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Chapter 31: Problem 67

31.67. You want to double the resonance angular frequency of a series \(R-L-C\) 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 L and C by a factor of 1/2.

Step by step solution

01

Understand the Problem

The task is to double the resonance angular frequency of an RLC circuit, which is determined by the formula \( \omega_0 = \frac{1}{\sqrt{LC}} \). Only certain circuit elements should be changed by the same factor to achieve this.
02

Identify Elements Affecting Resonance Frequency

In the formula \( \omega_0 = \frac{1}{\sqrt{LC}} \), the resonance frequency \( \omega_0 \) depends on the inductance \( L \) and the capacitance \( C \) of the RLC circuit. The resistance \( R \) does not affect \( \omega_0 \).
03

Determine the Change Needed for Frequency

To double the resonance angular frequency, the new frequency \( 2\omega_0 = \frac{1}{\sqrt{L'C'}} \) must be achieved. This implies \( 2 = \sqrt{\frac{1}{L'C'}} \cdot \sqrt{LC} \). Simplifying gives \( LC = \frac{L'C'}{4} \).
04

Solve for Required Factor Change

To satisfy \( LC = \frac{L'C'}{4} \), and knowing \( L' = kL \) and \( C' = kC \), substitute to get \( k^2 \cdot LC = \frac{1}{4}LC \). This equation gives \( k^2 = \frac{1}{4} \), so \( k = \frac{1}{2} \).
05

Conclusion

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

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Key Concepts

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

Resonance Angular Frequency
In a series RLC circuit, the resonance angular frequency is a crucial concept. It represents the frequency at which the circuit naturally oscillates with minimal impedance. The formula for resonance angular frequency is given by \[ \omega_0 = \frac{1}{\sqrt{LC}} \]where:
  • \( \omega_0 \) is the resonance angular frequency.
  • \( L \) is the inductance.
  • \( C \) is the capacitance.
The resonance angular frequency is independent of the resistance in the circuit. Thus, any changes aimed at adjusting \( \omega_0 \) are typically made by altering the inductance \( L \) and the capacitance \( C \). By adjusting these elements, the "tuning" of the circuit can be finely controlled, which is a crucial aspect in various electronic applications, such as radios and tuners.
Inductance and Capacitance
Inductance and capacitance are two fundamental elements in an RLC circuit that determine how the circuit responds to AC signals. **Inductance**:
  • Represented by \( L \).
  • Measures the circuit's opposition to changes in current.
  • Stores energy in a magnetic field when current flows through a coil.
**Capacitance**:
  • Represented by \( C \).
  • Measures the circuit's ability to store an electrical charge.
  • Stores energy in an electric field between two plates.
Both these elements work together in a circuit to influence the resonance angular frequency. Changing either of them can dramatically alter how the circuit behaves. If both are reduced by the same factor, the overall natural frequency of the circuit increases, as seen in the exercise where halving both \( L \) and \( C \) doubles \( \omega_0 \).
Series Circuit Analysis
Series circuit analysis is a method used to evaluate complex circuits by focusing on how components are connected along a single path. In RLC circuits, the resistors (\( R \)), inductors (\( L \)), and capacitors (\( C \)) are connected in series. The current flowing through the circuit is the same at any point, simplifying the analysis:
  • **Impedance**: In series circuits, total impedance \( Z \) adds up since all components share the current. \[ Z = R + j(\omega L - \frac{1}{\omega C}) \]
  • **Voltage**: The total voltage across the circuit is the sum of the voltages across each component.
  • **Resonance**: Occurs when the reactance of the inductor equals the reactance of the capacitor.
Series RLC circuits are often analyzed using phasor diagrams and complex numbers to simplify calculations. At resonance, the impedance is minimized, and the circuit behaves almost like it only contains the resistance \( R \), with current reaching its maximum value.
Circuit Element Modification
Modifying circuit elements is essential in tuning the performance of electrical circuits. Suppose you want to alter a series RLC circuit to change its resonance angular frequency. In that case, adjustments will be made primarily to the inductance \( L \) and capacitance \( C \), as these elements directly affect the frequency:
  • **Doubling Frequency**: To double the resonance angular frequency, both \( L \) and \( C \) need to be reduced by half. As demonstrated, the factor should be \( \frac{1}{2} \).
  • **Reason**: This is because the angular frequency is inversely proportional to the square root of the product of \( L \) and \( C \).
Adjusting these elements is a common practice in circuit design, allowing for precise control over how a circuit interacts with an alternating current. Modifications can be theoretical, like using varied component values, or practical, like switching out physical components in a circuit to achieve the desired characteristics.

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Most popular questions from this chapter

3.1. The plate on the back nf a certain computer scanner says that the unit draws 0.34 \(\mathrm{A}\) of current from a \(120-\mathrm{V}, 60-\mathrm{Hz}\) line. Find (a) the root-mean-square current, (b) the current amplitude, (c) the average current; (d) the average square of the current.

31\. 27. (a) Show that for an \(L-R-C\) series circuit the power factor is equal to \(R / Z\) . (Hint: Use the phasor diagram; see Fig. \(31.13 b . )\) (b) Show that for any ac circuit, not just one containing pure resistance only, the average power delivered by the voltage source is given by \(P_{\mathrm{av}}=I_{\mathrm{rms}}^{2} R\)

31.73. In an \(L-R-C\) series circuit the current is given by \(i=I \cos \omega t .\) The voltage amplitudes for the resistor, inductor, and capacitor are \(V_{R}, V_{L},\) and \(V_{C}\) (a) Show that the instantaneous power into the resistor is \(p_{R}=V_{R} I \cos ^{2} \omega t=\frac{1}{2} V_{R} I(1+\cos 2 \omega t) .\) What does this expression give for the average power into the resistor? (b) Show that the instantaneous power into the inductor is \(p_{L}=\) \(-V_{L}\) Isin \(\omega t \cos \omega t=-\frac{1}{2} V_{L} I \sin 2 \omega t .\) What does this expression give for the average power into the inductor?(c) Show that the instantaneous power into the capacitor is \(p_{C}=V_{C}\) Isin\omegat cos \(\omega t=\) \(\frac{1}{2} V_{C}\) Isin \(2 \omega t .\) What does this expression give for the average power into the capacitor? (d) The instantancous power delivered by the source is shown in Section 31.4 to be \(p=V I \cos \omega t\) (cos \(\phi \cos \omega t-\) sin \(\phi \sin \omega t ) .\) Show that \(p_{R}+p_{L}+p_{C}\) equals \(p\) at each instant of time.

\(\operatorname{An} L-R-C\) series circuit consists of a \(2.50-\mu \mathrm{F}\) capacitor, a 5.00-mH inductor, and a \(75.0-\Omega\) resistor connected across an ac source of voltage amplitude 15.0 V having variable frequency. (a) Under what circumstances is the average power delivered to the circuit equal to \(\frac{1}{2} V_{\mathrm{rms}} I_{\mathrm{rma}} ?\) (b) Under the conditions of part (a), what is the average power delivered to each circuit element and what is the maximum current through the capacitor?

31\. 46. A circuit consists of a resistor and a capacitor in series with an ac source that supplies an rms voltage of 240 \(\mathrm{V}\) . At the frequency of the source the reactance of the capacitor is 50.0\(\Omega\) . The rms current in the circuit is 3.00 A. What is the average power supplied by the source?
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