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An LC circuit consists of a 20.0 -mH inductor and a \(0.500-\mu \mathrm{F}\) capacitor. If the maximum instantaneous current is \(0.100 \mathrm{A},\) what is the greatest potential difference across the capacitor?

Short Answer

Expert verified
The greatest potential difference across the capacitor in the described LC circuit is 20.0 Volt.

Step by step solution

01

Calculate the maximum energy in the inductor

The maximum energy stored in the inductor can be calculated using the formula: \(U = \frac{1}{2}L I_{max}^2\), so let's substitute the given values into this formula: \(U = \frac{1}{2} \times 20 \times 10^{-3} H \times (0.1 A)^2 = 0.0001 J\)
02

Calculate the maximum voltage across the capacitor

By equating the energies in both sides, this energy is also equal to the maximum energy stored in the capacitor: \(U=\frac{1}{2}C V_{max}^2\). Solving for \(V_{max}\) we have \(V_{max}=\sqrt{\frac{2U}{C}}\). Substitute the known values into this equation to calculate \(V_{max}\): \(V_{max}=\sqrt{\frac{2 \times 0.0001 J} {0.500 \times 10^{-6} F}} = 20.0 V\)

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Key Concepts

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

Inductor Energy
Inductors, often referred to as coils or solenoids, play a crucial role in LC circuits by storing energy in the form of a magnetic field. When current flows through an inductor, it creates a magnetic field around it, and energy is stored in this field. The amount of energy (\(U\text{inductor}\)) can be calculated using the formula: \(U = \frac{1}{2}LI^2\) where \(L\) is the inductance in henrys (H) and \(I\) is the current in amperes (A).

In the context of our LC circuit example, the maximum energy stored in a 20.0 mH inductor with a maximum instantaneous current of 0.100 A is: \(U = \frac{1}{2} \times 20 \times 10^{-3} H \times (0.1 A)^2 = 0.0001 J\). This simple calculation leads us to understand how an inductor's capacity to store energy directly impacts the behavior of the entire circuit, particularly in the case of oscillations.
Capacitor Voltage
A capacitor in an LC circuit is tasked with storing energy in an electric field between its plates, with voltage being a measure of the electric potential energy per unit charge. The maximum voltage (\(V_{max}\text{capacitor}\)) can be found with the equation \(V_{max} = \sqrt{\frac{2U}{C}}\), where \(U\) represents the energy in joules (J) and \(C\) is the capacitance in farads (F).

For our example, we already determined the energy stored in the inductor, which will equal the energy in the capacitor at maximum voltage due to energy conservation. Therefore, by plugging in the inductor's energy (0.0001 J) and the capacitance of the capacitor (0.500 μF), we find a maximum voltage across the capacitor of 20.0 V. This voltage is crucial for predicting how the electric field between the plates changes over time.
Electromagnetic Oscillations
Electromagnetic oscillations in an LC circuit occur due to the continuous energy exchange between the inductor and the capacitor. The inductor releases its stored magnetic energy to the capacitor, converting it to electric potential energy. This energy is then released back to the inductor, and the cycle repeats. The circuit, absent any resistance, will oscillate indefinitely.

The frequency of these oscillations is determined by the values of the inductance and capacitance and is given by \(f = \frac{1}{2\pi\sqrt{LC}}\). The oscillations in an LC circuit are analogous to a mechanical system's mass-spring oscillations, where energy similarly transfers between kinetic and potential forms. Understanding these oscillations is crucial, as they form the foundation of many applications in communication and signal processing.
Energy Conservation in LC Circuits
In an ideal LC circuit, energy conservation is a key principle, stating that the total energy in the circuit remains constant over time, provided there is no resistance to dissipate it. This energy manifests in two forms: magnetic energy in the inductor and electric potential energy in the capacitor. The hand-off of energy between these two components is what creates the electromagnetic oscillations.

Using the law of conservation of energy, we equate the energy in the inductor with that in the capacitor (\(\frac{1}{2}L I^2 = \frac{1}{2}C V^2\)). In our previous solution, we saw how the maximum energy stored in the inductor was used to calculate the maximum voltage across the capacitor, illustrating this principle. Recognizing the way energy is conserved allows students to better predict circuit behavior and understand the underpinning physics of LC oscillations.

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Most popular questions from this chapter

This problem extends the reasoning of Section \(26.4,\) Problem \(26.37,\) Example \(30.6,\) and Section 32.3. (a) Consider a capacitor with vacuum between its large, closely spaced, oppositely charged parallel plates. Show that the force on one plate can be accounted for by thinking of the electric field between the plates as exerting a "negative pressure" equal to the energy density of the electric field. (b) Consider two infinite plane sheets carrying electric currents in opposite directions with equal linear current densities \(J_{s} .\) Calculate the force per area acting on one sheet due to the magnetic field created by the other sheet. (c) Calculate the net magnetic field between the sheets and the field outside of the volume between them. (d) Calculate the energy density in the magnetic field between the sheets. (e) Show that the force on one sheet can be accounted for by thinking of the magnetic field between the sheets as exerting a positive pressure equal to its energy density. This result for magnetic pressure applies to all current configurations, not just to sheets of current.

Calculate the inductance of an \(L C\) circuit that oscillates at \(120 \mathrm{Hz}\) when the capacitance is \(8.00 \mu \mathrm{F}.\)

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A self-induced emf in a solenoid of inductance \(L\) changes in time as \(\boldsymbol{\varepsilon}=\boldsymbol{\varepsilon}_{0} e^{-k t} .\) Find the total charge that passes through the solenoid, assuming the charge is finite.

A \(1.00-\mathrm{mH}\) inductor and a \(1.00-\mu \mathrm{F}\) capacitor are connected in series. The current in the circuit is described by \(I=20.0 t,\) where \(t\) is in seconds and \(I\) is in amperes. The capacitor initially has no charge. Determine (a) the voltage across the inductor as a function of time, (b) the voltage across the capacitor as a function of time, and (c) the time when the energy stored in the capacitor first exceeds that in the inductor.

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