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You want to double the resonance angular frequency of an L-R-C series circuit by changing only the pertinent circuit elements all by the same factor. (a) Which ones should you change? (b) By what factor should you change them?

Short Answer

Expert verified
Change both L and C by a factor of \( \frac{1}{2} \).

Step by step solution

01

Understand Resonance Frequency

The resonance angular frequency \( \omega_0 \) of an L-R-C series circuit is given by the formula \( \omega_0 = \frac{1}{\sqrt{LC}} \). Here, \( L \) is the inductance and \( C \) is the capacitance. Resistance \( R \) does not affect the resonance frequency.
02

Set the Desired Outcome

To double the resonance angular frequency \( \omega_0' = 2\omega_0 \), we require \( \frac{1}{\sqrt{L'C'}} = 2\times\frac{1}{\sqrt{LC}} \).
03

Express the New Values

Assume the new inductance and capacitance are changed by the same factor \( x \), i.e., \( L' = xL \) and \( C' = xC \).
04

Substitute and Solve for x

Plug these expressions into \( \frac{1}{\sqrt{L'C'}} = 2\times\frac{1}{\sqrt{LC}} \) to get \[ \frac{1}{\sqrt{x^2LC}} = 2\times\frac{1}{\sqrt{LC}}. \] Simplifying gives \( \frac{1}{x} = 2 \), hence \( x = \frac{1}{2}. \)
05

Conclusion on Changes Needed

To double the resonance angular frequency, both the inductance \( L \) and capacitance \( C \) should be changed by a factor of \( \frac{1}{2} \).

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Key Concepts

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

L-R-C Circuit
An L-R-C circuit is composed of three integral components: an inductor (L), a resistor (R), and a capacitor (C). These components are connected in series to form an electric circuit that can store and transfer energy among themselves in oscillatory motion. One of the most captivating aspects of an L-R-C circuit is its ability to resonate, or oscillate naturally at a specific frequency known as the resonance frequency.

The resonance frequency is particularly important because it is the frequency at which the system oscillates with the greatest amplitude. It's determined by the balance between the inductive reactance and capacitive reactance, while resistance plays no direct role in setting this frequency. An L-R-C circuit finds its application in radios, televisions, and other electronics where tuning to a frequency is essential.

Understanding the resonance behaviors of L-R-C circuits offers insights into how frequencies are manipulated in various devices, offering control over electronic signals and currents.
Inductance
Inductance is a property of an electrical conductor by which a change in current flowing through it induces an electromotive force (EMF). In an L-R-C circuit, the inductor regulates how quickly electric current can build up and subsequently decay.

Mathematically, inductance is expressed in henries ( H ) and directly influences the circuit's resonance frequency. The inductor's coil stores magnetic energy when current passes through it. As a part of the L-R-C circuit, adjusting the inductance can significantly influence the circuit's behavior. For instance, doubling the inductance while keeping other parameters constant will decrease the resonance frequency. This inverse relationship is crucial when setting up circuits to respond to certain frequencies.

It's noteworthy that while inductance plays a predominant role in the resonance frequency, it doesn't have any significant impact on energy dissipation, which is largely dominated by the resistance component in the circuit.
Capacitance
Capacitance refers to the capability of a system to store an electric charge. It is determined by the amount of electric charge stored per unit change in electric potential. In simple terms, it’s the ability of a circuit to hold charge, and in an L-R-C circuit, the capacitor does just that.

Similar to inductance, capacitance is a pivotal element in determining the resonance frequency of the circuit. It is measured in farads (F), and changes in capacitance will inversely affect the resonance frequency. A larger capacitance decreases the resonance frequency, making it vital to adjust when aiming for specific oscillation properties.

The capacitor holds energy in an electric field and releases it when needed. Understanding how capacitance works in tandem with inductance helps in fine-tuning the circuit's response to various input frequencies, enhancing the precise control of electronic devices and their effective functioning.

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Most popular questions from this chapter

The power of a certain \(\mathrm{CD}\) player operating at 120 \(\mathrm{V} \mathrm{rms}\) is 20.0 \(\mathrm{W}\) . Assuming that the CD player behaves like a pure resistor, find (a) the maximum instantaneous power; (b) the rms current; \((\mathrm{c})\) the resistance of this player.

In an \(L-R-C\) series circuit, \(R=400 \Omega, L=0.350 \mathrm{H},\) and \(C=0.0120 \mu \mathrm{F}\) (a) What is the resonance angular frequency of the circuit? (b) The capacitor can withstand a peak voltage of 550 \(\mathrm{V}\) . If the voltage source operates at the resonance frequency, what maximum voltage amplitude can it have if the maximum capacitor voltage is not exceeded?

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 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?

A Step-Down Transformer. A transformer connected to a \(120-\mathrm{V}(\mathrm{rms})\) ac line is to supply 12.0 \(\mathrm{V}(\mathrm{rms})\) 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 \(120-\mathrm{V}\) 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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