/*! 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 76 A \(10 \mathrm{~V}\) battery is ... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(10 \mathrm{~V}\) battery is connected to a series of \(n\) capacitors, each of capacitance \(2.0 \mu \mathrm{F}\). If the total stored energy is \(25 \mu \mathrm{J}\), what is \(n\) ?

Short Answer

Expert verified
The number of capacitors, \(n\), is 4.

Step by step solution

01

Understand the Formula for Energy Stored in a Capacitor

The energy stored in a capacitor is given by the formula:\[U = \frac{1}{2} C V^2\]where \(U\) is the energy, \(C\) is the capacitance, and \(V\) is the voltage. In this problem, we want to find the number of capacitors \(n\), given that each capacitor has a capacitance of \(2.0 \, \mu \mathrm{F}\) and the total energy is \(25 \, \mu \mathrm{J}\).
02

Determine the Equivalent Capacitance

Since the capacitors are in series, the equivalent capacitance \(C_{eq}\) for \(n\) capacitors, each of capacitance \(2.0 \, \mu \mathrm{F}\), is given by:\[\frac{1}{C_{eq}} = n \left( \frac{1}{2.0 \, \mu \mathrm{F}} \right)\]Therefore,\[C_{eq} = \frac{2.0 \, \mu \mathrm{F}}{n}\]
03

Substitute the Known Values into the Energy Formula

We know that the total energy \(U = 25 \, \mu \mathrm{J}\) and the voltage \(V = 10 \, \mathrm{V}\). Substitute these along with the expression for \(C_{eq}\) into the energy formula:\[25 \, \mu \mathrm{J} = \frac{1}{2} \left( \frac{2.0 \, \mu \mathrm{F}}{n} \right) (10 \, \mathrm{V})^2\]
04

Solve for the Number of Capacitors \(n\)

Plugging in the numbers:\[25 \times 10^{-6} \mathrm{J} = \frac{1}{2} \times \left( \frac{2.0 \times 10^{-6} \mathrm{F}}{n} \right) \times 100\]Simplifying gives:\[25 \times 10^{-6} = \frac{100 \times 10^{-6}}{n}\]\[25 = \frac{100}{n}\]Solving for \(n\) gives:\[n = \frac{100}{25} = 4\]
05

Confirm the Accuracy of Calculation

Let's confirm the calculation by recalculating the energy with \(n = 4\):The equivalent capacitance:\[C_{eq} = \frac{2.0 \, \mu \mathrm{F}}{4} = 0.5 \, \mu \mathrm{F} = 0.5 \times 10^{-6} \mathrm{F}\]Energy calculation:\[U = \frac{1}{2} \times 0.5 \times 10^{-6} \mathrm{F} \times (10^2) = 25 \times 10^{-6} \mathrm{J}\]The calculated energy matches the given, confirming that \(n = 4\).

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Key Concepts

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

Energy Stored in a Capacitor
When a voltage is applied across a capacitor, it stores energy in an electric field. This stored energy can be calculated using the formula:\[U = \frac{1}{2} C V^2\]where:
  • \(U\) is the energy stored (in joules),
  • \(C\) is the capacitance (in farads), and
  • \(V\) is the voltage (in volts).
Understanding this relationship is crucial. It tells us that energy stored increases with both higher capacitance and higher applied voltage. Increasing either parameter will result in more stored energy. Conversely, for a fixed energy, if the voltage is higher, the capacitance must be lower, or vice versa. This relationship is fundamental in circuits and gives insights into how capacitors behave when connected to different voltages. It's important to remember that the energy stored is proportional to the square of the voltage, meaning small increases in voltage cause larger increases in stored energy. This quadratic dependency should always be taken into account when designing circuits involving capacitors.
Equivalent Capacitance
In a series connection of capacitors, the equivalent capacitance is not simply the sum of individual capacitances. Instead, the reciprocal of the equivalent capacitance is the sum of the reciprocals of each individual capacitance. To calculate it for \(n\) capacitors in series, each with capacitance \(C\):\[\frac{1}{C_{eq}} = n \left( \frac{1}{C}\right)\]Therefore,\[C_{eq} = \frac{C}{n}\]This means that adding more capacitors in series decreases the total, or equivalent, capacitance. This is the opposite of what happens when capacitors are connected in parallel, where the capacitance increases with more capacitors. Understanding equivalent capacitance is important for predicting how the series of capacitors will behave in a circuit. When capacitors are added in series, the equivalent capacitor behaves like one with a smaller capacitance, which can affect the way energy is stored and delivered in the circuit.
Capacitor Networks
Capacitors are often used in networks, which are combinations of capacitors arranged in series, parallel, or a combination of both. Networks can be designed to achieve desired capacitance values that suit specific electrical circuit requirements. In series networks, as discussed, the total capacitance is less than any single capacitor in the combination. This is useful when you want a lower capacitance value than available capacitors offer. On the other hand, parallel networks are different:
  • The capacitance adds up, resulting in a larger total capacitance.
  • This setup is useful when you require higher capacitance.
When designing circuits, it's important to understand that capacitor networks allow flexibility. Altering the arrangement of capacitors can optimize the circuit performance for specific applications, such as in tuning radios, smoothing voltage spikes, or storing larger amounts of energy temporarily. Comprehending how to manipulate these networks enables engineers to create more efficient and versatile electronic systems.

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Most popular questions from this chapter

This story begins with Problem 60 in Chapter \(23 .\) As part of the investigation of the biscuit factory explosion, the electric potentials of the workers were measured as they emptied sacks of chocolate crumb powder into the loading bin, stirring up a cloud of the powder around themselves. Each worker had an electric potential of about \(7.0 \mathrm{kV}\) relative to the ground, which was taken as zero potential. (a) Assuming that each worker was effectively a capacitor with a typical capacitance of \(200 \mathrm{pF}\), find the energy stored in that effective capacitor. If a single spark between the worker and any conducting object connected to the ground neutralized the worker, that energy would be transferred to the spark. According to measurements, a spark that could ignite a cloud of chocolate crumb powder, and thus set off an explosion, had to have an energy of at least \(150 \mathrm{~mJ}\). (b) Could a spark from a worker have set off an explosion in the cloud of powder in the loading bin? (The story continues with Problem 60 in Chapter 26.)

A potential difference of \(300 \mathrm{~V}\) is applied to a series connection of two capacitors of capacitances \(C_{1}=2.00 \mu \mathrm{F}\) and \(C_{2}=8.00 \mu \mathrm{F}\). What are (a) charge \(q_{1}\) and (b) potential difference \(V_{1}\) on capacitor 1 and (c) \(q_{2}\) and (d) \(V_{2}\) on capacitor \(2 ?\) The charged capacitors are then disconnected from each other and from the battery. Then the capacitors are reconnected with plates of the same signs wired together (the battery is not used). What now are (e) \(q_{1}\), (f) \(V_{1},\left(\right.\) g) \(q_{2}\), and (h) \(V_{2}\) ? Suppose, instead, the capacitors charged in part (a) are reconnected with plates of opposite signs wired together. What now are (i) \(q_{1},(\mathrm{j}) V_{1},(\mathrm{k}) q_{2}\), and (1) \(V_{2}\) ?

A \(2.0 \mu\) F capacitor and a \(4.0 \mu \mathrm{F}\) capacitor are connected in parallel across a \(300 \mathrm{~V}\) potential difference. Calculate the total energy stored in the capacitors.

Figure \(25-35\) shows a variable "air gap" capacitor for manual tuning. Alternate plates are connected together; one group of plates is fixed in position, and the other group is capable of rotation. Consider a capacitor of \(n=8\) plates of alternating polarity, each plate having area \(A=1.25 \mathrm{~cm}^{2}\) and separated from adjacent plates by distance \(d=3.40 \mathrm{~mm} .\) What is the maximum capacitance of the device?

How many \(1.00 \mu \mathrm{F}\) capacitors must be connected in parallel to store a charge of \(1.00 \mathrm{C}\) with a potential of \(110 \mathrm{~V}\) across the capacitors?

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