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Chapter 31: Problem 21

In an oscillating \(L C\) circuit with \(C=64.0 \mu \mathrm{F}\), the current is given by \(i=(1.60) \sin (2500 t+0.680),\) where \(t\) is in seconds, \(i\) in amperes, and the phase constant in radians. (a) How soon after \(t=0\) will the current reach its maximum value? What are (b) the inductance \(L\) and (c) the total energy?

Short Answer

Expert verified
(a) Solve \(2500t + 0.680 = \frac{\pi}{2}\), (b) use \(L = \frac{1}{(2500)^2 \times 64 \times 10^{-6}}\), (c) calculate \(U = \frac{1}{2} C V^2\) or \(U = \frac{1}{2} L I^2\).

Step by step solution

01

Identify the Maximum Current Condition

The current function is given by: \(i = (1.60) \sin(2500t + 0.680)\).\ To reach the maximum current, the value inside the sine function must be \(\frac{\pi}{2}\), where \(\sin\left(\frac{\pi}{2}\right) = 1\).
02

Solve for Time to Maximum Current

Set \(2500t + 0.680 = \frac{\pi}{2}\).\ Solving for \(t\), we get:\[2500t + 0.680 = \frac{\pi}{2}\]\[2500t = \frac{\pi}{2} - 0.680\]\[t = \frac{\frac{\pi}{2} - 0.680}{2500}\]
03

Calculate Inductance from Angular Frequency

The angular frequency \(\omega\) in the current equation is \(2500\), where \(\omega = \frac{1}{\sqrt{LC}}\).\ Rearrange to find \(L\):\[L = \frac{1}{\omega^2 C}\]Substitute \(\omega = 2500\) and \(C = 64.0 \times 10^{-6} \ \mathrm{F}\) to find \(L\).
04

Calculate Total Energy in the Circuit

The total energy \(U\) stored in the circuit can be calculated using either the electric or magnetic formula:\[U = \frac{1}{2} C V^2\]or\[U = \frac{1}{2} L I^2\]First determine \(V\) or \(I\) maximum using \(I = 1.60\ A\) (current amplitude) and substitute into the formula to find \(U\).

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Key Concepts

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

Inductance
An oscillating LC circuit consists of an inductor (L) and a capacitor (C) working together to create electrical oscillations. One of the critical elements that determine the behavior of an LC circuit is the inductance, which is represented by \( L \). Inductance is a property of an inductor that measures its ability to store energy in a magnetic field. In mathematical terms, it is defined by the relationship between the current change in the circuit and the electromotive force (EMF) induced.
To find the inductance in an LC circuit, the angular frequency (\( \omega \)) is a key parameter. This is because \( \omega \) is related to both inductance and capacitance by the formula:
\[\omega = \frac{1}{\sqrt{LC}}\]Rearranging the formula to solve for \( L \):
\[L = \frac{1}{\omega^2 C}\]Given \( \omega = 2500 \) rad/s and \( C = 64.0 \times 10^{-6} \) F, one can substitute these values into the formula to compute the inductance.
Current Amplitude
The current amplitude in an LC circuit refers to the maximum value of the oscillating current described by the current function. It gives us the peak current flowing through the circuit. In the given problem, the current is defined as:
\(i = (1.60) \sin(2500t + 0.680)\)
Here, the amplitude is the coefficient of the sine function, which is 1.60 A. This value indicates the maximum strength of current that flows through the circuit during one complete cycle of oscillation.
  • The amplitude tells us how strong the source is energizing the circuit.
  • It's crucial for calculating other important parameters like energy.
Understanding the role of current amplitude is essential for analyzing how circuits perform under different conditions.
Angular Frequency
Angular frequency is a measure of how rapidly the oscillations in an LC circuit occur. It is denoted by \( \omega \) and is related to the frequency \( f \) by the equation:
\[\omega = 2\pi f\]For an LC circuit, \( \omega \) is more conveniently expressed in relation to the inductance \( L \) and capacitance \( C \):
\[\omega = \frac{1}{\sqrt{LC}}\]In the exercise given, the angular frequency is \( 2500 \) rad/s, dictating how quickly the oscillations happen in terms of radians per second.
  • High \( \omega \) implies faster oscillations.
  • It determines the rate at which energy is transferred between the inductor and capacitor.
Angular frequency is crucial in timing and energy calculations within the circuit.
Total Energy in Circuit
The total energy in an LC circuit can be calculated from the energy stored either in the electric field of the capacitor or the magnetic field of the inductor. This energy is retained in the form of alternating current and voltage oscillations within the circuit.
The formulas to calculate this energy are:
\[U = \frac{1}{2} C V^2\] (using capacitance)
\[U = \frac{1}{2} L I^2\] (using inductance)Using either formula requires knowledge of the maximum voltage \( V \) across the capacitor or the current amplitude \( I \) in the circuit. For this circuit, you can compute energy using the known current amplitude \( I = 1.60 \) A.
  • The energy remains constant and oscillates between the inductor and capacitor.
  • Understanding total energy helps in analyzing the circuit's efficiency.
Energy calculations reveal how well the circuit sustains and utilizes energy through each oscillation cycle.

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Most popular questions from this chapter

In a series oscillating \(R L C\) circuit, \(R=16.0 \Omega, C=\) \(31.2 \mu \mathrm{F}, L=9.20 \mathrm{mH},\) and \(\mathscr{C}_{m}=\mathscr{E}_{m} \sin \omega_{d} t\) with \(\mathscr{E}_{m}=45.0 \mathrm{~V}\) and \(\omega_{d}=3000 \mathrm{rad} / \mathrm{s} .\) For time \(t=0.442 \mathrm{~ms}\) find \((\mathrm{a})\) the rate \(P_{g}\) at which energy is being supplied by the generator, (b) the rate \(P_{C}\) at which the energy in the capacitor is changing, (c) the rate \(P_{L}\) at which the energy in the inductor is changing, and (d) the rate \(P_{R}\) at which energy is being dissipated in the resistor. (e) Is the sum of \(P_{C}, P_{L}\) and \(P_{R}\) greater than, less than, or equal to \(P_{g} ?\)

A variable capacitor with a range from 10 to \(365 \mathrm{pF}\) is used with a coil to form a variable-frequency \(L C\) circuit to tune the input to a radio. (a) What is the ratio of maximum frequency to minimum frequency that can be obtained with such a capacitor? If this circuit is to obtain frequencies from \(0.54 \mathrm{MHz}\) to \(1.60 \mathrm{MHz}\), the ratio computed in (a) is too large. By adding a capacitor in parallel to the variable capacitor, this range can be adjusted. To obtain the desired frequency range, (b) what capacitance should be added and (c) what inductance should the coil have?

89 SSM For a sinusoidally driven series \(R L C\) circuit, show that over one complete cycle with period \(T\) (a) the energy stored in the capacitor does not change; (b) the energy stored in the inductor does not change; (c) the driving emf device supplies energy \(\left(\frac{1}{2} T\right) \mathscr{E}_{m} I \cos \phi ;\) and (d) the resistor dissipates energy \(\left(\frac{1}{2} T\right) R I^{2}\). (e) Show that the quantities found in (c) and (d) are equal.

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.

For a certain driven series \(R L C\) circuit, the maximum generator emf is \(125 \mathrm{~V}\) and the maximum current is \(3.20 \mathrm{~A}\). If the current leads the generator emf by 0.982 rad, what are the (a) impedance and (b) resistance of the circuit? (c) Is the circuit predominantly capacitive or inductive?
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