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Chapter 23: Problem 38

In a series \(\mathrm{RCL}\) circuit the generator is set to a frequency that is not the resonant frequency. This nonresonant frequency is such that the ratio of the inductive reactance to the capacitive reactance of the circuit is observed to be 5.36. The resonant frequency is 225 Hz. What is the frequency of the generator?

Short Answer

Expert verified
The generator frequency is approximately 328.5 Hz.

Step by step solution

01

Understanding Reactances in RLC Circuit

In an RLC circuit, the inductive reactance \(X_L\) is given by \(X_L = 2\pi fL\), and the capacitive reactance \(X_C\) is given by \(X_C = \frac{1}{2\pi fC}\). The resonant frequency occurs when these reactances are equal, i.e., \(X_L = X_C\).
02

Using Given Reactance Ratio

We know from the problem that the ratio of inductive reactance to capacitive reactance is 5.36. Thus, \(\frac{X_L}{X_C} = 5.36\). Substituting the expressions for \(X_L\) and \(X_C\), we have \(\frac{2\pi f L}{\frac{1}{2\pi f C}} = 5.36\).
03

Simplifying the Reactance Ratio Equation

Simplify the equation from Step 2: \(\frac{2\pi f L}{\frac{1}{2\pi f C}} = 5.36\) can be rearranged to \(4\pi^2 f^2 LC = 5.36\).
04

Identify Resonant Condition Equation

At resonance, the resonant frequency \(f_0\) is given by \(f_0 = \frac{1}{2\pi\sqrt{LC}}\). Here \(f_0 = 225 \,\text{Hz}\) is provided.
05

Express LC Using Resonant Frequency Equation

Using \(f_0 = 225 \,\text{Hz}\), express \(LC\) in terms of \(f_0\): \(LC = \frac{1}{(2\pi \times 225)^2}\). Substitute this into the modified reactance ratio equation.
06

Solving for Generator Frequency

Substitute \(LC\) from Step 5 into the equation from Step 3: \(4\pi^2 f^2 \left(\frac{1}{(2\pi \times 225)^2}\right) = 5.36\). Solve for \(f\): \(f^2 = 5.36 \times (225)^2\), \(f = \sqrt{5.36} \times 225\). Calculate \(f\).
07

Calculate Actual Frequency

Compute \(f\): Plug in values to get \(f = \sqrt{5.36} \times 225 = 1.46 \times 225 = 328.5\,\text{Hz}\) approximately.

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Key Concepts

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

Inductive Reactance
Inductive reactance is a measure of how much an inductor resists the flow of alternating current (AC) due to its inductance. It is an important concept when analyzing RLC circuits. The formula for inductive reactance is given by:
\[X_L = 2\pi f L\]
Here, \(X_L\) represents the inductive reactance, \(f\) is the frequency of the AC source, and \(L\) is the inductance of the inductor.
  • It depends directly on the frequency \(f\), meaning that as the frequency increases, so does the reactance.
  • The higher the inductive reactance, the more an inductor will oppose changes in current through it.
Understanding inductive reactance is crucial for solving circuits at different frequencies, particularly in determining how voltage and current behave in the presence of inductors.
Capacitive Reactance
Capacitive reactance is a measure of how much a capacitor resists the flow of alternating current (AC). It plays a crucial role in AC circuit analysis, especially in RLC circuits. The formula for capacitive reactance is defined as:
\[X_C = \frac{1}{2 \pi f C}\]
Where \(X_C\) is the capacitive reactance, \(f\) is the frequency of the AC signal, and \(C\) is the capacitance.
  • Capacitive reactance is inversely related to both frequency and capacitance.
  • As the frequency of the applied AC signal goes up, the capacitive reactance decreases.
  • This means capacitors offer less resistance to higher frequencies.
In RLC circuits, capacitive reactance is often balanced with inductive reactance to achieve resonance, where the total impedance is minimized.
Resonant Frequency
The resonant frequency in an RLC circuit is the frequency at which the circuit resonates, allowing AC signals to pass through with minimal impedance. This phenomenon occurs when the inductive and capacitive reactances are equal, effectively canceling each other out. The formula for resonant frequency \(f_0\) is:
\[f_0 = \frac{1}{2\pi\sqrt{LC}}\]
  • At resonant frequency, energy oscillates between the inductor and capacitor, leading to max current flow in the circuit.
  • It is a critical point for designing circuits like filters and tuners.
Choosing the correct resonant frequency is essential in applications like radio receivers, where specific frequencies need to be selected and amplified while filtering out others.

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Most popular questions from this chapter

A \(63.0-\mu \mathrm{F}\) capacitor is connected to a generator operating at a low frequency. The rms voltage of the generator is \(4.00 \mathrm{V}\) and is constant. A fuse in series with the capacitor has negligible resistance and will burn out when the rms current reaches \(15.0 \mathrm{A}\). As the generator frequency is increased, at what frequency will the fuse burn out?

A \(40.0-\mu \mathrm{F}\) capacitor is connected across a \(60.0-\mathrm{Hz}\) generator. An inductor is then connected in parallel with the capacitor. What is the value of the inductance if the rms currents in the inductor and capacitor are equal?

A capacitor is connected across the terminals of an ac generator that has a frequency of \(440 \mathrm{Hz}\) and supplies a voltage of \(24 \mathrm{V} .\) When a second capacitor is connected in parallel with the first one, the current from the generator increases by 0.18 A. Find the capacitance of the second capacitor.

Part \(a\) of the figure shows a heterodyne metal detector being used. As part \(b\) of the figure illustrates, this device utilizes two capacitor/ inductor oscillator circuits, A and B. Each produces its own resonant frequency, \(f_{0 \mathrm{A}}=1 /\left[2 \pi\left(L_{\mathrm{A}} C\right)^{1 / 2}\right]\) and \(f_{0 \mathrm{B}}=1 /\left[2 \pi\left(L_{\mathrm{B}} C\right)^{1 / 2}\right]\). Any difference between these frequencies is detected through earphones as a beat frequency \(\mid f_{0 \mathrm{B}}-\) \(f_{0 A} \mid .\) In the absence of any nearby metal object, the inductances \(L_{\mathrm{A}}\) and \(L_{\mathrm{B}}\) are identical. When inductor \(\mathrm{B}\) (the search coil) comes near a piece of metal, the inductance \(L_{\mathrm{B}}\) increases, the corresponding oscillator frequency \(f_{\mathrm{oB}}\) decreases, and a beat frequency is heard. Suppose that initially each inductor is adjusted so that \(L_{\mathrm{B}}=L_{\mathrm{A}},\) and each oscillator has a resonant frequency of \(855.5 \mathrm{kHz}\). Assuming that the inductance of search coil \(\mathrm{B}\) increases by \(1.000 \%\) due to a nearby piece of metal, determine the beat frequency heard through the earphones.

An ac generator has a frequency of \(7.5 \mathrm{kHz}\) and a voltage of \(39 \mathrm{V}\). When an inductor is connected between the terminals of this generator, the current in the inductor is \(42 \mathrm{mA}\). What is the inductance of the inductor? \(?\)
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