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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 mH and \(C\) = 3.50 \(\mu\)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 7546.5 rad/s. The reactance of each element is 37.73 ohms.

Step by step solution

01

Understanding Reactance Equality

The reactance of an inductor is given by the formula \( X_L = \omega L \) and the reactance of a capacitor is given by \( X_C = \frac{1}{\omega C} \), where \( \omega \) is the angular frequency. To find the angular frequency where they have the same reactance, set the two expressions equal: \( \omega L = \frac{1}{\omega C} \).
02

Solving for Angular Frequency

To solve for \( \omega \), we equate and rearrange the equation: \( \omega^2 = \frac{1}{LC} \). Thus, the angular frequency is given by \( \omega = \frac{1}{\sqrt{LC}} \).
03

Calculating Angular Frequency

Given \( L = 5.00 \text{ mH} = 5.00 \times 10^{-3} \text{ H} \) and \( C = 3.50 \mu\text{F} = 3.50 \times 10^{-6} \text{ F} \), substitute these values into the equation: \( \omega = \frac{1}{\sqrt{(5.00 \times 10^{-3})(3.50 \times 10^{-6})}} \).
04

Computing the Angular Frequency Numerically

Calculate \( \omega \):\[ \omega = \frac{1}{\sqrt{17.5 \times 10^{-9}}} = \frac{1}{\sqrt{17.5} \times 10^{-4.5}} \]Evaluate this to find \( \omega \approx 7546.5 \text{ rad/s} \).
05

Calculating Reactance

Once \( \omega \) is known, compute the reactance using either formula, e.g., \( X_L = \omega L = 7546.5 \times 5.00 \times 10^{-3} \). Thus, \( X_L = 37.73 \text{ ohms} \).
06

Confirming Reactance Equality

Calculate \( X_C \) as well to verify equality: \( X_C = \frac{1}{\omega C} = \frac{1}{7546.5 \times 3.50 \times 10^{-6}} = 37.73 \text{ ohms} \). Both reactances match, confirming the solution.

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Key Concepts

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

Angular Frequency
Angular frequency is a concept that describes how many times a cycle occurs in a unit of time, specifically in a circular motion. It is typically represented by the symbol \( \omega \) and is measured in radians per second. Unlike regular frequency, which counts the number of cycles per second in Hertz, angular frequency considers the rotational aspect by using radians.

In the context of inductors and capacitors, angular frequency is a critical component because it affects the reactance of these components. For inductors, the reactance (\( X_L \)) is directly proportional to angular frequency, as given by the formula \( X_L = \omega L \). On the other hand, the reactance of capacitors (\( X_C \)) is inversely related, described by \( X_C = \frac{1}{\omega C} \).

If you want the reactance of both an inductor and a capacitor to be equal, you can set these two equations equal to each other, leading to \( \omega = \frac{1}{\sqrt{LC}} \). By doing so, you define a special angular frequency at which the reactance of both elements is identical.
Inductance
Inductance, represented by the symbol \( L \), measures an inductor's ability to store energy in a magnetic field when an electric current flows through it. It is measured in Henries (H). This property arises from the inductor's coil winding which generates a magnetic field opposing changes in current.

A key factor about inductors is their reactance, often called inductive reactance, which signifies the opposition to current changes at a particular frequency. This opposition, expressed as \( X_L = \omega L \), means that as the angular frequency \( \omega \) increases, so does the reactance. Therefore, the reactance of an inductor is dependent on both operation frequency and its inductance value.
  • Higher inductance results in higher reactance.
  • Increased frequency leads to increased reactance.
When an inductor and a capacitor are placed in a circuit, they can resonate at a specific frequency, where their reactances equal. This relationship plays a crucial role in tuning frequencies within electronic circuits like radio transmitters and receivers.
Capacitance
Capacitance, denoted by the symbol \( C \), refers to a capacitor's ability to store charge in an electric field. It is measured in Farads (F). A capacitor consists of two conductive plates separated by an insulating material, or dielectric, which prevents current from flowing directly between them.

The reactance of a capacitor is inversely proportional to both the angular frequency and the capacitance, as illustrated by the formula \( X_C = \frac{1}{\omega C} \). This relationship implies that the reactance of a capacitor decreases with an increase in angular frequency. Hence, capacitors allow more current to pass at higher frequencies.
  • Lower capacitance results in higher reactance.
  • Higher frequency reduces reactance, allowing easier current flow.
By balancing capacitance and angular frequency, capacitors can be used alongside inductors to create tuning elements in electronic applications, similar to radio tuners, which is why frequency and capacitance are essential concepts in electrical circuits.

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Most popular questions from this chapter

An inductor with \(L\) = 9.50 mH is connected across an ac source that has voltage amplitude 45.0 V. (a) What is the phase angle \(\phi\) for the source voltage relative to the current? Does the source voltage lag or lead the current? (b) What value for the frequency of the source results in a current amplitude of 3.90 A?

The wiring for a refrigerator contains a starter capacitor. A voltage of amplitude 170 V and frequency 60.0 Hz applied across the capacitor is to produce a current amplitude of 0.850 A through the capacitor. What capacitance \(C\) is required?

In an \(L-R-C\) series circuit the source is operated at its resonant angular frequency. At this frequency, the reactance X\(_C\) of the capacitor is 200 \(\Omega\) and the voltage amplitude across the capacitor is 600 V. The circuit has \(R\) = 300 \(\Omega\). What is the voltage amplitude of the source?

A coil has a resistance of 48.0 \(\Omega\). At a frequency of 80.0 Hz the voltage across the coil leads the current in it by 52.3\(^\circ\). Determine the inductance of the coil.

A transformer connected to a 120-V (rms) ac line is to supply 12.0 V (rms) to a portable electronic device. The load resistance in the secondary is 5.00 \(\Omega\). (a) What should the ratio of primary to secondary turns of the transformer be? (b) What rms current must the secondary supply? (c) What average power is delivered to the load? (d) What resistance connected directly across the 120-V line would draw the same power as the transformer? Show that this is equal to 5.00 \(\Omega\) times the square of the ratio of primary to secondary turns.

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