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Chapter 11: Problem 5

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 \(\omega = 7.55 \times 10^3 \, \text{rad/s}\), and the reactance of each element is approximately \(37.75 \, \Omega\).

Step by step solution

01

Understanding Reactance

Reactance is the opposition to the change in current by inductors and capacitors. For an inductor with inductance \(L\), the inductive reactance \(X_L\) is \(X_L = \omega L\). For a capacitor with capacitance \(C\), the capacitive reactance \(X_C\) is \(X_C = \frac{1}{\omega C}\), where \(\omega\) is the angular frequency.
02

Setting Reactances Equal

Since we want the reactances to be equal, set \(X_L = X_C\). This gives us the equation \(\omega L = \frac{1}{\omega C}\). From this equation, we can solve for \(\omega\).
03

Solving for Angular Frequency

Rearrange the equation from Step 2: \(\omega^2 = \frac{1}{LC}\). Solve for \(\omega\): \(\omega = \sqrt{\frac{1}{LC}}\). This gives the angular frequency where the reactances are equal.
04

Substitute Given Values

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

Calculate Angular Frequency

Calculate \(\omega\): \[\omega = \sqrt{\frac{1}{(5.00 \times 10^{-3})(3.50 \times 10^{-6})}}\] Simplify and calculate to find \(\omega\).
06

Calculate Numerical Value of Reactance

Once \(\omega\) is found, calculate the reactance using \(X_L = \omega L\) or \(X_C = \frac{1}{\omega C}\). Both should be equal as they occur at the same angular frequency.

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Key Concepts

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

Reactance
Reactance is a fundamental concept in alternating current (AC) circuits involving inductors and capacitors. It describes how these components resist changes in current. While resistance applies both to DC and AC circuits, reactance is unique to AC.

Key points to understand about reactance:
  • Reactance depends on the frequency of the AC current.
  • There are two types of reactance: inductive reactance and capacitive reactance.
  • Reactance can be thought of as the "AC equivalent" of resistance.
The unit of reactance is the ohm (惟), the same as resistance, but it represents a different kind of opposition to current flow.

In circuits with both inductors and capacitors, reactance plays a crucial role in determining how the circuit behaves at different frequencies.
Inductive Reactance
Inductive reactance, denoted as \(X_L\), arises when an inductor is present in an AC circuit. An inductor opposes changes in current because of the magnetic field it creates.

Key characteristics of inductive reactance:
  • The formula for inductive reactance is \(X_L = \omega L\), where \(\omega\) is the angular frequency and \(L\) is the inductance in henrys.
  • As frequency increases, inductive reactance also increases. This means inductors oppose high-frequency currents more than low-frequency ones.
  • Inductive reactance causes the current to lag behind the voltage by 90 degrees in phase.
Understanding inductive reactance helps in designing circuits that manage or exploit these frequency-dependent properties.
Capacitive Reactance
Capacitive reactance, symbolized as \(X_C\), emerges in circuits with capacitors and provides an opposition to changes in voltage.

Key aspects of capacitive reactance:
  • The formula is \(X_C = \frac{1}{\omega C}\), where \(\omega\) is the angular frequency and \(C\) is the capacitance in farads.
  • Unlike inductive reactance, capacitive reactance decreases as the frequency increases. Therefore, capacitors pass higher frequency currents more easily than lower ones.
  • It causes the current to lead the voltage by 90 degrees in phase.
When working with capacitive reactance, it's important to consider how the frequency affects capacitive circuits' behavior.
Resonance Frequency
In LC circuits, the resonance frequency is a key concept where the inductive and capacitive reactances are equal. At this frequency, the reactive components effectively cancel each other out.

For an LC circuit:
  • The resonance frequency occurs when \(X_L = X_C\), leading to \(\omega L = \frac{1}{\omega C}\).
  • This simplifies to \(\omega^2 = \frac{1}{LC}\), and the resonance frequency \(\omega\) can be calculated as \(\omega = \sqrt{\frac{1}{LC}}\).
  • At resonance, the circuit has a purely resistive impedance, and the total impedance is minimized.
Resonance is utilized in many applications like radios, where it helps in selecting the desired frequency.
LC Circuit
An LC circuit consists of an inductor (L) and a capacitor (C) connected together, and it's often used in oscillating circuits.

Important notes about LC circuits:
  • They are fundamental building blocks in radio frequency applications and filters.
  • The exchange of energy between the capacitor and inductor can cause oscillations at the circuit's resonance frequency.
  • In a parallel LC circuit, at resonance, the impedance is maximized, while in a series LC circuit, the impedance is minimized.
Understanding how LC circuits work is crucial for mastering AC circuit design and analysis, especially in telecommunications.
Angular Frequency Calculation
Angular frequency \(\omega\) is a central concept in alternating current (AC) circuit analysis and is related to how rapidly oscillations occur in a circuit.

To calculate angular frequency, especially when equalizing reactances, use the formula:
  • For two reactive components to have the same reactance: \(\omega = \sqrt{\frac{1}{LC}}\).
  • This formula is derived when equating the inductive and capacitive reactances: \(\omega L = \frac{1}{\omega C}\).
  • Angular frequency is measured in radians per second (rad/s).
Angular frequency helps in understanding the dynamic behavior of circuits, especially those involving resonant frequencies.

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Most popular questions from this chapter

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 \(53^{\circ}\). Determine the inductance of the coil.

You have a special light bulb with a very delicate wire filament. The wire will break if the current in it ever exceeds \(2 \mathrm{~A}\), even for an instant. What is the largest root-mean-square current you can run through this bulb?

An \(L-R-C\) series circuit has \(C=5 \mu \mathrm{F}, L=0.8 \mathrm{H}\),and source voltage amplitude \(V-56.0 \mathrm{~V} .\) The source is operated at the resonance frequency of the circuit. If the voltage across the capacitor has amplitude \(80.0 \mathrm{~V}\), what is the value of \(R\) for the resistor in the circuit?

An \(L-R-C\) series circuit is constructed using a \(200-\Omega\) resistor, a \(20-\mu \mathrm{F}\) capacitor, and an \(8.00-\mathrm{mH}\) inductor, all connected across an ac source having a variable frequency and a voltage amplitude of \(25.0 \mathrm{~V} .\) (a) At what angular frequency will the impedance be smallest, and what is the impedance at this frequency? (b) At the angular frequency in part (a), what is the maximum current through the inductor? (c) At the angular frequency in part (a), find the potential difference across the ac source, the resistor, the capacitor, and the inductor at the instant that the current is equal to one-half its greatest positive value. (d) In part (c), how are the potential differences across the resistor, inductor, and capacitor related to the potential difference across the ac source?

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