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Chapter 31: Problem 27

Analyzing an \(L \cdot R-C\) Circuit. You have a \(200-\Omega\) resistor, a \(0.400-\mathrm{H}\) inductor, a \(5.00-\mu \mathrm{F}\) capacitor, and a variable-frequency ac source with an amplitude of 3.00 \(\mathrm{V} .\) You connect all four elements together to form a series circuit. (a) At what frequency will the current in the circuit be greatest? What will be the current amplitude at this frequency? (b) What will be the current amplitude at an angular frequency of 400 \(\mathrm{rad} / \mathrm{s} ?\) At this frequency, will the source voltage lead or lag the current?

Short Answer

Expert verified
(a) Maximum current at 356 Hz with amplitude 0.015 A. (b) At 400 rad/s, current amplitude is 0.011 A; voltage leads current.

Step by step solution

01

Understanding Resonance

In a series LCR circuit, the current is greatest when the impedance is minimum. This occurs at the resonant frequency. The resonant frequency \( f_0 \) is calculated by the formula \( f_0 = \frac{1}{2\pi\sqrt{LC}} \), where \( L \) is the inductance and \( C \) is the capacitance.
02

Calculating Resonant Frequency

Given \( L = 0.400 \) H and \( C = 5.00 \times 10^{-6} \) F, substitute these values into the formula: \[ f_0 = \frac{1}{2\pi\sqrt{0.400 \times 5.00 \times 10^{-6}}} \]Calculate \( f_0 \) to find the resonant frequency for maximum current.
03

Calculating Current Amplitude at Resonance

At resonance, the impedance is just the resistance \( R \), which is \( 200 \) \( \Omega \). The current amplitude \( I_0 \) at resonance is calculated by \( I_0 = \frac{V}{R} \), where \( V = 3.00 \) V is the amplitude of the source voltage. Calculate \( I_0 \).
04

Analyzing the Circuit at 400 rad/s

The angular frequency \( \omega = 400 \) rad/s. Calculate the impedance \( Z \) using the formula: \[ Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2} \]Substitute \( R = 200 \) \( \Omega \), \( \omega = 400 \) rad/s, \( L = 0.400 \) H, and \( C = 5.00 \times 10^{-6} \) F into the formula to find \( Z \).
05

Calculating Current Amplitude at 400 rad/s

The current amplitude \( I \) is given by \( I = \frac{V}{Z} \). Using \( V = 3.00 \) V and the \( Z \) from Step 4, calculate the new current amplitude at \( \omega = 400 \) rad/s.
06

Determine Phase Difference at 400 rad/s

The phase angle \( \phi \) is determined by \( \tan \phi = \frac{\omega L - \frac{1}{\omega C}}{R} \). If \( \phi > 0 \), the voltage leads the current; if \( \phi < 0 \), the voltage lags the current. Calculate \( \phi \) using the given and calculated values.

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Key Concepts

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

Resonant Frequency
In an RLC circuit, the resonant frequency is crucial because it represents the frequency at which the circuit's impedance is minimized and current is maximized. Essentially, it's the frequency at which the reactive effects of the inductor and capacitor cancel each other out. This balance allows the circuit to operate more efficiently, with minimal energy loss.
To determine the resonant frequency, we use the formula:
\[ f_0 = \frac{1}{2\pi\sqrt{LC}} \]Where:
  • \( L \) is the inductance (in henries, H)
  • \( C \) is the capacitance (in farads, F)
This calculation helps identify the precise frequency for maximum current flow in the circuit, highlighting the importance of both inductance and capacitance in tuning the circuit's performance.
Impedance Calculation
Impedance in an RLC circuit is the total opposition to the flow of alternating current, combining both resistance and reactance into a single value. It dictates how much current flows for a given voltage.
The formula to calculate impedance \( Z \) is:
\[ Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2} \]Where:
  • \( R \) is the resistance in ohms (\( \Omega \))
  • \( \omega \) is the angular frequency in rad/s
  • \( L \) is the inductance in henries
  • \( C \) is the capacitance in farads
Impedance combines the direct resistance with the contributions from the inductor and capacitor, depending on the frequency of the source. Calculating impedance is key to understanding how a circuit will behave at any particular frequency, especially one other than the resonant frequency.
Phase Difference
Phase difference in an RLC circuit is the angular difference between the source voltage and the circuit current. This concept is crucial for understanding power delivery and efficiency in AC circuits.
The phase angle \( \phi \) can be determined using:
\[ \tan \phi = \frac{\omega L - \frac{1}{\omega C}}{R} \]
  • A positive \( \phi \) indicates that the source voltage leads the current.
  • A negative \( \phi \) shows that the voltage lags behind the current.
Phase difference can have a significant impact on the performance and efficiency of the circuit, especially under conditions that aren't at resonance. Understanding the phase relationship helps in designing circuits that are efficient and effective for their intended use.
Current Amplitude
Current amplitude refers to the maximum current that flows through an RLC circuit for a given voltage source. It varies with the frequency of the AC source due to the circuit's impedance changes.
To find the current amplitude \( I \) at resonance or any specific frequency, use:
\[ I = \frac{V}{Z} \]Where:
  • \( V \) is the amplitude of the supply voltage
  • \( Z \) is the impedance of the whole circuit
At resonant frequency, because impedance is mostly comprised of resistance, the amplitude is largest. As frequency changes away from resonance, reactance plays a bigger role, affecting the current amplitude. Thus, calculating current amplitude at different frequencies helps in circuit design and understanding how the circuit will perform across various situations.
Angular Frequency
Angular frequency \( \omega \) is an essential parameter in the analysis of RLC circuits, representing how quickly the sinusoidal source voltage oscillates. It is related to the frequency by the equation:
\[ \omega = 2\pi f \]Where \( f \) is the frequency in hertz (Hz).
  • It measures in radians per second (rad/s).
  • This metric helps in determining the reactance of inductors and capacitors within the circuit.
Understanding angular frequency is critical, as it allows one to explore how variations in source frequency influence the circuit's impedance and operating point. This understanding is vital to manipulating the behavior of RLC circuits effectively.

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Most popular questions from this chapter

You have a \(200-\Omega\) resistor, a \(0.400-\mathrm{H}\) inductor, and a \(6.00-\mu \mathrm{F}\) capacitor. Suppose you take the resistor and inductor and make a series circuit with a voltage source that has voltage amplitude 30.0 \(\mathrm{V}\) and an angular frequency of 250 \(\mathrm{rad} / \mathrm{s}\) . (a) What is the impedance of the circuit? (b) What is the current amplitude? (c) What are the voltage amplitudes across the resistor and across the inductor? (d) What is the phase angle \(\phi\) of the source voltage with respect to the current? Does the source voltage lag or lead the current? (e) Construct the phasor diagram.

An \(L-R-C\) series circuit is connected to an ac source of constant voltage amplitude \(V\) and variable angular frequency \(\omega\) (a) Show that the current amplitude, as a function of \(\omega,\) is $$I=\frac{V}{\sqrt{R^{2}+(\omega L-1 / \omega C)^{2}}}$$ (b) Show that the average power dissipated in the resistor is $$P=\frac{V^{2} R / 2}{R^{2}+(\omega L-1 / \omega C)^{2}}$$ (c) Show that \(I\) and \(P\) are both maximum when \(\omega=1 / \sqrt{L C},\) the resonance frequency of the circuit. (d) Graph \(P\) as a function of \(\omega\) for \(V=100 \mathrm{V}, R=200 \Omega, L=2.0 \mathrm{H},\) and \(C=0.50 \mu \mathrm{F}\) . Compare to the light purple curve in Fig. \(31.19 .\) Discuss the behavior of \(I\) and \(P\) in the limits \(\omega=0\) and \(\omega \rightarrow \infty\) .

A sinusoidal current \(i=I \cos \omega t\) has an rms value \(I_{\mathrm{rms}}=\)2.10 A. (a) What is the current amplitude? (b) The current is passed through a full-wave rectifier circuit. What is the rectified average current? (c) Which is larger: \(I_{\text { ms or }} I_{\text { rav }} ?\) Explain, using graphs of \(i^{2}\) and of the rectified current.

An \(L-R-C\) series circuit draws 220 \(\mathrm{W}\) from a \(120-\mathrm{V}\) (rms), 50.0 -Hz ac line. The power factor is 0.560 , and the source voltage leads the current. (a) What is the net resistance \(R\) of the circuit? (b) Find the capacitance of the series capacistor that will result in a power factor of unity when it is added to the original circuit. (c) What power will then be drawn from the supply line?

A Radio Inductor. You want the current amplitude through a \(0.450-\mathrm{mH}\) inductor (part of the circuitry for a radio receiver) to be 2.60 \(\mathrm{mA}\) when a sinusoidal voltage with amplitude 12.0 \(\mathrm{V}\) is applied across the inductor. What frequency is required?
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