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A capacitor with initial charge \(q_{0}\) is discharged through a resistor. What multiple of the time constant \(\tau\) gives the time the capacitor takes to lose (a) the first one-third of its charge and (b) two-thirds of its charge?

Short Answer

Expert verified
(a) 0.405\(\tau\); (b) 1.099\(\tau\).

Step by step solution

01

Understanding the Problem

The problem involves a capacitor discharging through a resistor. We need to find the multiple of the time constant \( \tau \) when the capacitor has lost a specific amount of its initial charge, \( q_0 \).
02

Using the Discharge Equation

The charge \( q(t) \) on a capacitor at time \( t \) during discharge can be described by \( q(t) = q_0 e^{-t/\tau} \). For this problem, we need to rearrange this equation to find \( t \) in terms of \( \tau \).
03

Finding Time for Losing One-third Charge

To find the time when the capacitor has lost one-third of its charge, set \( q(t) = \frac{2}{3}q_0 \). Then, \( \frac{2}{3}q_0 = q_0 e^{-t/\tau} \). Simplify to find \( e^{-t/\tau} = \frac{2}{3} \). Taking the natural logarithm gives: \( -\frac{t}{\tau} = \ln\left(\frac{2}{3}\right) \) thus, \( t = -\tau \ln\left(\frac{2}{3}\right) \).
04

Calculating the Multiple of Time Constant for One-third Charge

From the equation \( t = -\tau \ln\left(\frac{2}{3}\right) \), compute the multiple \( x \) of the time constant \( t = x\tau \). Thus, \( x = - \ln\left(\frac{2}{3}\right) \). Calculate \( x \approx 0.405 \).
05

Finding Time for Losing Two-thirds Charge

To find time when the capacitor has lost two-thirds of its charge, set \( q(t) = \frac{1}{3}q_0 \). Consequently, \( \frac{1}{3}q_0 = q_0 e^{-t/\tau} \). Simplify to \( e^{-t/\tau} = \frac{1}{3} \). Taking the natural logarithm gives: \( -\frac{t}{\tau} = \ln\left(\frac{1}{3}\right) \), resulting in \( t = -\tau \ln\left(\frac{1}{3}\right) \).
06

Calculating the Multiple of Time Constant for Two-thirds Charge

Now calculate the multiple \( x \) of the time constant \( t = x\tau \) from \( t = -\tau \ln\left(\frac{1}{3}\right) \). Therefore, \( x = -\ln\left(\frac{1}{3}\right) \). Calculate \( x \approx 1.099 \).

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Key Concepts

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

Time Constant
The time constant, often represented by the Greek letter \( \tau \), is a crucial parameter in the study of capacitors. It gives us an indication of how fast a capacitor can discharge or charge up to a significant level. The time constant is defined mathematically as \( \tau = RC \), where \( R \) is the resistance Through which the capacitor discharges, and \( C \) is the capacitance of the capacitor. In the case of discharging, the time constant represents the time it takes for a capacitor to discharge to about 37% of its initial charge. Conversely, it's also the time needed to charge a capacitor to about 63% of its total capacity.Understanding the time constant helps us predict charging and discharging trends in circuits. It gives practical insights into how quickly a system will respond to changes. This, in turn, helps engineers design circuits that better meet their needs.
Exponential Decay
Exponential decay describes how quantities decrease over time at a rate proportional to their current value. In the context of a discharging capacitor, it means that the rate at which the charge decreases is not constant but instead decreases as time passes. The formula \( q(t) = q_0 e^{-t/\tau} \) illustrates this concept in capacitor discharge. Here, \( q(t) \) is the charge at time \( t \), \( q_0 \) is the initial charge, and \( \tau \) is the time constant. The negative exponent, \( -t/\tau \), signifies decay. This is why the curve of charge against time is steep initially and flattens out as the charge approaches zero. This is why it is called exponential decay; because the decrease slows down as time goes on.Exponential decay is important in many fields, not just electronics. You’ll see it in populations, radioactive decay, and even cooling processes. Recognizing patterns of exponential decay can help comprehend how systems evolve over time.
Natural Logarithm
Natural logarithms are a mathematical function that plays a vital role in solving problems involving exponential growth or decay. When dealing with equations involving exponential decay, such as \( q(t) = q_0 e^{-t/\tau} \), natural logarithms provide a pathway to solving for unknowns like time.The function \( \ln(x) \) is the natural logarithm of \( x \). It's the inverse operation of taking the exponential function, \( e^x \). In the context of capacitor discharge, if you know \( q(t) = A \times q_0 \), you can take the natural logarithm of both sides to solve for time, \( t \), using the formulae derived from the exponential decay expression.The natural logarithm is critical when you need to "undo" an exponential function. It allows you to isolate variables in the exponent, crucial for finding exact values of \( t \) in discharge problems. This isn't just a trick of mathematics—it's a powerful tool that lets you translate real-world decay patterns into manageable calculations.

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Most popular questions from this chapter

Two identical batteries of emf \(\mathscr{E}=12.0 \mathrm{~V}\) and internal resistance \(r=0.200 \Omega\) are to be connected to an external resistance \(R\), either in parallel (Fig. \(27-50\) ) or in series (Fig. \(27-51\) ). If \(R=2.00 r,\) what is the current \(i\) in the external resistance in the (a) parallel and (b) series arrangements? (c) For which arrangement is \(i\) greater? If \(R=r / 2,00\), what is \(i\) in the external resistance in the (d) parallel arrangement and (e) series arrangement? (f) For which arrangement is \(i\) greater now?

A car battery with a \(12 \mathrm{~V} \mathrm{emf}\) and an internal resistance of \(0.040 \Omega\) is being charged with a current of \(50 \mathrm{~A}\). What are (a) the potential difference \(V\) across the terminals, (b) the rate \(P_{\underline{r}}\) of energy dissipation inside the battery, and (c) the rate \(P_{\mathrm{emf}}\) of energy conversion to chemical form? When the battery is used to supply 50 A to the starter motor, what are (d) \(V\) and (e) \(P_{r} ?\)

A wire of resistance \(5.0 \Omega\) is connected to a battery whose emf 8 is \(2.0 \mathrm{~V}\) and whose internal resistance is \(1.0 \Omega .\) In \(2.0 \mathrm{~min},\) how much energy is (a) transferred from chemical form in the battery, (b) dissipated as thermal energy in the wire, and (c) dissipated as thermal energy in the battery?

A \(120 \mathrm{~V}\) power line is protected by a 15 A fuse. What is the maximum number of 500 W lamps that can be simultaneously operated in parallel on this line without "blowing" the fuse because of an excess of current?

A simple ohmmeter is made by connecting a \(1.50 \mathrm{~V}\) flashlight battery in series with a resistance \(R\) and an ammeter that reads from 0 to \(1.00 \mathrm{~mA},\) as shown in Fig. \(27-59 .\) Resistance \(R\) is adjusted so that when the clip leads are shorted together, the meter deflects to its full- scale value of \(1.00 \mathrm{~mA}\). What external resistance across the leads results in a deflection of (a) \(10.0 \%,(\mathrm{~b}) 50.0 \%\) and (c) \(90.0 \%\) of full scale? (d) If the ammeter has a resistance of \(20.0 \Omega\) and the internal resistance of the battery is negligible, what is the value of \(R ?\)

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