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At \(t=0\), let \(Q=Q_{0}\), and \(I=0\) in an \(L C\) circuit. (a) At the first moment when the energy is shared equally by the inductor and the capacitor, what is the charge on the capacitor? (b) How much time has elapsed (in terms of the period \(T\) )?

Short Answer

Expert verified
(a) Charge on capacitor: \(\frac{Q_0}{\sqrt{2}}\); (b) Time elapsed: \(\frac{T}{4}\).

Step by step solution

01

Understand the Energy Distribution

In an LC circuit, the energy is given by the sum of the energies stored in the capacitor and the inductor. Initially, the energy is fully stored in the capacitor as electrostatic energy, \(\frac{1}{2}CV^2\), where \(Q=CV\) and \(Q=Q_0\). We need to find the point where the energy is equally shared between the capacitor and the inductor.
02

Write the Energy Equation

When sharing energy equally, the energy in the capacitor is \(\frac{1}{4}CV^2\) and the energy in the inductor is \(\frac{1}{4}LI^2\). Since the total energy should be \(\frac{1}{2}CV^2=\frac{1}{2}\frac{Q_0^2}{C}\), equate \(\frac{1}{4}CV^2 + \frac{1}{4}LI^2 = \frac{1}{2}\frac{Q_0^2}{C}\).
03

Find the Charge on the Capacitor

The energy on the capacitor at this point is \(\frac{1}{4}\frac{Q^2}{C}\). Setting \(\frac{1}{4}\frac{Q^2}{C} = \frac{1}{4}\frac{Q_0^2}{C}\), we can solve for \(Q\). This simplifies to \(Q = \frac{Q_0}{\sqrt{2}}\) as the charge on the capacitor when energy is equally shared.
04

Find the Period T

The natural frequency of oscillation of the system is \(\omega = \frac{1}{\sqrt{LC}}\), and the period \(T = \frac{2\pi}{\omega} = 2\pi\sqrt{LC}\). We will use this in the next step to find the time elapsed.
05

Determine Time Elapsed

Since the system oscillates with a period \(T\), the time to first equal energy sharing corresponds to a quarter of the period due to the quarter of an oscillation it takes to complete a full energy shift from being fully on the capacitor to being equally shared. Thus, the elapsed time is \(\frac{T}{4}\).

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Key Concepts

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

Energy Distribution in LC Circuits
In an LC circuit, energy alternates between the capacitor and the inductor. Initially, all the energy resides in the capacitor as electrostatic energy. This energy is given by the expression \( \frac{1}{2}CV^2 \), where \( C \) is the capacitance, and \( V \) is the voltage across the capacitor. At the outset, the charge \( Q \) satisfies \( Q = CV \) and equals \( Q_0 \).
Once the circuit begins to oscillate, the energy is transferred back and forth between kinetic energy in the inductor, given by \( \frac{1}{2}LI^2 \), and electrostatic energy in the capacitor. Finding the charge at the point of equal energy sharing involves understanding this distribution.
When energy is shared equally, each component holds half of the system's total energy. The total energy initially was entirely in the capacitor, thus \( \frac{1}{2}\frac{Q_0^2}{C} \). At the shared point, both components have \( \frac{1}{4}CV^2 \) of energy.
Charge Sharing in LC Circuits
To find the exact value of charge on the capacitor when energy is equally shared with the inductor, we solve \( \frac{1}{4}\frac{Q^2}{C} = \frac{1}{4}\frac{Q_0^2}{C} \). By simplifying this equation, we determine \( Q = \frac{Q_0}{\sqrt{2}} \).
At this moment, both the inductor and the capacitor have the same amount of energy, confirming that the system's energy is balanced. This conclusion is pivotal for understanding the movement of energy within an LC circuit. Such understanding assists in calculating transitional points of charge and energy state within oscillating systems.
Oscillation Period in LC Circuits
The oscillation period of an LC circuit is determined by its natural frequency, defined by \( \omega = \frac{1}{\sqrt{LC}} \). The period \( T \) is the time taken to complete one full oscillation, given by \( T = 2\pi\sqrt{LC} \).
When considering the time elapse for energy to become equally distributed between the inductor and capacitor, which is asked in this context, it's noted to occur at a quarter of a full cycle. Hence, the elapsed time is \( \frac{T}{4} \).
This means that from the moment energy starts in the capacitor, it takes just a quarter of the full period for the energy to transfer and balance between both elements, illustrating the cyclic nature of charge and energy in LC circuits.

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Most popular questions from this chapter

(II) At \(t = 0 ,\) let \(Q = Q _ { 0 } ,\) and \(I = 0\) in an \(L C\) circuit. (a) At the first moment when the energy is shared equally by the inductor and the capacitor, what is the charge on the capacitor? (b) How much time has elapsed (in terms of the period \(T ) ?\)

(I) For a \(120 - \mathrm { V } , 60 - \mathrm { Hz }\) voltage, a current of 70\(\mathrm { mA }\) passing through the body for 1.0 s could be lethal. What must be the impedance of the body for this to occur?

An LRC series circuit with \(R=150 \Omega, L=25 \mathrm{mH}\), and \(C=2.0 \mu \mathrm{F}\) is powered by an ac voltage source of peak voltage \(V_{0}=340 \mathrm{~V}\) and frequency \(f=660 \mathrm{~Hz}\). (a) Determine the peak current that flows in this circuit. (b) Determine the phase angle of the source voltage relative to the current. (c) Determine the peak voltage across \(R\) and its phase angle relative to the source voltage. ( \(d\) ) Determine the peak voltage across \(L\) and its phase angle relative to the source voltage. (e) Determine the peak voltage across \(C\) and its phase angle relative to the source voltage.

(a) Show that the self-inductance \(L\) of a toroid (Fig. 31\()\) of radius \(r _ { 0 }\) containing \(N\) loops each of diameter \(d\) is \(L \approx \frac { \mu _ { 0 } N ^ { 2 } d ^ { 2 } } { 8 r _ { 0 } }\) if \(r _ { 0 } \gg d .\) Assume the field is uniform inside the toroid; is this actually true? Is this result consistent with \(L\) for a solenoid? Should it be? (b) Calculate the inductance \(L\) of a large toroid if the diameter of the coils is 2.0\(\mathrm { cm }\) and the diameter of the whole ring is 66\(\mathrm { cm }\) . Assume the field inside the toroid is uniform. There are a total of 550 loops of wire.

(II) The frequency of the ac voltage source (peak voltage \(V_{0}\) ) in an \(L R C\) circuit is tuned to the circuit's resonant frequency \(f_{0}=1 /(2 \pi \sqrt{L C}) \cdot(a)\) Show that the peak voltage across the capacitor is \(\left.V_{C D}=V_{0} T_{0} / 2 \pi \tau\right),\) where \(T_{0}\left(=1 / f_{0}\right)\) is the period of the resonant frequency and \(\tau=R C\) is the time constant for charging the capacitor \(C\) through a resistor \(R .(b)\) Define \(\beta=T_{0} /(2 \pi \tau)\) so that \(V_{C 0}=\beta V_{0} .\) Then \(\beta\) is the \(^{4}\) amplification" of the source voltage across the capacitor. If a particular \(L R C\) circuit contains a \(2.0-n \mathrm{~F}\) capacitor and has a resonant frequency of \(5.0 \mathrm{kHz},\) what value of \(R\) will yield \(\beta=125 ?\)

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