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Chapter 23: Problem 39

Suppose that you have a number of capacitors. Each is identical to the capacitor that is already in a series \(\mathrm{RCL}\) circuit. How many of these additional capacitors must be inserted in series in the circuit so the resonant frequency triples?

Short Answer

Expert verified
Add 8 additional capacitors in series.

Step by step solution

01

Understanding the Resonant Frequency Equation

The resonant frequency of an RCL circuit is given by the formula \( f_0 = \frac{1}{2\pi \sqrt{LC}} \), where \( L \) is the inductance and \( C \) is the capacitance. We need to triple this frequency.
02

Expressing the New Frequency

To triple the resonant frequency \( f_0 \), the new frequency \( f_1 \) can be expressed as \( f_1 = 3f_0 \). Substituting this into the resonant frequency formula gives \( 3f_0 = \frac{1}{2\pi \sqrt{L(C/n)}} \), where \( n \) is the new number of capacitors in series.
03

Modify the Capacitance in Series

In a series circuit, the total capacitance \( C_t \) is \( \frac{C}{n} \) for \( n \) identical capacitors. The new capacitance should make the expression for the resonant frequency satisfy \( 3f_0 = \frac{1}{2\pi \sqrt{L(C/n)}} \).
04

Setting Up the Equation for 'n'

We have \( 3f_0 = \frac{1}{2\pi \sqrt{L(C/n)}} \), and we also know \( f_0 = \frac{1}{2\pi \sqrt{LC}} \). By equating and solving \( \frac{1}{2\pi \sqrt{LC}} \times 3 = \frac{1}{2\pi \sqrt{L(C/n)}} \). Simplifying gives \( n = \frac{1}{9} \).
05

Solving for Additional Capacitors

Since \( n \) is the number for total capacitors needed in series, calculate the additional capacitors by subtracting 1 from the value derived for \( n \). Since \( n \) must equal 9 to get a value of \( 1/9 \) in the base equation and still maintain integer units, this means you need to add 8 additional capacitors.

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Key Concepts

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

Capacitance in Series
When capacitors are placed in series, their overall effect on the circuit is that the total capacitance decreases. This happens because the electric charge each capacitor can store is spread across more components. As a rule of thumb, the total capacitance of capacitors in series is given by the reciprocal of the sum of the reciprocals of the individual capacitances. For identical capacitors, this simplifies to dividing the capacitance of one capacitor by the number of all capacitors. Let's use simpler math. If you connect three identical capacitors of capacitance $C$, the total capacitance, $C_t$, becomes $C/3$. This is because the total capacitance $C_t$ for $n$ identical capacitors in series is $C/n$.
  • Total capacitance decreases with more capacitors in series.
  • Overall voltage rating increases but energy stored decreases.
Understanding this concept is crucial when analyzing how the capacitive part of the RCL circuit impacts its overall behavior.
RCL Circuit Analysis
An RCL circuit involves Resistance (R), Capacitance (C), and Inductance (L). Such circuits are essential in electronics due to their reactive properties. One of the key features of an RCL circuit is its ability to resonate at a particular frequency, known as the resonant frequency.The resonant frequency, \(f_0\), is determined by the formula: \[f_0 = \frac{1}{2\pi \sqrt{LC}}\] where \(L\) represents inductance and \(C\) stands for capacitance. At this frequency, the circuit behaves effectively like a pure resistance circuit, minimizing reactance.
  • At resonance, the impedance is at its lowest, a significant factor for tuning circuits.
  • The relationship between inductance and capacitance is pivotal for defining the resonant frequency.
Understanding these dynamics allows one to correctly predict how adding or modifying components like capacitors will impact overall circuit performance.
Electronics Problem Solving
Solving electronics problems, like adjusting the resonant frequency of RCL circuits, often requires breaking down complex ideas into manageable steps. Many problems involve relationships between circuit components such as resistors, capacitors, and inductors鈥攁ll of which determine the circuit's behavior. For instance, if we want to triple the resonant frequency, as in the original exercise, we deduce the necessary change in components. From the equation relating resonant frequency to circuit parameters, adjusting the number of capacitors changes the frequency significantly.
  • Mathematical reasoning lets you bridge theoretical concepts with practical outcomes.
  • Problem-solving in electronics often requires iterative thinking鈥攖he process may involve multiple steps to arrive at a desired solution.
By mastering these methods and equations, students develop strong troubleshooting skills, ideal for tackling wide-ranging electronics challenges.

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Most popular questions from this chapter

A tank circuit in a radio transmitter is a series RCL circuit connected to an antenna. The antenna broadcasts radio signals at the resonant frequency of the tank circuit. Suppose that a certain tank circuit in a shortwave radio transmitter has a fixed capacitance of \(1.8 \times 10^{-11} \mathrm{F}\) and a variable inductance. If the antenna is intended to broadcast radio signals ranging in frequency from \(4.0 \mathrm{MHz}\) to \(9.0 \mathrm{MHz},\) find the (a) minimum and (b) maximum inductance of the tank circuit.

A series \(\mathrm{RCL}\) circuit contains a \(47.0-\Omega\) resistor, a \(2.00-\mu \mathrm{F}\) capacitor, and a \(4.00-\mathrm{mH}\) inductor. When the frequency is \(2550 \mathrm{Hz},\) what is the power factor of the circuit?

At what frequency (in Hz) are the reactances of a \(52-\mathrm{mH}\) inductor and a \(76-\mu \mathrm{F}\) capacitor equal?

A capacitor is connected to an ac generator that has a frequency of 3.4 \(\mathrm{kHz}\) and produces a voltage of \(2.0 \mathrm{V}\). The current in the capacitor is \(35 \mathrm{mA} .\) When the same capacitor is connected to a second ac generator that has a frequency of \(5.0 \mathrm{kHz}\), the current in the capacitor is \(85 \mathrm{mA}\). What voltage does the second generator produce?

When only a resistor is connected across the terminals of an ac generator (112 V) that has a fixed frequency, there is a current of \(0.500 \mathrm{A}\) in the resistor. When only an inductor is connected across the terminals of this same generator, there is a current of \(0.400 \mathrm{A}\) in the inductor. When both the resistor and the inductor are connected in series between the terminals of this generator, what are (a) the impedance of the series combination and (b) the phase angle between the current and the voltage of the generator?
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