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\(A 15.0 \mu F\) capacitor is charged by a 150.0 V power supply, then disconnected from the power and connected in scries with a \(0.280 \mathrm{mH}\) inductor. Calculate: (a) the oscillation frequency of the circuit: (b) the energy stored in the capacitor at time \(t=0 \mathrm{~ms}\) (the moment of connection with the inductor); (c) the energy stored in the inductor at \(t=1.30 \mathrm{~ms}\)

Short Answer

Expert verified
The oscillation frequency is calculated from the value of inductance and capacitance. Initially, all the energy is stored in the capacitor, and it can be calculated using the capacitance and the initial voltage. Over time, the energy oscillates between the capacitor and inductor. To find the energy in the inductor at a given time, find how many complete half cycles have occurred by that time, and the energy in the inductor will either be the same as the initial energy in the capacitor or zero, depending on whether the number of half cycles is odd or even respectively.

Step by step solution

01

Calculate the oscillation frequency

The angular frequency (ω) for a LC circuit can be given by \(ω = 1/\sqrt{LC}\). Here, \(L = 0.280 \, mH = 0.280 \times 10^{-3} H\) and \(C = 15.0 \, μF = 15.0 \times 10^{-6} F\). Plugging these values in the formula, we get: \(ω = 1/\sqrt{(0.280 \times 10^{-3}) (15.0 \times 10^{-6})}\)
02

Calculate the energy stored in the capacitor at t = 0

When the capacitor is fully charged and connected to the inductor, all the energy in the circuit is stored in the capacitor. This can be given by \(E_C = 0.5CV^2\), where \(V\) is the voltage of the power supply. Here, \(V = 150.0 V\), and using the value of \(C\) from above we get: \(E_C = 0.5 (15.0 \times 10^{-6}) (150.0)^2\)
03

Calculate the energy stored in the inductor at t = 1.30 ms

The energy in the circuit oscillates between the inductor and the capacitor. After half a period (i.e., \(t = T/2\)), all the energy is stored in the inductor. So we first calculate the period of oscillation (\(T = 2π/ω\)), then check how many complete half cycles have taken place by time \(t = 1.30 ms\). The energy in the inductor at time \(t\) will be the same as the original energy in the capacitor if \(t\) is an odd multiple of \(T/2\), and zero otherwise.

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Key Concepts

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

Capacitor Energy
The energy stored in a capacitor is a critical concept in understanding the dynamics of an oscillating LC circuit. A capacitor, a component that holds electric charge, stores energy in the form of an electric field. We measure this energy using the formula \(E_C = \frac{1}{2}CV^2\) where \(E_C\) is the energy in joules, \(C\) is the capacitance in farads, and \(V\) is the voltage across the capacitor in volts.

At time \(t=0\) ms in our circuit, when the capacitor is disconnected from the power supply and connected to the inductor, all of the electrical energy supplied by the power supply is stored within the capacitor. This initial energy plays a key role in the subsequent oscillation of energy between the capacitor and the inductor once the circuit begins to operate.
Inductor Energy
An inductor, on the other hand, stores energy in a magnetic field when current flows through it. The expression to calculate the energy stored in an inductor is \(E_L = \frac{1}{2}LI^2\), where \(E_L\) is the energy in joules, \(L\) is the inductance in henrys, and \(I\) is the current in amperes.

Since the oscillating LC circuit periodically transfers energy between the capacitor and the inductor, at certain pivotal moments, all of the circuit's energy is housed within the inductor's magnetic field. Knowing when exactly these moments occur functionally depends on the period of the oscillation and can be calculated using the oscillation cycle of the LC circuit.
Oscillating LC Circuit
In an oscillating LC circuit, energy is continuously exchanged between the capacitor and the inductor. In the textbook exercise, the initial energy is wholly stored in the capacitor when connected to the inductor. As the circuit starts oscillating, this energy shuttles back and forth between the electric field of the capacitor and the magnetic field of the inductor.

At time \(t=1.30\) ms, to find the energy in the inductor, we must determine the point in the oscillation cycle we are observing. Ideally, at an exact half-oscillation period, all the energy would be found in the inductor. However, since time does not correspond exactly to a half-period, there could be partial energy transfer between the inductor and capacitor, which must be calculated accordingly.
Angular Frequency Calculation
Angular frequency, denoted by \(\omega\), is fundamental in describing the oscillation of the LC circuit. It is calculated using the formula \(\omega = \frac{1}{\sqrt{LC}}\), where \(L\) is inductance and \(C\) is capacitance. This value is crucial as it determines how fast the energy oscillates between the capacitor and inductor.

Once we have \(\omega\), finding the natural frequency of oscillation \(f\) is straightforward as \(f = \frac{\omega}{2\pi}\). The angular frequency helps us understand the timing of energy shifts within the circuit and allows us to determine the circuit's resonant frequency, where it will naturally oscillate at with no external interference.

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Most popular questions from this chapter

CALC A coil has 400 turns and self-inductance \(7.50 \mathrm{mH}\) The current in the coil varies with time according to \(i=\) \((680 \mathrm{~mA}) \cos (\pi t / 0.0250 \mathrm{~s}) .\) (a) What is the maximum emf induced in the coil? (b) What is the maximum average flux through each turn of the coil? (c) At \(t=0.0180 \mathrm{~s}\), what is the magnitude of the induced emf?

DATA To investigate the properties of a large industrial solenoid, you connect the solenoid and a resistor in series with a battery. Switches allow the battery to be replaced by a short circuit across the solenoid and resistor. Therefore Fig. 30.11 applies, with \(R=R_{\text {ext }}+R_{L}\). where \(R_{L}\) is the resistance of the solenoid and \(R_{\text {ext }}\) is the resistance of the series resistor. With switch \(S_{2}\) opcn, you close switch \(S_{1}\) and kecp it closed until the current \(i\) in the solenoid is constant (Fig. 30.11 ). Then you close \(S_{2}\) and open \(S_{1}\) simultaneously, using a rapid- response switching mechanism. With high-speed clectronics you measure the time \(t_{\text {balf }}\) that it takes for the current to decrease to half of its initial value. You repeat this measurement for several values of \(R_{\text {ext }}\) and obtain these results: \begin{tabular}{l|lllllll} \(R_{\text {ext }}(\Omega)\) & 3.0 & 4.0 & 5.0 & 6.0 & 7.0 & 8.0 & 10.0 & 12.0 \\\ \hline\(t_{\text {half }}(\mathrm{s})\) & 0.735 & 0.654 & 0.589 & 0.536 & 0.491 & 0.453 & 0.393 & 0.347 \end{tabular} (a) Graph your data in the form of \(1 / t_{\text {halt }}\) versus \(R_{e x l}\). Explain why the data points plotted this way fall close to a straight line. (b) Use your graph from part (a) to calculate the resistance \(R_{L}\) and inductance \(L\) of the solenoid. (c) If the current in the solenoid is \(20.0 \mathrm{~A}\), how much energy is stored there? At what rate is electrical energy being dissipated in the resistance of the solenoid?

\(\operatorname{An} L-C\) circuit containing an \(80.0 \mathrm{ml}\) inductor and a \(1.25 \mathrm{nF}\) capacitor oscillates with a maximum current of 0.750 A. Calculate: (a) the maximum charge on the capacitor and (b) the oscillation frequency of the circuit. (c) Assuming the capacitor had its maximum charge at time \(t=0,\) calculate the energy stored in the inductor after \(2.50 \mathrm{~ms}\) of oscillation.

I-C Oscillations. A capacitor with capacitance \(6.00 \times 10^{-5} \mathrm{~F}\) is charged by connecting it to a \(12.0 \mathrm{~V}\) battery. The capacitor is disconnected from the battery and connected across an inductor with \(L=1.50 \mathrm{H}\). (a) What are the angular frequency \(\omega\) of the electrical os cillations and the period of these ascillations (the time for one oscillation)? (b) What is the initial charge on the capacitor? (c) Ilow much energy is initially stored in the capacitor? (d) What is the charge on the capacitor \(0.0230 \mathrm{~s}\) after the connection to the inductor is made? Interpret the sign of your answer. (c) At the time given in part (d), what is the current in the inductor? Interpret the sign of your answer. (f) At the time given in part (d), how much electrical energy is stored in the capacitor and how much is stored in the inductor?

A charged capacitor with \(C=590 \mu \mathrm{F}\) is connected in series to an inductor that has \(L=0.330 \mathrm{H}\) and negligible resistance. At an instant when the current in the inductor is \(i=2.50 \mathrm{~A},\) the current is increasing at a ratc of \(d i / d t=73.0 \mathrm{~A} / \mathrm{s}\). During the current oscillations. what is the maximum voltage across the capacitor?

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