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A \(500.0 \Omega\) resistor is connected in series with a capacitor. What must be the capacitance of the capacitor to produce a time constant of \(2.00 \mathrm{~s} ?\)

Short Answer

Expert verified
The capacitance is 4000 \(\mu F\).

Step by step solution

01

Understand the Time Constant Formula

The time constant \( \tau \) in an RC circuit is given by \( \tau = R \times C \), where \( R \) is the resistance and \( C \) is the capacitance. We need to use this formula to find the capacitance.
02

Identify Given Values

The resistance \( R = 500.0 \Omega \) and the time constant \( \tau = 2.00 \mathrm{~s} \) are given. We need to find the capacitance \( C \).
03

Rearrange the Formula to Solve for Capacitance

Rearrange the time constant formula to solve for capacitance: \( C = \frac{\tau}{R} \).
04

Calculate the Capacitance

Substitute the known values into the formula: \( C = \frac{2.00 \mathrm{~s}}{500.0 \Omega} = 0.00400 \mathrm{~F} \).
05

Convert Capacitance to More Practical Units

Since capacitance is often measured in microfarads (\(\mu F\)), convert the answer: \( 0.00400 \mathrm{~F} = 4000 \mu \mathrm{F} \).

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Key Concepts

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

Resistance
In electronics, resistance is a critical factor in determining how much current flows through a circuit. It is measured in ohms (Ω). Resistance opposes the flow of electric current, causing energy to be lost as heat. For an intuitive understanding, think of resistance as the narrowness of a path; a narrow path (high resistance) makes it harder for water (current) to flow through, compared to a wider path (low resistance).

When dealing with an RC circuit, the resistance element is vital because it, along with the capacitance, establishes the time constant. The formula for the time constant \( \tau = R \times C \) shows how resistance directly influences the amount of time it takes for a capacitor to charge or discharge. This means higher resistance results in a larger time constant, and conversely, lower resistance results in a smaller time constant. Understanding resistance in this context helps in grasping how modifying resistance can control timing functions in circuits.
Capacitance
Capacitance is a measure of a component's ability to store electrical energy. It is quantified in farads (F), but for most practical purposes, microfarads (μF) or picofarads (pF) are used because a single farad is quite large. The capacitor stores energy in an electric field between its plates, and its ability to do so depends on its physical characteristics and materials.

In the RC circuit, capacitance, like resistance, plays a crucial role in determining the time constant. The time constant, represented by \( \tau = R \times C \), measures how quickly a capacitor can charge up to about 63% of the total voltage applied to it. It’s important to understand that larger capacitance means a capacitor will take longer to charge and discharge, reflecting in a longer time constant. This characteristic makes capacitors very useful in applications requiring precise time delays and filtering signals.
Series Circuit
A series circuit is a simple electrical circuit where components are arranged end to end in a single path for the electric current.

In a series circuit, the same current flows through each component because there is only one path for the current to take. In this type of circuit, resistors and capacitors can have significant interactions, as seen with the RC circuit's time constant concept.

By adding components in series, you effectively increase the total resistance and influence the overall behavior of the circuit, including timing characteristics like the time constant. This is why understanding series circuits is essential for predicting how changing one component, such as the capacitance, affects the entire circuit's functionality. Learning through exercises like calculating an RC circuit's capacitance in series helps in comprehensively grasping these foundational electrical concepts.

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Most popular questions from this chapter

The resistance of the body varies from approximately \(500 \mathrm{k} \Omega\) (when it is very dry) to about \(1 \mathrm{k} \Omega\) (when ?? it is wet). The maximum safe current is about \(5.0 \mathrm{~mA}\). At \(10 \mathrm{~mA}\) or above, muscle contractions can occur that may be fatal. What is the largest potential difference that a person can safely touch if his body is wet? Is this result within the range of common household voltages?

This procedure is not recommended! You'll see why after you work the problem. You are on an aluminum ladder that is standing on the ground, trying to fix an electrical connection with a metal screwdriver that has a metal handle. Your body is wet because you are sweating from the exertion; therefore, it has a resistance of \(1.0 \mathrm{k} \Omega .\) (a) If you accidentally touch the "hot" wire connected to the \(120 \mathrm{~V}\) line, how much current will pass through your body? Is this amount enough to be dangerous? (The maximum safe current is about \(5 \mathrm{~mA} .\) ) (b) How much electric power is delivered to your body?

The battery for a certain cell phone is rated at \(3.70 \mathrm{~V}\). According to the manufacturer it can produce \(3.15 \times 10^{4} \mathrm{~J}\) of electrical energy, enough for 5.25 h of operation, before needing to be recharged. Find the average current that this cell phone draws when turned on.

Three identical resistors are connected in series. When a certain potential difference is applied across the combination, the total power dissipated is 27 W. What power would be dissipated if the three resistors were connected in parallel across the same potential difference?

A \(12.4 \mu \mathrm{F}\) capacitor is connected through a \(0.895 \mathrm{M} \Omega\) resistor to a constant potential difference of \(60.0 \mathrm{~V}\). (a) Compute the charge on the capacitor at the following times after the connections are made: \(0,5.0 \mathrm{~s}, 10.0 \mathrm{~s}, 20.0 \mathrm{~s},\) and \(100.0 \mathrm{~s} .\) (b) Compute the charging currents at the same instants. (c) Graph the results of parts (a) and (b) for \(t\) between 0 and \(20 \mathrm{~s}\)

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