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Chapter 28: Problem 8

When the capacitor voltage in an undriven \(L C\) circuit reaches zero, why don't the oscillations stop?

Short Answer

Expert verified
Oscillations in an undriven \(L C\) circuit do not stop when capacitor voltage reaches zero because the energy stored in the inductor's magnetic field continues the oscillation by recharging the capacitor with the opposite polarity. The energy keeps oscillating back and forth between the inductor and the capacitor, creating a current flow which keeps the circuit oscillating.

Step by step solution

01

Understand the LC Circuit

A basic \(L C\) circuit consists of an inductor \(L\) and a capacitor \(C\) connected in series or parallel. When the circuit is driven or undriven, it forms electrical oscillations due to the energy kept in the inductor and capacitor. The energy in the system changes back and forth between the inductor, which stores energy in a magnetic field whenever there is current flow, and the capacitor which stores energy in an electric field when there is charge across it.
02

Capacitor Voltage Shift

As soon as the oscillation commences, the current charging the capacitor decreases until it gets to zero. This occurs when the capacitor has been fully charged and cannot take on more charges. At this point, the voltage across the capacitor is maximum, and the current flow in the circuit is zero
03

Inductor Takes over

When the capacitor voltage reaches zero it does not mean the end of the oscillation. Because the circuit is oscillating, the inductor now begins to release its stored energy into the circuit, which creates a current and recharges the capacitor with the opposite polarity.
04

Continuation of Oscillation

The energy keeps moving back and forth between the inductor and the capacitor perpetuating the oscillations, even when the voltage across either the inductor or the capacitor (or both) momentarily reaches zero. Therefore, even if the capacitor voltage is zero, the energy in the circuit is in the inductor's magnetic field that continues the oscillation.

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Most popular questions from this chapter

A series \(R L C\) circuit has \(R=18 \mathrm{k} \Omega, L=20 \mathrm{mH},\) and resonates at \(4.0 \mathrm{kHz}\). (a) What's the capacitance? (b) Find the circuit's impedance at resonance and (c) at \(3.0 \mathrm{kHz}\)

For \(R L C\) circuits in which the resistance isn't too high, the \(Q\) factor may be defined as the ratio of the resonant frequency to the difference between the two frequencies where the power dissipated in the circuit is half the power dissipated at resonance. Using suitable approximations, show that this definition leads to \(Q=\omega_{0} L / R,\) with \(\omega_{0}\) the resonant frequency.

For safety, medical equipment connected to patients is often powered by an isolation transformer, whose primary is connected to 120 -V AC power and whose secondary delivers \(120-\mathrm{V}\) power. What's the turns ratio of such a transformer?

An industrial electric motor runs at \(208 \mathrm{V}\) rms and \(400 \mathrm{Hz}\). What are (a) the peak voltage and (b) the angular frequency?

A series \(R L C\) circuit with \(R=47 \Omega, L=250 \mathrm{mH},\) and \(C=\) \(4.0 \mu \mathrm{F}\) is connected across a sine-wave generator whose peak output voltage is independent of frequency. Find the frequency range over which the peak current will exceed half its value at resonance.
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