/*! 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 62 A 5000 -pF capacitor is charged ... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 62

A 5000 -pF capacitor is charged to 100 V and then quickly connected to an \(80-\mathrm{mH}\) inductor. Determine (a) the maximum energy stored in the magnetic field of the inductor, (b) the peak value of the current, and (c) the frequency of oscillation of the circuit.

Short Answer

Expert verified
(a) The maximum energy stored in the magnetic field of inductor is \(E_L = 2.5\times10^{-5}\,\mathrm{J}\); (b) The peak value of the current is \(I_\mathrm{peak} = 7.07\times10^{-4}\,\mathrm{A}\); (c) The frequency of oscillation of the circuit is \(f = 7.96\times10^{4}\,\mathrm{Hz}\).

Step by step solution

01

Calculate the energy stored in the capacitor

Since energy stored in the capacitor (E_c) is given by the formula: \( E_c = \frac{1}{2} * C * V^2 \) Where: \(C\) is the capacitance (in Farads), and \(V\) is the voltage across the capacitor (in Volts) Given that \(C = 5000\,\mathrm{pF} = 5\times10^{-9}\,\mathrm{F}\) and \(V = 100\,\mathrm{V}\), we can now calculate the energy stored in the capacitor: \(E_c = \frac{1}{2} * (5\times10^{-9}) * (100)^2\)
02

Determine the energy stored in the magnetic field of the inductor

The energy stored in the magnetic field of the inductor (E_L) is equal to the energy stored in the capacitor, as the energy is transferred from the capacitor to the inductor: \(E_L = E_c\) Calculate \(E_L\): \(E_L = \frac{1}{2} * (5\times10^{-9}) * (100)^2\)
03

Calculate the peak current in the circuit

The peak current value (I_peak) can be calculated using the energy formula for an inductor: \( E_L = \frac{1}{2} * L * I_\mathrm{peak}^2 \) Where: \(L\) is the inductance (in Henry), and \(I_\mathrm{peak}\) is the peak current (in Amps) Given that \(L = 80\,\mathrm{mH} = 0.08\,\mathrm{H}\), we can now solve for the peak current: \( I_\mathrm{peak} = \sqrt{\frac{2 * E_L}{L}} \) \( I_\mathrm{peak} = \sqrt{\frac{2 * \frac{1}{2} * (5\times10^{-9}) * (100)^2}{0.08}} \)
04

Determine the frequency of oscillation in the circuit

The frequency of oscillation (\(f\)) in a LC circuit can be calculated using the formula: \( f = \frac{1}{2 * \pi * \sqrt{L * C}} \) Where \(L\) is the inductance (in Henry), and \(C\) is the capacitance (in Farads) Given that \(L = 0.08\,\mathrm{H}\) and \(C = 5\times10^{-9}\,\mathrm{F}\), we can now calculate the frequency of oscillation: \( f = \frac{1}{2 * \pi * \sqrt{(0.08) * (5\times10^{-9})}} \) Now that we have calculated the energy stored in the magnetic field of the inductor, the peak current, and the frequency of oscillation, we can answer the questions: (a) The maximum energy stored in the magnetic field of inductor is \(E_L = \frac{1}{2} * (5\times10^{-9}) * (100)^2\,\mathrm{J}\); (b) The peak value of the current is \(I_\mathrm{peak} = \sqrt{\frac{2 * \frac{1}{2} * (5\times10^{-9}) * (100)^2}{0.08}}\,\mathrm{A}\); (c) The frequency of oscillation of the circuit is \(f = \frac{1}{2 * \pi * \sqrt{(0.08) * (5\times10^{-9})}}\,\mathrm{Hz}\).

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Key Concepts

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

Capacitor Energy Calculation
In an LC circuit, a capacitor stores electrical energy by accumulating charge on its plates. The energy stored in a capacitor can be found using the formula:
\[ E_c = \frac{1}{2} C V^2 \]
where:
  • \(C\) is the capacitance in Farads
  • \(V\) is the voltage across the capacitor in Volts
For instance, if we have a 5000-picofarad (pF) capacitor charged to 100 Volts, first convert the capacitance to Farads (\(1\,\mathrm{pF} = 10^{-12}\,\mathrm{F}\)). Thus,\[C = 5000\,\mathrm{pF} = 5\times10^{-9}\,\mathrm{F}\].
Plugging the values into the formula, we calculate:
\[ E_c = \frac{1}{2} \times 5\times10^{-9} \times (100)^2 \]
This results in the energy stored in the capacitor. Understanding this energy calculation is crucial for comprehending how capacitors work in energy storage applications.
Inductor Energy Calculation
In an LC circuit, when the capacitor discharges its stored energy, it passes through the inductor. This energy is stored in the inductor as a magnetic field. The energy content of the inductor, \(E_L\), mirrors that of the capacitor due to conservation of energy in the oscillator system, provided no resistance or other losses are present. The use of the equation:
\[ E_L = \frac{1}{2} L I^2 \]
where:
  • \(L\) is the inductance in Henrys
  • \(I\) is the current in Amperes
helps us find the maximum magnetic energy. Here, the energy calculated previously, \(E_c\), becomes equal to \(E_L\).
For example if the inductance \(L\) is 80 milliHenrys (\(\mathrm{mH}\)), which converts to \(0.08\,\mathrm{H}\), then solve for the peak current \(I\) using:
\[ I = \sqrt{\frac{2E_L}{L}} \]
This peak current illustrates the maximum flow possible at the peak of the energy exchange from capacitor to inductor.
Oscillation Frequency
The frequency of oscillation characterizes how quickly the energy shuttles back and forth between the capacitor and the inductor. Calculating oscillation frequency in an LC circuit requires the formula:
\[ f = \frac{1}{2 \pi \sqrt{L C}} \]
where:
  • \(L\) is the inductance in Henrys
  • \(C\) is the capacitance in Farads
For our given example: \(L = 0.08\,\mathrm{H}\) and \(C = 5\times10^{-9}\,\mathrm{F}\).
Plug these values into the equation to find the frequency:
\[ f = \frac{1}{2 \pi \sqrt{(0.08) (5\times10^{-9})}} \]
This frequency tells us how fast the LC circuit naturally oscillates, playing a fundamental role in tuning circuits and signal processing applications, where certain frequencies need to be selected or filtered.

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Most popular questions from this chapter

Two long, parallel wires carry equal currents in opposite directions. The radius of each wire is \(a,\) and the distance between the centers of the wires is \(d .\) Show that if the magnetic flux within the wires themselves can be ignored, the self-inductance of a length \(l\) of such a pair of wires is $$L=\frac{\mu_{0} l}{\pi} \ln \frac{d-a}{a}$$. (Hint: Calculate the magnetic flux through a rectangle of length \(l\) between the wires and then use \(L=N \Phi / I\).)

Show that \(N \Phi_{\mathrm{m}} / I\) and \(\varepsilon /(d I / d t), \quad\) which are both expressions for self-inductance, have the same units.

In a damped oscillating circuit the energy is dissipated in the resistor. The \(Q\) -factor is a measure of the persistence of the oscillator against the dissipative loss. (a) Prove that for a lightly damped circuit the energy, \(U,\) in the circuit decreases according to the following equation. $$\frac{d U}{d t}=-2 \beta U, \quad \text { where } \beta=\frac{R}{2 L}$$. (b) Using the definition of the \(Q\) -factor as energy divided by the loss over the next cycle, prove that \(Q\) -factor of a lightly damped oscillator as defined in this problem is $$Q \equiv \frac{U_{\text {begin }}}{\Delta U_{\text {one cycle }}}=\frac{1}{R} \sqrt{\frac{L}{C}}$$.

A solenoid with \(4 \times 10^{7}\) turns/m has an iron core placed in it whose magnetic susceptibility is \(4.0 \times 10^{3}\). (a) If a current of \(2.0 \mathrm{A}\) flows through the solenoid, what is the magnetic field in the iron core? (b) What is the effective surface current formed by the aligned atomic current loops in the iron core? (c) What is the self-inductance of the filled solenoid?

An emf of \(9.7 \times 10^{-3} \mathrm{V}\) is induced in a coil while the current in a nearby coil is decreasing at a rate of \(2.7 \mathrm{A} /\) s. What is the mutual inductance of the two coils?
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