Problem 3
Question
In a certain oscillating \(L C\) circuit, the total energy is converted from electrical energy in the capacitor to magnetic energy in the inductor in \(1.50 \mu\) s. What are (a) the period of oscillation and (b) the frequency of oscillation? (c) How long after the magnetic energy is a maximum will it be a maximum again?
Step-by-Step Solution
Verified Answer
(a) 3.00 μs; (b) 333 kHz; (c) 1.50 μs
1Step 1: Understanding the Problem
An LC circuit oscillates energy between the electric field of a capacitor and the magnetic field of an inductor. We're informed that it takes 1.50 μs for energy to completely shift from the capacitor to the inductor.
2Step 2: Determine the Relationship
In a complete cycle for an LC circuit, energy shifts from the capacitor to the inductor and then back from the inductor to the capacitor. In this situation, energy shifting one full period between these states takes place over twice the time it takes for energy to shift in one direction alone.
3Step 3: Calculate the Period of Oscillation (T)
Since it takes 1.50 μs for half a cycle (energy completely moving to the inductor from the capacitor), the complete cycle or period of oscillation is twice this time. Thus, \( T = 2 \times 1.50 \times 10^{-6} \) s, which equals \( 3.00 \times 10^{-6} \) s.
4Step 4: Calculate the Frequency of Oscillation (f)
The frequency of oscillation is the reciprocal of the period. Using the formula \( f = \frac{1}{T} \), we find \( f = \frac{1}{3.00 \times 10^{-6}} \) Hz, which results in approximately \( 3.33 \times 10^{5} \) Hz.
5Step 5: Determine the Duration for Energy Re-mounting to Maximum
Since the magnetic energy becomes maximum every half period, it will be maximum again after a period of \( 1.50 \times 10^{-6} \) s, similar to the time it takes for energy to shift from the capacitor to the inductor.
Key Concepts
Period of OscillationFrequency of OscillationEnergy Transfer in LC Circuits
Period of Oscillation
The period of oscillation in an LC circuit is essentially the time it takes for the circuit to complete one full cycle of energy transfer. Energy in an LC circuit toggles between electrical energy in the capacitor and magnetic energy in the inductor. In this exercise, it's given that the energy transfers entirely to the inductor in 1.50 μs. Since this represents only half of the complete cycle, it takes twice this time for a full cycle. Thus, the period of oscillation, denoted by \( T \), is calculated as:
\[ T = 2 \times 1.50 \times 10^{-6} \text{ s} = 3.00 \times 10^{-6} \text{ s} \]
This period \( T \) represents the total time for a complete oscillation cycle, meaning the energy shifts entirely from the capacitor over to the inductor, and back again.
\[ T = 2 \times 1.50 \times 10^{-6} \text{ s} = 3.00 \times 10^{-6} \text{ s} \]
This period \( T \) represents the total time for a complete oscillation cycle, meaning the energy shifts entirely from the capacitor over to the inductor, and back again.
Frequency of Oscillation
To understand the frequency of oscillation in an LC circuit, we recognize that it indicates how many times the circuit completes its energy cycle per second. The frequency, denoted by \( f \), is the reciprocal of the period of oscillation \( T \). Knowing our period from earlier as \( 3.00 \times 10^{-6} \text{ s} \), the calculation for frequency is as follows:
\[ f = \frac{1}{T} = \frac{1}{3.00 \times 10^{-6} \text{ s}} \]
Upon computing, we get:
\[ f = \frac{1}{T} = \frac{1}{3.00 \times 10^{-6} \text{ s}} \]
Upon computing, we get:
- \( f \approx 3.33 \times 10^{5} \text{ Hz} \)
Energy Transfer in LC Circuits
In an LC circuit, energy swaps back and forth between the capacitor and the inductor. This oscillation forms the basis for how these circuits conserve energy. At maximum, the capacitor stores all the circuit's energy as electrical energy, resulting in zero current. Conversely, when this energy is fully transferred to the inductor, it exists as magnetic energy.
This exercise specifies the initial time for energy transfer as 1.50 μs for a half cycle, from the capacitor to the inductor. After another 1.50 μs (making it a full cycle of 3.00 μs), the circuit returns to its initial energy state.
This exercise specifies the initial time for energy transfer as 1.50 μs for a half cycle, from the capacitor to the inductor. After another 1.50 μs (making it a full cycle of 3.00 μs), the circuit returns to its initial energy state.
- Cycle completes when energy returns fully to the capacitor.
- Energy fluctuation follows harmonic oscillation patterns.
Other exercises in this chapter
Problem 1
An oscillating \(L C\) circuit consists of a \(75.0 \mathrm{mH}\) inductor and a \(3.60 \mu \mathrm{F}\) capacitor. If the maximum charge on the capacitor is \(
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The frequency of oscillation of a certain \(L C\) circuit is \(200 \mathrm{kHz}\). At time \(t=0,\) plate \(A\) of the capacitor has maximum positive charge. At
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What is the capacitance of an oscillating \(L C\) circuit if the maximum charge on the capacitor is \(1.60 \mu \mathrm{C}\) and the total energy is \(140 \mu \m
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In an oscillating \(L C\) circuit, \(L=1.10 \mathrm{mH}\) and \(C=4.00 \mu \mathrm{F}\). The maximum charge on the capacitor is \(3.00 \mu \mathrm{C}\). Find th
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