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Suppose we have a series resonant circuit for which \(B=15 \mathrm{kHz}, f_{0}=300 \mathrm{kHz}\), and \(C=663.1 \mathrm{pF}\). Determine the values of \(L\) and \(R\).

Short Answer

Expert verified
Inductance \(L \approx 42.5 \mathrm{\mu H}\) and Resistance \(R \approx 2.5 \mathrm{\Omega}\).

Step by step solution

01

Understand the Problem

In this problem, we need to determine the values of inductance \(L\) and resistance \(R\) in a series resonant circuit. We are given the bandwidth \(B\), the resonant frequency \(f_0\), and the capacitance \(C\). We're going to use the known formulas for resonant circuits to find \(L\) and \(R\).
02

Use the Quality Factor Formula

A key relationship in resonant circuits involves the quality factor \(Q\), given by \(Q = \frac{f_0}{B}\). Calculate \(Q\) using the provided resonant frequency \(f_0 = 300 \mathrm{kHz}\) and bandwidth \(B = 15 \mathrm{kHz}\):\[Q = \frac{300 \mathrm{kHz}}{15 \mathrm{kHz}} = 20\]
03

Calculate Inductance L

For a series resonant circuit, the inductance \(L\) can be found using the resonant frequency formula: \(f_0 = \frac{1}{2\pi\sqrt{LC}}\). Rearrange the formula to solve for \(L\):\[L = \frac{1}{(2\pi f_0)^2 C}\]Given \(f_0 = 300 \mathrm{kHz}\) and \(C = 663.1 \mathrm{pF}\), calculate \(L\):\[L = \frac{1}{(2\pi \times 300 \times 10^3)^2 \times 663.1 \times 10^{-12}}\]Computing the above gives:\[L \approx 42.5 \mathrm{\mu H}\]
04

Calculate Resistance R

The resistance \(R\) in a series resonant circuit is linked with \(Q\) through the relationship \(Q = \frac{1}{R} \sqrt{\frac{L}{C}}\). Rearrange to solve for \(R\):\[R = \frac{1}{Q} \sqrt{\frac{L}{C}}\]Using \(Q = 20\), \(L = 42.5 \mathrm{\mu H}\), and \(C = 663.1 \mathrm{pF}\), calculate \(R\):\[R = \frac{1}{20} \sqrt{\frac{42.5 \times 10^{-6}}{663.1 \times 10^{-12}}}\]Compute the expression to find:\[R \approx 2.5 \mathrm{\Omega}\]
05

Verify the Results

Let’s ensure the calculated values of \(L\) and \(R\) meet the problem's requirements. The formulas and calculations align with the theory of resonant circuits, ensuring these are the correct values.

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Key Concepts

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

Inductance Calculation
Inductance is a core component of resonant circuits, acting as one of the main elements responsible for storing and transferring energy. In a series resonant circuit, the inductance (L) is always connected in series with the capacitance (C). Together, they determine the circuit's ability to resonate at a specific frequency. To calculate the inductance (L) necessary for a given resonant frequency and capacitance, we use the formula:\[f_0 = \frac{1}{2\pi\sqrt{LC}}\]Rearanging this equation helps us solve for the inductance:\[L = \frac{1}{(2\pi f_0)^2 C}\]This formula is crucial for designing circuits that need to resonate at a specific frequency, ensuring efficient energy transfer. When substituting the given \(f_0 = 300\, \text{kHz}\) and \(C = 663.1\,\text{pF}\), we find the inductance:\[L \approx 42.5 \, \mu \text{H}\]Understanding how to calculate L allows designers to manipulate circuit characteristics such as frequency response and bandwidth effectively.
Quality Factor
The quality factor, often symbolized as Q, is a dimensionless parameter that describes how underdamped a resonant circuit is. It also signifies the sharpness of the resonance peak in the frequency response of the circuit. A higher Q indicates a sharper peak, which means the circuit is more selective in frequency. In the context of series resonant circuits, Q can be calculated using the formula:\[Q = \frac{f_0}{B}\]With \(f_0 = 300 \, \text{kHz}\) and \(B = 15 \, \text{kHz}\), you calculate:\[Q = \frac{300 \, \text{kHz}}{15 \, \text{kHz}} = 20\]This value of Q indicates that the circuit will have a moderate selectivity, being neither too broad nor too narrow in its frequency acceptance. Having a detailed understanding of Q helps in predicting how the circuit will perform, especially in filter design where frequency selectivity is paramount.
Circuit Bandwidth
The concept of circuit bandwidth is closely tied to both the resonant frequency and the quality factor. Bandwidth refers to the range of frequencies over which the circuit can operate effectively. In resonant circuits, it is defined as the difference between the lower and upper -3dB frequencies, the points at which the power drops to half of its peak value. For a given quality factor Q and resonant frequency \(f_0\), the bandwidth B is given by:\[B = \frac{f_0}{Q}\]The bandwidth of series resonant circuits is significant because it determines how selective the circuit is in terms of filtering one frequency from others. With a calculated bandwidth,\[B = 15 \, \text{kHz}\]it illustrates how the circuit will only pass a narrow band of frequencies around the resonant frequency (300 \, \text{kHz}). Engineers can manipulate bandwidth to allow circuits to either filter out unwanted signals or to focus narrowly on a target frequency range.
Resonant Frequency
Resonant frequency is the frequency at which the inductive and capacitive reactances in a circuit cancel each other out, resulting in a pure resistive circuit where maximum current flows. For series resonant circuits, this frequency is a hallmark of its design and is essential for energy transmission and communication systems. The resonant frequency (\(f_0\)) is determined by:\[f_0 = \frac{1}{2\pi\sqrt{LC}}\]A perfectly tuned circuit will resonate at this frequency, allowing maximum power transfer and minimal impedance. In our example with \(f_0 = 300 \, \text{kHz}\), it means the circuit is designed to be most effective at this frequency, which in practice is crucial for applications like radio transmitters, receivers, and filters where precision is required.Understanding the concept of resonant frequency is invaluable for designing and analyzing circuits that must perform efficiently under specific frequency conditions.

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Most popular questions from this chapter

List the frequencies in hertz for which the transfer function of a filter can be determined given that the input to the filter is $$ \begin{aligned} v_{\text {in }}(t)=& 2+3 \cos (1000 \pi t)+3 \sin (2000 \pi t) \\ &+\cos (3000 \pi t) \mathrm{V} \end{aligned} $$ and the output is $$ \begin{aligned} v_{\text {out }}(t)=& 3+2 \cos \left(1000 \pi t+30^{\circ}\right) \\ &+4 \sin (3000 \pi t) \mathrm{V} \end{aligned} $$ Compute the transfer function for each of these frequencies.

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