/*! 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 67 A capacitor is charged to a pote... [FREE SOLUTION] | 91Ó°ÊÓ

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A capacitor is charged to a potential of 12.0 \(\mathrm{V}\) and is then connected to a voltmeter having an internal resistance of 3.40 \(\mathrm{M\Omega} .\) After a time of 4.00 \(\mathrm{s}\) the voltmeter reads 3.0 \(\mathrm{V}\) What are (a) the capacitance and (b) the time constant of the circuit?

Short Answer

Expert verified
Capacitance is approximately 847 nF, and time constant is 2.88 s.

Step by step solution

01

Understand the Problem

The problem involves a charged capacitor connected to a voltmeter with a known resistance. We need to find the capacitance and the time constant of the circuit as the capacitor discharges.
02

Use the Formula for Voltage During Discharge

During capacitor discharge, the voltage across the capacitor is given by the formula: \[ V = V_0 \cdot e^{-\frac{t}{RC}} \]where \(V\) is the voltage at time \(t\), \(V_0\) is the initial voltage, \(R\) is the resistance, and \(C\) is the capacitance.
03

Plug Values into the Formula

Substitute the known values: \( V = 3.0 \, V, \; V_0 = 12.0 \, V, \; R = 3.40 \, M\Omega = 3.40 \times 10^6 \, \Omega, \; \text{and} \; t = 4.00 \, s \) into the formula:\[ 3.0 = 12.0 \cdot e^{-\frac{4.00}{3.40 \times 10^6 \cdot C}} \]
04

Solve for the Exponential Term

To isolate the exponential term, divide both sides by 12.0:\[ \frac{3.0}{12.0} = e^{-\frac{4.00}{3.40 \times 10^6 \cdot C}} \]This simplifies to:\[ \frac{1}{4} = e^{-\frac{4.00}{3.40 \times 10^6 \cdot C}} \]
05

Apply Natural Logarithm

To solve for the exponent, take the natural logarithm of both sides:\[ \ln\left(\frac{1}{4}\right) = -\frac{4.00}{3.40 \times 10^6 \cdot C} \]Then, calculate \( \ln\left(\frac{1}{4}\right) \approx -1.38629 \).
06

Solve for Capacitance

Rearrange the equation to solve for \( C \):\[ C = \frac{-4.00}{3.40 \times 10^6 \cdot (-1.38629)} \]Calculate \( C \) to find the capacitance.
07

Calculate the Value of Capacitance

Compute \( C \):\[ C \approx \frac{4.00}{3.40 \times 10^6 \cdot 1.38629} \]Upon calculation, \( C \approx 8.47 \times 10^{-7} \text{ F} \) or \( 847 \text{ nF} \).
08

Use the Definition of the Time Constant

The time constant \( \tau \) is defined as:\[ \tau = R \cdot C \]With \( R = 3.40 \times 10^6 \) and \( C = 8.47 \times 10^{-7} \) F.
09

Calculate the Time Constant

Compute \( \tau \):\[ \tau = 3.40 \times 10^6 \times 8.47 \times 10^{-7} \]This results in \( \tau \approx 2.88 \, s \).

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Key Concepts

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

Capacitance Calculation
Capacitance is a measure of a capacitor's ability to store charge. It is calculated by analyzing the rate at which a capacitor discharges its stored energy over time. In the original problem, we used the formula for voltage during the discharge of a capacitor:- This formula is: \[ V = V_0 \cdot e^{-\frac{t}{RC}} \] Here, \( V \) is the voltage across the capacitor at time \( t \), \( V_0 \) the initial voltage, \( R \) the resistance, and \( C \) the capacitance.
When we want to calculate the capacitance, we substitute the known values of voltage, time, and resistance, then solve for \( C \). It is important to take the natural logarithm to help isolate the exponent and make the computation manageable.
In our case, we found \( C = 8.47 \times 10^{-7} \text{ F} \), which indicates a fairly small capacitor, as capacitance is often measured in microfarads (\(\mu\text{F}\)) or nanofarads (\(\text{nF}\)). Understanding capacitance is crucial as it determines how much charge can be stored under a given voltage.
Time Constant
The time constant \( \tau \) of an RC circuit gives insight into how quickly the capacitor discharges. It is defined simply as \( \tau = R \cdot C \).- The time constant is a pivotal concept because it determines the rate of exponential decay, where approximately 63.2% of the total charge is discharged in one time constant.
For our example, we calculated the time constant as \( \tau = 3.40 \times 10^6 \times 8.47 \times 10^{-7} \approx 2.88 \, \text{s} \). This indicates that nearly two-thirds of the charge is lost after around 2.88 seconds.
Understanding the time constant helps in predicting how an actual circuit will behave over time, influencing designs of timers and various signal processing applications.
RC Circuit
RC circuits integrate resistors (\(R\)) and capacitors (\(C\)) to manage electrical signals. They are fundamental components in electronics used for filtering, timing, and stabilization applications.- In an RC circuit, the resistor controls the rate of charge and discharge of the capacitor. When a capacitor is charged, the current gradually stops until the capacitor voltage equals the source voltage, determined by \( V = Q/C \), where \( Q \) is the charge.
The formula for voltage during discharge, \( V = V_0 \cdot e^{-\frac{t}{RC}} \), captures this dynamic behavior, demonstrating how the capacitor's voltage decreases exponentially.
Recognizing the behavior of RC circuits is crucial for designing circuits that need specific timing and stability, such as in integrators or differentiators used for processing signals.
Voltmeter Resistance
A voltmeter measures the voltage across two points and has its own internal resistance. Often, high internal resistance is preferred to minimize the impact on the circuit.- In our exercise, the voltmeter's resistance is given as 3.40 \( \text{M}\Omega \). When connecting a voltmeter to a circuit, this resistance becomes part of the total circuit resistance, influencing the discharge rate of a capacitor.
It is crucial to consider voltmeter resistance in calculations to ensure accurate readings and analyses of circuit behaviors, especially in delicate measurements where precision matters.
High resistance in a voltmeter ensures minimal current flow through it, reducing the loading effect, which could otherwise skew the voltage readings.

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Most popular questions from this chapter

Treatment of heart failure. A heart defibrillator is used to enable the heart to start beating if it has stopped. This is done by passing a large current of 12 A through the body at 25 \(\mathrm{V}\) for a very short time, usually about 3.0 \(\mathrm{ms}\) (a) What power does the defibrillator deliver to the body, and (b) how much energy is transferred?

A 6.00 \(\mu \mathrm{F}\) capacitor that is initially uncharged is connected in series with a 4500\(\Omega\) resistor and a 500 \(\mathrm{V}\) emf source with negligible internal resistance. Just after the circuit is completed, what are (a) the voltage drop across the capacitor, (b) the voltage drop across the resistor, (c) the charge on the capacitor, and (d) the current through the resistor? (e) A long time after the circuit is completed (after many time constants), what are the values of the preceding four quantities?

A 4600\(\Omega\) resistor is connected across a charged 0.800 \(\mathrm{nF}\) capacitor. The initial current through the resistor, just after the connection is made, is measured to be 0.250 A. (a) What magnitude of charge was initially on each plate of this capacitor? (b) How long after the connection is made will it take before the charge is reduced to 1\(/ e\) of its maximum value?

With a 1500 \(\mathrm{M\Omega}\) resistor across its terminals, the terminal voltage of a certain battery is 2.50 \(\mathrm{V}\) . With only a 5.00\(\Omega\) ? resistor across its terminals, the terminal voltage is 1.75 \(\mathrm{V}\) . (a) Find the internal emf and the internal resistance of this battery. (b) What would be the terminal voltage if the 5.00\(\Omega\) resistor were replaced by a 7.00 \(\Omega\) resistor?

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