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An LCR series circuit with \(100 \Omega\) resistance is connected to an \(A C\) source of \(200 V\) and angular frequency 300 radians/sec. When only the capacitance is removed The current lags behind the voltage by \(60^{\circ} .\) When only the inductance is removed the current leads the voltage \(60^{\circ}\). Calculate the current and the power dissipated in the LCR circuit.

Short Answer

Expert verified
The current is approximately 2.0 A, and the power dissipated is about 400 W.

Step by step solution

01

Understand the problem

The problem describes a series LCR circuit with a known resistance, connected to an AC source. We are given conditions for when the capacitance and the inductance are removed, indicating phase differences between the current and voltage.
02

Analyze the circuit with removed capacitance

When the capacitance is removed, the circuit consists of resistance (R) and inductance (L). The current lags the voltage by \(60^{\circ}\), which implies the phase angle \(\phi = 60^{\circ} = \frac{\pi}{3}\). We use the formula \(\phi = \tan^{-1}\left(\frac{\omega L}{R}\right)\) to calculate the inductive reactance \(\omega L\).
03

Calculate Inductive Reactance \(\omega L\)

Given \(\phi = \frac{\pi}{3}\), we rewrite the equation as \(\tan\left(\frac{\pi}{3}\right) = \sqrt{3} = \frac{\omega L}{R}\). Solving for \(\omega L\), we have \(\omega L = R\sqrt{3} = 100 \times \sqrt{3}\). Calculate \(\omega L\).
04

Analyze the circuit with removed inductance

When the inductance is removed, the current leads the voltage by \(60^{\circ}\), which implies the phase angle \(\phi = -60^{\circ} = -\frac{\pi}{3}\). We use the formula \(\phi = \tan^{-1}\left(\frac{-1}{\omega C R}\right)\) to calculate the capacitive reactance \(\omega C\).
05

Calculate Capacitive Reactance \(\omega C\)

Given \(\phi = -\frac{\pi}{3}\), the equation is \(\tan\left(-\frac{\pi}{3}\right) = -\sqrt{3} = \frac{-1}{\omega CR}\). This gives us \(\omega CR = \frac{1}{\sqrt{3}}\), solving for \(\omega C\) gives \(\omega C = \frac{1}{100\sqrt{3}}\). Calculate \(\omega C\).
06

Determine total impedance in the LCR circuit

In a series LCR circuit, the total impedance \(Z\) is given by \(Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}\). Substitute the found values for \(\omega L\), \(\omega C\), and \(R = 100 \Omega\) to calculate \(Z\).
07

Calculate current in the LCR circuit

Using Ohm's Law for AC circuits, the current \(I\) is given by \(I = \frac{V}{Z}\), where \(V = 200 V\). Substitute the calculated total impedance \(Z\) to find the current \(I\).
08

Calculate power dissipated in LCR circuit

In an AC circuit, the power dissipated (real power \(P\)) is given by \(P = I^2 R\), where \(I\) is the current through the resistor and \(R\) is the resistance. Calculate the power using the found current value.

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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 key concept when analyzing AC circuits, especially those containing inductors. It refers to the opposition that an inductor presents to the change in current. Unlike resistance, which is constant, inductive reactance depends on the frequency of the AC source and the inductance value.
Given the formula for inductive reactance, \(X_L = \omega L\), where \(\omega\) is the angular frequency and \(L\) is the inductance, you can see that the reactance increases with frequency. Therefore, as the frequency of the AC source increases, the inductor blocks more current, increasing its opposition to change.
  • Inductive reactance causes the current to lag behind the voltage in phase.
  • In the exercise, when only the inductance is in the circuit, the phase angle of 60° indicates significant inductive reactance.
  • Calculating \(\omega L\) involves using the relationship \(\tan(\phi) = \frac{\omega L}{R}\).
Understanding this concept helps in predicting how an inductor will behave in different scenarios within AC circuits.
Capacitive Reactance
Capacitive reactance is the measure of a capacitor's opposition to changes in voltage in an AC circuit. Like inductive reactance, capacitive reactance depends on the frequency of the applied voltage but is inversely related.
The formula for capacitive reactance is given by \(X_C = \frac{1}{\omega C}\), where \(\omega\) is the angular frequency and \(C\) is the capacitance. As the frequency increases, the capacitive reactance decreases, allowing more current to flow through the capacitor.
  • Capacitive reactance causes the current to lead the voltage in phase.
  • In the exercise, with only the capacitor, the current leads the voltage by 60°, showing a strong capacitive effect.
  • Use the relation \(\tan(\phi) = \frac{-1}{\omega C R}\) to calculate \(\omega C\).
Gaining a firm understanding of capacitive reactance is crucial for comprehending the dynamic behavior of AC circuits with capacitors.
Impedance Calculation
Impedance is a fundamental concept in AC circuits that combines resistance, inductive reactance, and capacitive reactance into a single measure of opposition to current. Unlike simple resistance in DC circuits, impedance is a complex quantity.
Impedance is calculated using the formula \(Z = \sqrt{R^2 + (X_L - X_C)^2}\), where \(R\) is the resistance, and \(X_L\) and \(X_C\) are the inductive and capacitive reactances, respectively.
  • The impedance determines how much current the AC source will drive through the circuit.
  • It is essential to consider both reactances' contributions, as they can cancel each other out partially or entirely, depending on their values.
In the given exercise, calculating impedance involved plugging in the known values of resistance, inductive reactance, and capacitive reactance.
Phase Angle
The phase angle in an AC circuit describes the phase difference between the voltage and the current. This angle is crucial because it affects how efficiently the circuit uses power.
Phase angle \(\phi\) can be calculated with the formula \(\phi = \tan^{-1}\left(\frac{X_L - X_C}{R}\right)\). The sign and magnitude of the phase angle indicate whether the circuit behaves more dominantly inductive or capacitive.
  • A positive phase angle means the circuit is mainly inductive, resulting in the current lagging behind the voltage.
  • A negative phase angle indicates dominant capacitive behavior, with current leading the voltage.
  • In balanced circuits, where inductive and capacitive reactances are equal, the phase angle becomes zero.
Understanding phase angles helps in predicting the circuit's performance and efficiency in power consumption.
AC Circuit Power
In an AC circuit, the power consumed is not only based on resistance as in DC circuits. It also depends on the phase difference between current and voltage.
Real power, also known as active power \(P\), is given by \(P = I^2 R\), where \(I\) is the current. However, it is worth noting that in AC circuits, power is usually calculated as \(P = VI\cos(\phi)\), where \(V\) and \(I\) are the voltage and current amplitudes, respectively, and \(\cos(\phi)\) is the power factor.
  • The power factor \(\cos(\phi)\) indicates the efficiency with which the circuit converts electric power into useful work.
  • A high power factor means most of the energy is being converted effectively into useful power.
  • In the exercise, power dissipation is calculated once the current is determined using the impedance value.
Grasping the concept of AC circuit power is essential for understanding how circuits are utilized in real-world electrical systems.

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Most popular questions from this chapter

A series circuit has an impedance of \(60.0 \Omega\) and a power factor of \(0.8\) at \(50.0 \mathrm{~Hz}\). The source voltage lags the current. (a) What circuit element, an inductor or a capacitor, should be placed in series with the circuit to raise its power factor? (b) What size element will raise the power factor to unity?

You have a special light bulb with a very delicate wire filament. The wire will break if the current in it ever exceeds \(2 \mathrm{~A}\), even for an instant. What is the largest root-mean-square current you can run through this bulb?

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

You have a \(200-\Omega\) resistor, a \(0.400-H\) 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.

In an \(L-R-C\) series circuit, the source has a voltage amplitude of \(120 \mathrm{~V}, R=80.0 \Omega\), and the reactance of the capacitor is \(480 \Omega .\) The voltage amplitude across the capacitor is \(360 \mathrm{~V}\). (a) What is the current amplitude in the circuit? (b) What is the impedance? (c) What two values can the reactance of the inductor have? (d) For which of the two values found in part (c) is the angular frequency less than the resonance angular frequency? Explain.

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