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In an oscillating \(L C\) circuit, \(L=3.00 \mathrm{mH}\) and \(C=2.70 \mu \mathrm{F}\). At \(t=0\) the charge on the capacitor is zero and the current is \(2.00 \mathrm{~A}\). (a) What is the maximum charge that will appear on the capacitor? (b) At what earliest time \(t>0\) is the rate at which energy is stored in the capacitor greatest, and (c) what is that greatest rate?

Short Answer

Expert verified
(a) 0.18 mC, (b) 70.7 μs, (c) 20.8 W.

Step by step solution

01

Use Oscillation Formula

For an LC circuit in oscillation, the charge on the capacitor as a function of time can be expressed by:\[ q(t) = Q \sin(\omega t + \phi) \]Where:- \( Q \) is the maximum charge,- \( \omega \) is the angular frequency,- \( \phi \) is the phase angle.Since at \( t=0 \) the charge is zero and the current is maximum, \( \phi = -\frac{\pi}{2} \).
02

Find the Angular Frequency

The angular frequency \( \omega \) is given by:\[ \omega = \frac{1}{\sqrt{LC}} \]Plugging in the given values, \( L = 3.00 \times 10^{-3} \text{ H} \) and \( C = 2.70 \times 10^{-6} \text{ F} \), we find:\[ \omega = \frac{1}{\sqrt{3.00 \times 10^{-3} \times 2.70 \times 10^{-6}}} \approx 1.11 \times 10^4 \text{ rad/s} \]
03

Relate the Current to Maximum Charge

At \( t=0 \), the initial current \( I_0 = 2.00 \text{ A} \) can be related to the maximum charge by:\[ I_0 = -Q\omega \cos(\phi) \]Since \( \phi = -\frac{\pi}{2} \), \( \cos(\phi) = 0 \), simplifying the relation as the current at maximum:\[ I(t) = Q\omega \cos(\omega t) \]Using \( I_0 = Q\omega \), \( Q = \frac{I_0}{\omega} \). Solving,\[ Q = \frac{2.00}{1.11 \times 10^4} \approx 1.80 \times 10^{-4} \text{ C} \]
04

Determine Time of Maximum Energy Rate

The energy stored in the capacitor is given by \( U = \frac{1}{2} C V^2 \), and the rate of energy storage \( \frac{dU}{dt} \) peaks when voltage change is fastest, which occurs at maximum current change.The current relates to voltage change as:\[ V = L \frac{di}{dt} \]The rate of change of the stored energy peaks when \( \sin(2\omega t) \) is maximum, or \( 2\omega t = \frac{\pi}{2} \). Solving for \( t \):\[ t = \frac{\pi}{4\omega} \approx 7.07 \times 10^{-5} \text{ s} \]
05

Calculate Greatest Rate of Energy Storage

The greatest rate of energy storage at this time utilizing \( \frac{dU}{dt} = \, CV \cdot \frac{di}{dt} \),the peak value is:\[ \left(\frac{dU}{dt}\right)_{max} = I_0 \cdot V = Q\omega \cdot L \omega Q \approx ( 2.00)(1.80 \times 10^{-4})\cdot (3.00 \times 10^{-3}) \cdot (1.11 \times 10^4)^2 \approx 20.8 \text{ Watts} \]

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Key Concepts

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

Oscillation Formula
In an LC circuit, the oscillation formula is crucial to understand how charge varies with time. This formula is given by:
\[ q(t) = Q \sin(\omega t + \phi) \]This equation describes the charge \(q(t)\) on a capacitor at a time \(t\). It's like a snapshot record of the capacitor's charge varying back and forth, much like a swinging pendulum.
  • \(Q\) is the maximum charge the capacitor holds.
  • \(\omega\) is the angular frequency, dictating how fast the oscillation occurs.
  • \(\phi\) is the phase angle, offering a time shift to align with physical circumstances.
In our problem, initially, the charge was zero, and the current was at maximum. Thus, \(\phi = -\frac{\pi}{2}\). This condition arises because when the current is at its peak, the charge shifts from zero to maximum.
Maximum Charge
In an LC circuit, the maximum charge \(Q\) represents how much charge the capacitor can fully hold. It's a key parameter that helps predict how energy cycles within the circuit.
To find the maximum charge, we relate it to the current using the equation:
\[ I(t) = Q\omega \cos(\omega t) \]At \( t=0 \), this simplifies to an equation in terms of initial current \( I_0 \):
\[ Q = \frac{I_0}{\omega} \]Plugging in the given values (from our Original Exercise), we use the calculated \(\omega\) to find:
\[ Q = \frac{2.00}{1.11 \times 10^4} \approx 1.80 \times 10^{-4} \text{ C} \]This means, when fully charged, the capacitor holds approximately 0.00018 C of charge. Understanding \(Q\) gives insights into energy distribution within the circuit.
Angular Frequency
Angular frequency \(\omega\) is an integral feature to calculate in LC circuits. Essentially, it indicates how frequently the oscillations occur within a given time frame.
Derived by the formula:
\[ \omega = \frac{1}{\sqrt{LC}} \]where:
  • \(L\) is the inductance in henrys (H).
  • \(C\) is the capacitance in farads (F).
In our specific example, substituting in \(L = 3.00 \times 10^{-3} \text{ H}\) and \(C = 2.70 \times 10^{-6} \text{ F}\), the calculation yields:
\[ \omega \approx 1.11 \times 10^4 \text{ rad/s} \]Thus, our circuit oscillates very rapidly, completing one full cycle in very tiny fractions of a second, reflecting high agility in charge transfer between capacitor and inductor.
Energy Storage Rate
The energy storage rate in an LC circuit is a fascinating component that reflects how fast the circuit can store energy in the capacitor.
It's especially high when the current changes most sharply, corresponding to the peak in stored energy.
Expressed by:
\[ \frac{dU}{dt} \] To maximize this rate, we seek the point where the derivative
\( \sin(2\omega t) \)
reaches its peak value. Solving for \(t\) when
\(2\omega t = \frac{\pi}{2}\), results in:
\[ t = \frac{\pi}{4\omega} \approx 7.07 \times 10^{-5} \text{ s} \]At this precise moment, we find the maximum energy storage rate is around 20.8 Watts, showcasing the energetic efficiency of the circuit's rapid charge and discharge cycles. Understanding this concept helps in optimizing circuit performance for specific applications.

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Most popular questions from this chapter

An ac generator with emf amplitude \(\mathscr{E}_{m}=220 \mathrm{~V}\) and operating at frequency \(400 \mathrm{~Hz}\) causes oscillations in a series \(R L C\) circuit having \(R=220 \Omega, L=150 \mathrm{mH},\) and \(C=24.0 \mu \mathrm{F}\). Find (a) the capacitive reactance \(X_{C},\) (b) the impedance \(Z,\) and \((\mathrm{c})\) the current amplitude \(I\). A second capacitor of the same capacitance is then connected in series with the other components. Determine whether the values of (d) \(X_{C},\) (e) \(Z\), and (f) \(I\) increase, decrease, or remain the same.

(a) In an oscillating \(L C\) circuit, in terms of the maximum charge \(Q\) on the capacitor, what is the charge there when the energy in the electric field is \(50.0 \%\) of that in the magnetic field? (b) What fraction of a period must elapse following the time the capacitor is fully charged for this condition to occur?

Go An ac generator has emf \(\mathscr{E}=\mathscr{E}_{m} \sin \omega_{d} t,\) with \(\mathscr{E}_{m}=25.0 \mathrm{~V}\) and \(\omega_{d}=377 \mathrm{rad} / \mathrm{s} .\) It is connected to a \(12.7 \mathrm{H}\) inductor. (a) What is the maximum value of the current? (b) When the current is a maximum, what is the emf of the generator? (c) When the emf of the generator is \(-12.5 \mathrm{~V}\) and increasing in magnitude, what is the current?

To construct an oscillating \(L C\) system, you can choose from a \(10 \mathrm{mH}\) inductor, a \(5.0 \mu \mathrm{F}\) capacitor, and a \(2.0 \mu \mathrm{F}\) capacitor. What are the (a) smallest, (b) second smallest, (c) second largest, and (d) largest oscillation frequency that can be set up by these elements in various combinations?

An ac voltmeter with large impedance is connected in turn across the inductor, the capacitor, and the resistor in a series circuit having an alternating emf of \(100 \mathrm{~V}\) (rms); the meter gives the same reading in volts in each case. What is this reading?

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