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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

(a) Compute the reactance of a 0.450-H inductor at frequencies of 60.0 Hz and 600 Hz. (b) Compute the reactance of a 2.50-\(\mu\)F capacitor at the same frequencies. (c) At what frequency is the reactance of a 0.450-H inductor equal to that of a 2.50-\(\mu\)F capacitor?

A transformer connected to a 120-V (rms) ac line is to supply 13,000 V (rms) 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 rms current in the secondary circuit exceeds 8.50 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 mA? (c) What current rating should the fuse in the primary circuit have?

In an \(L-R-C\) series circuit, the rms voltage across the resistor is 30.0 V, across the capacitor it is 90.0 V, and across the inductor it is 50.0 V. What is the rms voltage of the source?

A resistor, an inductor, and a capacitor are connected in parallel to an ac source with voltage amplitude \(V\) and angular frequency \(\omega\). Let the source voltage be given by \(v\) = V cos \(\omega\)t. (a) Show that each of the instantaneous voltages \(v_R\) , \(v_L\), and \(v_C\) at any instant is equal to \(v\) and that \(i\) = \(i_R\) + \(i_L\) + \(i_C\), where i is the current through the source and iR , iL, and iC are the currents through the resistor, inductor, and capacitor, respectively. (b) What are the phases of \(i_R\) , \(i_L\), and \(i_C\) with respect to v? Use current phasors to represent i, iR , iL, and iC. In a phasor diagram, show the phases of these four currents with respect to \(v\). (c) Use the phasor diagram of part (b) to show that the current amplitude I for the current i through the source is \(I =\sqrt{I_{R ^2} + (I_C - I_L)^2}\). (d) Show that the result of part (c) can be written as\( I = V/Z\), with \(1/Z = \sqrt{(1/R^2) + [\omega C - (1/\omega L)]^2}.\)

You plan to take your hair dryer to Europe, where the electrical outlets put out 240 V instead of the 120 V seen in the United States. The dryer puts out 1600 W at 120 V. (a) What could you do to operate your dryer via the 240-V line in Europe? (b) What current will your dryer draw from a European outlet? (c) What resistance will your dryer appear to have when operated at 240-V?

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