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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 \(L=5.00 \mathrm{mH}\) 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
Angular frequency is \(7.55 \times 10^3\) rad/s, and reactance is 37.75 \(\Omega\).

Step by step solution

01

Understand Reactance

The reactance of an inductor (inductive reactance) is given by \( X_L = \omega L \), where \( \omega \) is the angular frequency and \( L \) is the inductance. The reactance of a capacitor (capacitive reactance) is \( X_C = \frac{1}{\omega C} \), where \( C \) is the capacitance.
02

Equate Inductive and Capacitive Reactance

To find the angular frequency where the inductive and capacitive reactances are equal, set \( X_L = X_C \). This gives the equation: \[ \omega L = \frac{1}{\omega C} \].
03

Solve for Angular Frequency

Rearrange the equation from Step 2 to solve for \( \omega \): \[ \omega^2 = \frac{1}{LC} \]. Then find \( \omega \) by \( \omega = \sqrt{\frac{1}{LC}} \).
04

Calculate Angular Frequency

Substitute \( L = 5.00 \text{ mH} = 5.00 \times 10^{-3} \text{ H} \) and \( C = 3.50 \mu \text{F} = 3.50 \times 10^{-6} \text{ F} \) into the formula: \( \omega = \sqrt{\frac{1}{(5.00 \times 10^{-3})(3.50 \times 10^{-6})}} \).
05

Compute Angular Frequency Value

Perform the calculation: \[ \omega = \sqrt{\frac{1}{(5.00 \times 10^{-3})(3.50 \times 10^{-6})}} \approx \sqrt{\frac{1}{1.75 \times 10^{-8}}} = \sqrt{5.71 \times 10^{7}} \approx 7.55 \times 10^3 \text{ rad/s} \].
06

Find Reactance

Use the found \( \omega \) to calculate the reactance: \( X_L = \omega L = (7.55 \times 10^3)(5.00 \times 10^{-3}) \approx 37.75 \text{ } \Omega \) and \( X_C = \frac{1}{\omega C} = \frac{1}{(7.55 \times 10^3)(3.50 \times 10^{-6})} \approx 37.75 \text{ } \Omega \).

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Key Concepts

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

Inductive Reactance
Inductive reactance is an important concept in understanding how inductors behave in an AC circuit. An inductor, typically a coil of wire, stores energy in a magnetic field when electric current flows through it. The opposition to the change in current is measured by inductive reactance.
- The formula for inductive reactance is given by \( X_L = \omega L \), where \( \omega \) is the angular frequency in radians per second and \( L \) is the inductance in henries.
- Inductive reactance increases with the frequency of the AC signal. This means that as the frequency increases, the inductor will oppose the change in current more strongly.
- It acts like a sort of "resistance" that varies with frequency but does not dissipate power like a resistor, instead storing energy in the magnetic field.
Understanding inductive reactance is crucial when designing circuits, especially when aiming for a certain behavior at specific frequencies.
Capacitive Reactance
Capacitive reactance is the measure of a capacitor's opposition to the change in voltage in an AC circuit. A capacitor stores energy in an electric field between its plates, which impacts how it reacts to varying frequencies.
- The formula for capacitive reactance is defined as \( X_C = \frac{1}{\omega C} \), where \( \omega \) is the angular frequency and \( C \) is the capacitance in farads.
- Contrary to inductive reactance, capacitive reactance decreases as the frequency increases. Thus, at higher frequencies, a capacitor allows the AC signal to pass more easily.
- Just like inductive reactance, capacitive reactance does not consume power but temporarily stores energy, influencing how circuits behave at different frequencies.
Understanding capacitive reactance is essential when managing the flow and storage of energy within AC circuits, especially when tuning circuits to particular frequencies.
LC Circuit
An LC circuit, or inductor-capacitor circuit, is a fundamental electrical circuit that consists of an inductor (L) and a capacitor (C) connected together.
- Such circuits are used to create resonant circuits, where the inductive reactance and capacitive reactance can balance each other out at a specific frequency known as the resonant frequency.
- The resonant frequency \( \omega_0 \) of an LC circuit is given by the formula \( \omega_0 = \sqrt{\frac{1}{LC}} \), showing the interplay between the inductance and capacitance.
- When an LC circuit reaches its resonance, the energy oscillates between the magnetic field of the inductor and the electric field of the capacitor.
This oscillating energy can produce very high currents and voltages, making LC circuits extremely useful in applications like radio transmitters and receivers, where specific frequencies need to be targeted.

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Most popular questions from this chapter

A toroidal solenoid has 2900 closely wound turns, cross-sectional area \(0.450 \mathrm{cm}^{2},\) mean radius \(9.00 \mathrm{cm},\) and resistance \(R=2.80 \Omega .\) The variation of the magnetic field across the cross section of the solenoid can be neglected. What is the amplitude of the current in the solenoid if it is connected to an ac source that has voltage amplitude 24.0 \(\mathrm{V}\) and frequency 365 \(\mathrm{Hz}\) ?

(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 \(\mathrm{Hz} ?\)

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

In an \(L-R-C\) series circuit, the components have the following values: \(L=20.0 \mathrm{mH}, C=140 \mathrm{nF},\) and \(R=350 \Omega .\) The generator has an rms voltage of 120 \(\mathrm{V}\) and a frequency of 1.25 \(\mathrm{kHz}\) . Determine (a) the power supplied by the generator and (b) the power dissipated in the resistor.

The \(L \cdot R-C\) Parallel Circuit. A resistor, inductor, and 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 the instantaneous voltages \(v_{R}, v_{L},\) and \(v_{C}\) at any instant are each equal to \(v\) and that \(i=i_{R}+i_{L}+i_{C},\) where \(i\) is the current through the source and \(i_{R},\) \(i_{L},\) and \(i_{C}\) are the currents through the resistor, the inductor, and the capacitor, respectively. (b) What are the phases of \(i_{R}, i_{L},\) and \(i_{C}\) with respect to \(v ?\) Use current phasors to represent \(i, i_{R}, i_{L},\) and \(i_{C}\) 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 given by \(I=\sqrt{I_{R}^{2}+\left(I_{C}-I_{L}\right)^{2}}\) (d) Show that the result of part (c) can be written as \(I=V / Z,\) with \(1 / Z=\sqrt{1 / R^{2}+(\omega C-1 / \omega L)^{2}}\)

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