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(a) Show that for an \(L-R-C\) series circuit the power factor is equal to R/Z. (b) An \(L-R-C\) series circuit has phase angle -31.5\(^\circ\). The voltage amplitude of the source is 90.0 V. What is the voltage amplitude across the resistor?

Short Answer

Expert verified
(a) PF = R/Z. (b) Voltage across resistor is 77.15 V.

Step by step solution

01

Define Power Factor in L-R-C Circuit

In an L-R-C series circuit, the power factor (PF) is defined as the cosine of the phase angle \( \phi \) between the total voltage and the total current. Mathematically, \( \text{PF} = \cos(\phi) \).
02

Express Phase Angle with Circuit Parameters

The phase angle \( \phi \) in an L-R-C circuit can be expressed as \( \phi = \tan^{-1}\left( \frac{X_L - X_C}{R} \right) \), where \(X_L = \omega L\) is the inductive reactance, \(X_C = \frac{1}{\omega C}\) is the capacitive reactance, and \(R\) is the resistance.
03

Calculate Impedance Z

The impedance \( Z \) of an L-R-C series circuit is given by \( Z = \sqrt{R^2 + (X_L - X_C)^2} \). This formula represents the total opposition to the flow of current.
04

Relate R/Z to Cosine

To show \( \text{PF} = \frac{R}{Z} \), use the relationship \( \cos(\phi) = \frac{R}{Z} \). Since \(\text{PF} = \cos(\phi)\), we have \( \text{PF} = \frac{R}{\sqrt{R^2 + (X_L - X_C)^2}} \), confirming that the power factor equals \( \frac{R}{Z} \).
05

Calculate Voltage Amplitude Across Resistor

Given that the phase angle \( \phi \) is \(-31.5^\circ\) and the voltage amplitude of the source is 90.0 V, the voltage amplitude \( V_R \) across the resistor can be found using \( V_R = V \times \cos(\phi) \), where \(V\) is the total voltage of 90.0 V.
06

Compute Numerical Value for Voltage Across Resistor

Calculate \( \cos(-31.5^\circ) \) to obtain approximately 0.8572. Thus, \( V_R = 90.0 \times 0.8572 = 77.148 \) V. So, the voltage amplitude across the resistor is approximately 77.15 V.

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Key Concepts

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

Power Factor
In an L-R-C series circuit, the power factor is a crucial element to understand. It measures how effectively electrical power is being converted into useful work. The power factor is defined as the cosine of the phase angle \( \phi \) between the current and voltage. Hence, its formula is \( \text{PF} = \cos(\phi) \).
An important relationship in these circuits is that the power factor can also be expressed as \( \frac{R}{Z} \), where \( R \) stands for the resistance and \( Z \) stands for the impedance of the circuit. This relationship shows that the power factor depends on how much of the impedance is due to resistance.
  • High power factor: Closer to 1, indicating more power is being effectively utilized.
  • Low power factor: Means less efficiency, often due to more reactive components like inductors and capacitors.
The significance of a high power factor is efficiency; more of the electricity being consumed is turned into useful work rather than wasted. Correctly understanding and improving the power factor can lead to significant savings in energy bills and increased capacity of electrical systems.
Phase Angle
The phase angle \( \phi \) in an L-R-C series circuit tells us about the difference in phase between the current and the voltage. It's an indicator of the relationship between these two quantities, particularly whether they are in sync or not.
This angle can be calculated using the formula \( \phi = \tan^{-1}\left( \frac{X_L - X_C}{R} \right) \). Here, \( X_L = \omega L \) is the inductive reactance and \( X_C = \frac{1}{\omega C} \) is the capacitive reactance, with \( R \) as resistance.
  • A positive phase angle indicates that the circuit is more inductive.
  • A negative phase angle indicates more capacitive behavior.
For instance, in the exercise, the given phase angle is \(-31.5^\circ\). This negative phase angle indicates that the current leads voltage, characteristic of circuits with predominant capacitive reactance. Knowing the phase angle helps predict circuit behavior and optimize conditions for desired electrical performance.
Impedance
Impedance \( Z \) in a circuit combines resistance with reactance, representing the entire opposition to the flow of alternating current. The impedance of an L-R-C series circuit can be calculated with the formula:\[Z = \sqrt{R^2 + (X_L - X_C)^2}\]
Reactance is split into inductive reactance \( X_L \) and capacitive reactance \( X_C \) in this type of circuit. The difference \( X_L - X_C \) shows if the circuit is more inductively or capacitively reactive.
  • Impedance is lower when the circuit is at resonant frequency \( \omega_0 \), where \( X_L = X_C \).
  • Impedance increases as either inductive or capacitive reactance becomes dominant.
Understanding impedance is essential since it impacts how much current flows for a given voltage in the circuit. Moreover, it plays a vital role in determining the power factor, as seen in the previous section. Recognizing both resistance and reactance contributions will help in building more efficient and predictive electrical systems.

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Most popular questions from this chapter

When a solenoid is connected to a 48.0-V dc battery that has negligible internal resistance, the current in the solenoid is 5.50 A. When this solenoid is connected to an ac source that has voltage amplitude 48.0 V and angular frequency 20.0 rad/s, the current in the solenoid is 3.60 A. What is the inductance of this solenoid?

A series circuit has an impedance of 60.0 \(\Omega\) and a power factor of 0.720 at 50.0 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?

A resistor, an inductor, and a capacitor are connected in parallel to an ac source with voltage amplitude \(V\) and angular frequency \(\omega\). Let the source voltage be given by \(v\) = V cos \(\omega\)t. (a) Show that each of the instantaneous voltages \(v_R\) , \(v_L\), and \(v_C\) at any instant is equal to \(v\) and that \(i\) = \(i_R\) + \(i_L\) + \(i_C\), where i is the current through the source and iR , iL, and iC are the currents through the resistor, inductor, and capacitor, respectively. (b) What are the phases of \(i_R\) , \(i_L\), and \(i_C\) with respect to v? Use current phasors to represent i, iR , iL, and iC. In a phasor diagram, show the phases of these four currents with respect to \(v\). (c) Use the phasor diagram of part (b) to show that the current amplitude I for the current i through the source is \(I =\sqrt{I_{R ^2} + (I_C - I_L)^2}\). (d) Show that the result of part (c) can be written as\( I = V/Z\), with \(1/Z = \sqrt{(1/R^2) + [\omega C - (1/\omega L)]^2}.\)

An \(L-R-C\) series circuit has R = 500 \(\Omega\), L = 2.00 H, \(C\) = 0.500 \(\mu\)F, and \(V\) = 100 V. (a) For \(\omega\) = 800 rad/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 rad/s. (c) Repeat part (a) for \(\omega = 1250\space rad/s\).

At a frequency \(\omega_1\) the reactance of a certain capacitor equals that of a certain inductor. (a) If the frequency is changed to \(\omega_2\) = \(2\omega_1\), what is the ratio of the reactance of the inductor to that of the capacitor? Which reactance is larger? (b) If the frequency is changed to \(\omega3\)= \(\omega_1\)/3 , what is the ratio of the reactance of the inductor to that of the capacitor? Which reactance is larger? (c) If the capacitor and inductor are placed in series with a resistor of resistance R to form an \(L-R-C\) series circuit, what will be the resonance angular frequency of the circuit?

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