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31.60. An \(L-R-C\) series circuit has \(R=500 \Omega, L=2.00 \mathrm{H}\) \(C=0.500 \mu \mathrm{F},\) and \(V=100 \mathrm{V} .\) (a) For \(\omega=800 \mathrm{rad} / \mathrm{s}\) , calculate \(V_{R}, V_{L}, V_{C},\) and \(\phi .\) Using a single set of axes, graph \(v, v_{R}, v_{L},\) and \(v_{C}\) as functions of time. Include two cycles of \(v\) on your graph. (b) Repeat part \((a)\) for \(\omega=1000 \mathrm{rad} / \mathrm{s}\) . (c) Repeat part (a) for \(\omega=1250 \mathrm{rad} / \mathrm{s}\) .

Short Answer

Expert verified
Calculate reactances, impedance, and phase angle for each frequency; then find component voltages and graph over two cycles.

Step by step solution

01

Reactance of Components

Calculate the reactance of the inductance, which is given by the formula \( X_L = \omega L \). For part (a) with \( \omega = 800 \ \text{rad/s} \), \( L = 2.00 \ \text{H} \), \[ X_L = 800 \times 2.00 = 1600 \ \Omega.\]Similarly, calculate the reactance of the capacitance using \( X_C = \frac{1}{\omega C} \). For \( C = 0.500 \ \mu\text{F} = 0.500 \times 10^{-6} \ \text{F}\), \[ X_C = \frac{1}{800 \times 0.500 \times 10^{-6}} \approx 2500 \ \Omega.\]
02

Impedance Calculation

Calculate the total impedance \( Z \) of the circuit using the formula: \[ Z = \sqrt{R^2 + (X_L - X_C)^2}. \]Substitute the given values:\[ Z = \sqrt{500^2 + (1600 - 2500)^2} = \sqrt{500^2 + (-900)^2} = \sqrt{250000 + 810000} = \sqrt{1060000} \approx 1020 \ \Omega.\]
03

Phase Angle

The phase angle \( \phi \) is calculated using \( \phi = \tan^{-1}\left(\frac{X_L - X_C}{R}\right) \).\[ \phi = \tan^{-1}\left(\frac{1600 - 2500}{500}\right) = \tan^{-1}(-1.8) \approx -60.9^\circ.\]
04

Voltage Across Each Component

The voltage across the resistor \( V_R \) is found using \( V_R = IZ \cos\phi \) where \( I = \frac{V}{Z} \).\[ I = \frac{100}{1020} \approx 0.098 \ \text{A} \]\[ V_R = 0.098 \times 500 = 49 \ \text{V}. \]Similar calculations apply for \( V_L \) and \( V_C \):\[ V_L = IX_L = 0.098 \times 1600 = 156.8 \ \text{V}. \]\[ V_C = IX_C = 0.098 \times 2500 = 245 \ \text{V}. \]
05

Repeat for Other Angular Frequencies

Repeat Steps 1 to 4 for \( \omega = 1000 \ \text{rad/s} \) and \( \omega = 1250 \ \text{rad/s} \) using the same formulas:For \( \omega = 1000 \ \text{rad/s} \):* \( X_L = 2000 \ \Omega \), \( X_C = 2000 \ \Omega \), \( Z = 500 \ \Omega \), \( \phi = 0^\circ \).* \( V_R \approx 100 \ \text{V} \), \( V_L \approx 0 \ \text{V} \), \( V_C \approx 0 \ \text{V} \).For \( \omega = 1250 \ \text{rad/s} \):* \( X_L = 2500 \ \Omega \), \( X_C = 1600 \ \Omega \), \( Z \approx 1020 \ \Omega \), \( \phi = +60.9^\circ \).* \( V_R \approx 49 \ \text{V} \), \( V_L \approx 245 \ \text{V} \), \( V_C \approx 156.8 \ \text{V} \).
06

Graphing Voltage Over Time

Graph the voltages \( v, v_R, v_L, \) and \( v_C \) over two cycles considering \( v(t) = V_m\sin(\omega t) \) and phase shifts where required. For example, for the given \( \omega \) values, mark the time axis for the required period using \( T = \frac{2\pi}{\omega} \). Ensure to plot the overall supplied voltage and the individual component voltages according to their calculated magnitudes and phase angles.

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Key Concepts

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

Reactance in L-R-C Circuits
In an L-R-C circuit, reactance is a measure of how much the inductors and capacitors resist the flow of alternating current (AC). There are two types: inductive reactance and capacitive reactance. Inductive reactance, represented as \(X_L\), is governed by the formula \(X_L = \omega L\), where \(\omega\) is the angular frequency and \(L\) is the inductance. A higher angular frequency results in greater inductive reactance. Capacitive reactance, represented as \(X_C\), is given by \(X_C = \frac{1}{\omega C}\), where \(C\) is the capacitance. Unlike inductive reactance, capacitive reactance decreases with increasing angular frequency. Together, \(X_L\) and \(X_C\) determine how the circuit responds to changes in frequency.
Impedance in L-R-C Circuits
Impedance is the total opposition that a circuit presents to the flow of alternating current. It is a complex quantity, denoted by \(Z\), which combines resistance and reactance and can be calculated using the formula: \[Z = \sqrt{R^2 + (X_L - X_C)^2}\]. Here, \(R\) is the resistance, \(X_L\) is the inductive reactance, and \(X_C\) is the capacitive reactance. Impedance is crucial because it affects the current flow and the phase relationship between voltage and current in the circuit.
  • At resonance, where \(X_L = X_C\), the impedance is purely resistive and least, being equal to \(R\).
  • If \(X_L > X_C\), the circuit behaves more like an inductor.
  • If \(X_C > X_L\), the circuit behaves more like a capacitor.
Understanding impedance helps in designing circuits according to desired functionalities.
Phase Angle in L-R-C Circuits
The phase angle \(\phi\) in L-R-C circuits represents the difference in phase between the total voltage across and the total current through the circuit. It can be calculated using \(\phi = \tan^{-1}\left(\frac{X_L - X_C}{R}\right)\). This angle indicates how much the current is "leading" or "lagging" the voltage.
  • If \(\phi\) is positive, the circuit is inductive, meaning the current lags the voltage.
  • If \(\phi\) is negative, the circuit is capacitive, and the current leads the voltage.
  • If \(\phi = 0\), the circuit is at resonance, indicating that the current and voltage are in phase.
This phase relationship is important for applications like signal processing and power regulation.
Angular Frequency in Electrical Circuits
Angular frequency, denoted by \(\omega\), is a key parameter in alternating current circuits, representing how fast the alternating current is oscillating. It is related to the frequency \(f\) through the formula \(\omega = 2\pi f\). Angular frequency is pivotal because it affects the reactance of the circuit components:
  • Higher \(\omega\) increases \(X_L\) as \(X_L = \omega L\).
  • Higher \(\omega\) decreases \(X_C\) as \(X_C = \frac{1}{\omega C}\).
The choice of \(\omega\) in circuits can influence the voltage distribution across the inductor, resistor, and capacitor. Understanding angular frequency helps in tuning circuits to desired operating conditions, such as achieving resonance or minimizing power loss.

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Most popular questions from this chapter

31.9. (a) What is the reactance of a \(3.00-\mathrm{H}\) inductor at a frequency of 80.0 \(\mathrm{Hz}\) (b) What is the inductance of an inductor whose reactance is 120\(\Omega\) at 80.0 \(\mathrm{Hz}\) (c) What is the reactance of a \(4.00-\mu \mathrm{F}\) capacitor at a frequency of 80.0 \(\mathrm{Hz}\) (d) What is the capacitance of a capacitor whose reactance is 120\(\Omega\) at 80.0 \(\mathrm{Hz}\) ?

31\. A. Analyzing an \(L . R-C\) 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?

31.25. The power of a certain CD player operating at 120 \(\mathrm{V}\) rms is 20.0 \(\mathrm{W}\) . Assuming that the CD player behaves like a pure resistance, find (a) the maximum instantancous power; (b) the rms current; (c) the resistance of this player.

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.

31.62. A series circuit consists of a \(1.50-\mathrm{mH}\) inductor, a \(125-\Omega\) resistor, and a \(25.0-\mathrm{nF}\) capacitor connected across an ac source having an rms voltage of 35.0 \(\mathrm{V}\) and variable frequency. (a) At what angular frequency will the current amplitude be equal to \(\frac{1}{3}\) of its maximum possible value? (b) At the frequency in part (a) what are the current amplitude and the voltage amplitude across each of the circuit elements (including the ac source)?

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