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31.6. A capacitance \(C\) and an inductance \(L\) are operated at the same angular frequency. (a) At what angular frequency will they have the same reactance? (b) If \(L=5.00 \mathrm{mH}\) and \(C=3.50 \mu \mathrm{F}\) , what is the numerical value of the angular frequency in part (a), and what is the reactance of each element?

Short Answer

Expert verified
The angular frequency is approximately 758 rad/s, and the reactance of each element is about 3.79 ohms.

Step by step solution

01

Understand Reactance

Reactance is a measure of the opposition that a circuit element presents to the flow of AC current. For a capacitor, the reactance is given by \(X_C = \frac{1}{\omega C}\) and for an inductor, it is \(X_L = \omega L\), where \(\omega\) is the angular frequency.
02

Set Reactances Equal

Since the problem states that the reactances are equal, we set the reactance of the capacitor equal to the reactance of the inductor: \(\frac{1}{\omega C} = \omega L\).
03

Solve for Angular Frequency

Solve the equation \(\frac{1}{\omega C} = \omega L\) for \(\omega\). Multiply both sides by \(\omega C\) to obtain \(1 = \omega^2 LC\). Thus, \(\omega = \frac{1}{\sqrt{LC}}\).
04

Substitute Numerical Values

Use the given values \(L = 5.00 \times 10^{-3} \text{ H}\) and \(C = 3.50 \times 10^{-6} \text{ F}\) into \(\omega = \frac{1}{\sqrt{LC}}\). Calculate \(\omega = \frac{1}{\sqrt{(5.00 \times 10^{-3})(3.50 \times 10^{-6})}}\).
05

Calculate Reactance

Once \(\omega\) is calculated, use it to determine the reactance \(X\) of each element using \(X_L = \omega L\) and \(X_C = \frac{1}{\omega C}\). Verify that both reactances are equal as expected.

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Key Concepts

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

Capacitance
Capacitance is a fundamental property in electrical circuits that describes a component's ability to store electrical energy in an electric field. This property is mainly associated with capacitors and is represented by the symbol \( C \). Capacitors are devices that store energy in the form of an electric charge, widely used in electronic circuits for various applications such as filtering and energy storage.

Capacitors oppose changes in voltage across their plates. This opposition is quantified by their reactance, specifically capacitive reactance \( X_C \). The capacitive reactance is calculated by the formula:
  • \( X_C = \frac{1}{\omega C} \)
where \( \omega \) is the angular frequency of the alternating current (AC) supply, and \( C \) is the capacitance in farads. The higher the frequency, the smaller the reactance, allowing the current to pass more easily. On the other hand, at lower frequencies, the reactance is higher, impeding the current flow.
Inductance
Inductance is the property of a circuit, or component like an inductor, to resist any change in the flow of electrical current through it. It's often associated with coils of wire and is denoted by the symbol \( L \). Inductors store energy in a magnetic field when electrical current flows through them, making them crucial for tasks like filtering, tuning circuits, and power management.

Inductors oppose changes to the current with inductive reactance \( X_L \). It's expressed through the formula:
  • \( X_L = \omega L \)
Here, \( \omega \) is the same angular frequency as in capacitive reactance, and \( L \) is the inductance measured in henrys. Unlike capacitors, as the frequency of AC increases, the inductive reactance also increases, presenting more opposition to the current flow. This relationship is essential in creating circuits that handle different frequencies differently, such as filters or tuners.
Angular frequency
Angular frequency \( \omega \) is a measure of how quickly a periodic wave oscillates, a vital concept in the analysis of AC circuits. It is related to the ordinary frequency \( f \) (measured in hertz) by the formula:
  • \( \omega = 2\pi f \)
This expression allows you to convert between the conventional frequency and angular frequency effortlessly.

In circuits containing both capacitors and inductors, the angular frequency determines how these components interact with AC. Particularly, when dealing with reactance, \( \omega \) plays a central role. For both capacitance and inductance, it alters the reactance levels, dictating the degree of opposition each component presents to AC. Specifically, when the reactance of a capacitor matches the reactance of an inductor, it is often at a particular angular frequency where their effects balance each other out. This frequency is crucial in designing circuits like resonant LC circuits, where stable oscillations are needed.

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Most popular questions from this chapter

31.38. A Step-Up Transformer. A transformer connected to a \(120-\mathrm{V}(\mathrm{rms})\) ac line is to supply \(13,000 \mathrm{V}(\mathrm{ms})\) for a neon sign. To reduce shock hazard, a fuse is to be inserted in the primary circuit; the fuse is to blow when the ms current in the secondary circuit exceeds 8.50 \(\mathrm{mA}\) (a) What is the ratio of secondary to primary turns of the transformer? (b) What power must be supplied to the transformer when the rms secondary current is 8.50 \(\mathrm{mA} ?(\mathrm{c})\) What current rating should the fuse in the primary circuit have?

31.66. A resistance \(R,\) capacitance \(C,\) and inductance \(L\) are connected in series to a voltage source with amplitude \(V\) and variable angular frequency \(\omega\) . If \(\omega=\omega_{0}\) , the resonance angular frequency, find (a) the maximum current in the resistor; (b) the maximum voltage across the capacitor; (c) the maximum voltage across thinductor; (d) the maximum energy stored in the capacitor; (e) the maximum energy stored in the inductor. Give your answers in terms of \(R, C, L,\) and \(V .\)

31.10. A Radio Inductor. You want the current amplitude through a \(0.450-\mathrm{mH}\) inductor (part of the circuitry for a radio receiver) to be 2.60 \(\mathrm{mA}\) when a sinusoidal voltage with amplitude 12.0 \(\mathrm{V}\) is applied across the inductor. What frequency is required?

31.9. (a) What is the reactance of a \(3.00-\mathrm{H}\) inductor at a frequency of 80.0 \(\mathrm{Hz}\) (b) What is the inductance of an inductor whose reactance is 120\(\Omega\) at 80.0 \(\mathrm{Hz}\) (c) What is the reactance of a \(4.00-\mu \mathrm{F}\) capacitor at a frequency of 80.0 \(\mathrm{Hz}\) (d) What is the capacitance of a capacitor whose reactance is 120\(\Omega\) at 80.0 \(\mathrm{Hz}\) ?

31.59. In an \(L . R-C\) series circuit the magnitude of the phase angle is \(54.0^{\circ}\) , with the source voltage lagging the current. The reactance of the capacitor is 350\(\Omega\) , and the resistor resistance is \(180 \Omega .\) The average power delivered by the source is 140 \(\mathrm{W}\) . Find (a) the reactance of the inductor, \((\mathrm{b})\) the rms current; (c) the rms voltage of the source.

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