thirteen thin wires (resistivity \(=88.0 \times 10^{-8} \Omega \cdot \mathrm{m}\) ) embedded in the glass. The wires are connected in parallel to the \(12.0-\mathrm{V}\) battery, and each has a length of 1.30 \(\mathrm{m} .\) The defogger can melt \(2.10 \times 10^{-2} \mathrm{kg}\) of ice at \(0^{\circ} \mathrm{C}\) into water at \(0^{\circ} \mathrm{C}\) in two minutes. Assume that all the power delivered to the wires is used immediately to melt the ice. Find the cross-sectional area of each wire.

A battery has an internal resistance of 0.50\(\Omega .\) A number of identical light bulbs, each with a resistance of \(15 \Omega,\) are connected in parallel across the battery terminals. The terminal voltage of the battery is observed to be one-half the emf of the battery. How many bulbs are connected?

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