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Chapter 34: Problem 31

I A charged capacitor is connected across a \(10-\mathrm{k} \Omega\) resistor and allowed to discharge. The potential difference across the capacitor drops to \(0.37\) of its original value after a time of \(7.0 \mathrm{~s}\). What is the capacitance of the capacitor?

Short Answer

Expert verified
The capacitance of the capacitor is approximately 704 碌F.

Step by step solution

01

Understanding the Problem

We need to find the capacitance of a capacitor that discharges through a resistor. The potential difference across the capacitor drops to 37% (0.37) of its initial value after 7 seconds. The resistor value is given as 10k惟. We will use the discharge formula for capacitors.
02

Using the Capacitor Discharge Formula

The voltage across a discharging capacitor can be described by the formula: \[ V(t) = V_0 e^{-\frac{t}{RC}} \]where \( V(t) \) is the voltage at time \( t \), \( V_0 \) is the initial voltage, \( R \) is the resistance, and \( C \) is the capacitance.
03

Setting Up the Equation

Substitute the given values into the discharge formula: \[ 0.37V_0 = V_0 e^{-\frac{7}{10,000 \times C}} \]Since \( V_0 \) can be canceled from both sides, we have: \[ 0.37 = e^{-\frac{7}{10,000 \times C}} \]
04

Solving for Capacitance

Take the natural logarithm of both sides to solve for \( C \):\[ \ln(0.37) = -\frac{7}{10,000 \times C} \]Calculate \( \ln(0.37) \) which is approximately \(-0.994\). Now, solve for \( C \):\[ C = \frac{7}{10,000 \times -0.994} \]
05

Calculating the Capacitance

Substitute the calculated value of \( \ln(0.37) \) and compute:\[ C \approx \frac{7}{-9940} \approx 7.04 \times 10^{-4} \mathrm{~F} \] or 704 碌F.

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Key Concepts

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

Resistor-Capacitor Circuit
A resistor-capacitor (RC) circuit is a basic electronic circuit that consists of a resistor and a capacitor connected together. In this setup, the capacitor is first charged, accumulating electrical energy. When a pathway is provided via the resistor, the capacitor begins to discharge, slowly releasing its stored energy.

In a discharging phase, the resistor controls the flow of current. The amount of resistance impacts how swiftly the capacitor can release its charge. Simply put, higher resistance results in a slower discharge process. This unique interaction between the resistor and capacitor allows RC circuits to be utilized in various applications like timing devices and filters in electronic systems.

In practical terms, the discharge in an RC circuit results in a gradual drop of the potential difference (voltage) across the capacitor, showcasing a key principle of RC circuits.
Exponential Decay
Exponential decay is a concept that describes how a quantity reduces over time at a rate proportional to its current value. In the context of RC circuits, the discharge of a capacitor follows an exponential decay pattern. As the capacitor discharges, the voltage falls rapidly at first and then more slowly over time, forming a characteristic exponential curve.

This behavior is described by the formula:
\[ V(t) = V_0 e^{-\frac{t}{RC}} \]
Here, \(V(t)\) represents the voltage at time \(t\), \(V_0\) is the initial voltage, \(R\) is the resistance, and \(C\) is the capacitance. The exponential function \(e^{-\frac{t}{RC}}\) captures the essence of how the voltage decreases in a non-linear manner.

Understanding this pattern is crucial for predicting how long a capacitor will take to discharge to a certain voltage level and is widely applied in electronics to control circuits precisely.
Time Constant in RC Circuits
The time constant in RC circuits is a key concept that defines the exponential rate of charge and discharge within the circuit. It is denoted by the Greek letter \(\tau\) (tau), and is the product of the resistance \(R\) and capacitance \(C\):

\[ \tau = R \times C \]
The time constant represents the time it takes for the voltage across a capacitor to drop to about 37% of its initial value during discharge. It is a measure of the speed of the discharge process.

In our original exercise, we observed that the voltage dropped to 0.37 of its initial value after 7 seconds, indicating that 7 seconds is equivalent to one time constant for this particular circuit configuration.

Understanding the time constant is crucial because it lets engineers and students predict the behavior of RC circuits in real-world applications, such as calculating delay times and filtering signals in electronics.

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Most popular questions from this chapter

Two neighboring coils, \(A\) and \(B\), have 300 and 600 turns, respectively. A current of \(1.5 \mathrm{~A}\) in \(A\) causes \(1.2 \times 10^{-4} \mathrm{~Wb}\) to pass through \(A\) and \(0.90 \times 10^{-4} \mathrm{~Wb}\) to pass through \(B\). Determine \((a)\) the self-inductance of \(A,(b)\) the mutual inductance of \(A\) and \(B\), and \((c)\) the average induced emf in \(B\) when the current in \(A\) is interrupted in \(0.20 \mathrm{~s}\).

A coil of resistance \(15 \Omega\) and inductance \(0.60 \mathrm{H}\) is connected to a steady 120 -V power source. At what rate will the current in the coil rise \((a)\) at the instant the coil is connected to the power source, and \((b)\) at the instant the current reaches 80 percent of its maximum value? The effective driving voltage in the circuit is the \(120 \mathrm{~V}\) power supply minus the induced back emf, \(\mathrm{L}(\Delta \mathrm{i} / \Delta t)\). This equals the p.d. in the resistance of the coil: $$ 120 \mathrm{~V}-L \frac{\Delta i}{\Delta t}=i R $$ [This same equation can be obtained by writing the loop equation for the circuit of \(\underline{\text { Fig. } 34-2(a) \text { . In doing so, remember that the }}\) inductance acts as a back emf of value \(\mathrm{L}(\Delta i / \Delta t) .]\) (a) At the first instant, \(i\) is essentially zero. Then $$ \frac{\Delta i}{\Delta t}=\frac{120 \mathrm{~V}}{L}=\frac{120 \mathrm{~V}}{0.60 \mathrm{H}}=0.20 \mathrm{~mA} / \mathrm{s} $$ (b) The current reaches a maximum value of \((120 \mathrm{~V}) / R\) when the current finally stops changing (i.e., when \(\Delta i / \Delta t=0\) ). We are interested in the case when $$ i=(0.80)\left(\frac{120 \mathrm{~V}}{R}\right) $$ Substitution of this value for \(i\) in the loop equation gives $$ 120 \mathrm{~V}-L \frac{\Delta i}{\Delta t}=(0.80)\left(\frac{120 \mathrm{~V}}{R}\right) R $$ from which $$ \frac{\Delta i}{\Delta t}=\frac{(0.20)(120 \mathrm{~V})}{L}=\frac{(0.20)(120 \mathrm{~V})}{0.60 \mathrm{H}}=40 \mathrm{~A} / \mathrm{s} $$

[II] When a long iron-core solenoid is connected across a 6-V battery, the current rises to \(0.63\) of its maximum value after a time of \(0.75 \mathrm{~s}\). The experiment is then repeated with the iron core removed. Now the time required to reach \(0.63\) of the maximum is \(0.0025 \mathrm{~s}\). Calculate ( \(a\) ) the relative permeability of the iron and \((b) L\) for the aircore solenoid if the maximum current is \(0.5 \mathrm{~A}\).

A long air-core solenoid has cross-sectional area \(A\) and \(N\) loops of wire on its length \(d\). (a) Find its self-inductance. (b) What is its inductance if the core material has a permeability of \(\mu\) ? (a) We can write $$ |\varepsilon|=N\left|\frac{\Delta \Phi_{M}}{\Delta t}\right| \text { and } \quad|\varepsilon|=L\left|\frac{\Delta i}{\Delta t}\right| $$ Equating these two expressions for \(|\varepsilon|\) yields $$ L=N\left|\frac{\Delta \Phi_{M}}{\Delta i}\right| $$ If the current changes from zero to \(I\), then the flux changes from zero to \(\Phi_{M}\). Therefore, \(\Delta i=I\) and \(\Delta \Phi_{M}=\Phi_{M}\) in this case. The selfinductance, assumed constant for all cases, is then $$ L=N \frac{\Phi_{M}}{I}=N \frac{B A}{I} $$ But for an air-core solenoid, \(B=\mu_{0} \mathrm{nI}=\mu(N / d) I\). Substitution gives (b) If the material of the core has permeability \(\mu\) instead of \(\mu_{0}\), then \(B\), and therefore \(L\), will be increased by the factor \(\mu / \mu_{0}\). In that case, \(L=\mu N^{2} A / d\). An iron-core solenoid has a much higher self-inductance than an air-core solenoid has.

How much energy is stored in a 0.500-H inductor carrying a current of \(4.80 \mathrm{~A}\) ?
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