/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 A series \(\mathrm{RCL}\) circui... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 23: Problem 18

A series \(\mathrm{RCL}\) circuit contains a \(47.0-\Omega\) resistor, a \(2.00-\mu \mathrm{F}\) capacitor, and a \(4.00-\mathrm{mH}\) inductor. When the frequency is \(2550 \mathrm{Hz},\) what is the power factor of the circuit?

Short Answer

Expert verified
The power factor of the circuit is approximately 0.832.

Step by step solution

01

Calculate the Reactance of the Inductor

The reactance of the inductor is given by the formula \( X_L = 2 \pi f L \), where \( f \) is the frequency and \( L \) is the inductance. Plugging in the values: \( f = 2550 \, \text{Hz} \) and \( L = 4.00 \, \text{mH} = 4.00 \times 10^{-3} \, \text{H} \), we find:\[ X_L = 2 \pi \times 2550 \times 4.00 \times 10^{-3} \approx 64.06 \, \Omega. \]
02

Calculate the Reactance of the Capacitor

The reactance of the capacitor is given by the formula \( X_C = \frac{1}{2 \pi f C} \), where \( C \) is the capacitance. Substituting the given values: \( C = 2.00 \, \mu\text{F} = 2.00 \times 10^{-6} \, \text{F} \), we have:\[ X_C = \frac{1}{2 \pi \times 2550 \times 2.00 \times 10^{-6}} \approx 31.22 \, \Omega. \]
03

Determine the Impedance of the Circuit

The total impedance \( Z \) in an RCL circuit is calculated using the formula: \[ Z = \sqrt{R^2 + (X_L - X_C)^2}. \] Using \( R = 47.0 \, \Omega \), \( X_L = 64.06 \, \Omega \), and \( X_C = 31.22 \, \Omega \), we compute:\[ Z = \sqrt{47.0^2 + (64.06 - 31.22)^2} \approx 56.51 \, \Omega. \]
04

Calculate the Power Factor

The power factor \( \cos \phi \) is given by the ratio \( \frac{R}{Z} \). We already know \( R = 47.0 \, \Omega \) and \( Z = 56.51 \, \Omega \) from previous steps, so:\[ \cos \phi = \frac{47.0}{56.51} \approx 0.832. \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Reactance of Inductor
In an RCL circuit, the reactance of an inductor is an important concept that helps determine how an inductor will resist changes in current. The formula to calculate the reactance of the inductor is:
\[X_L = 2 \pi f L\]where:
  • \(X_L\) is the inductive reactance in ohms (\(\Omega\)),
  • \(f\) is the frequency in hertz (Hz),
  • \(L\) is the inductance in henrys (H).
The reactance, \(X_L\), increases with higher frequency and higher inductance. For the given exercise, with a frequency of \(2550\) Hz and an inductance of \(4.00\) mH, the calculated reactance is \(64.06 \, \Omega\).
This means the inductor opposes the alternating current with this amount of resistance at the specified frequency.
Reactance of Capacitor
The reactance of a capacitor in an RCL circuit is another critical factor that shows how the capacitor will impede the flow of alternating current. The formula for determining the reactance of the capacitor is:
\[X_C = \frac{1}{2 \pi f C}\]where:
  • \(X_C\) is the capacitive reactance in ohms (\(\Omega\)),
  • \(f\) is the frequency in hertz (Hz),
  • \(C\) is the capacitance in farads (F).
Capacitive reactance decreases with an increase in frequency. Therefore, capacitors are quite effective in filtering high-frequency signals. With a capacitance of \(2.00 \, \mu\text{F}\) and a frequency of \(2550\) Hz in the problem, we find the capacitive reactance to be approximately \(31.22 \, \Omega\).
This indicates the capacitor will reduce the AC flow by this amount.
Power Factor
Power factor in an RCL circuit helps us understand the efficiency of power usage. It represents the cosine of the phase angle \(\phi\) between the voltage and current. The formula for the power factor is:
\[\cos \phi = \frac{R}{Z}\]where:
  • \(\cos \phi\) is the power factor,
  • \(R\) is the resistance in ohms,
  • \(Z\) is the total impedance of the circuit in ohms.
A power factor of 1 indicates that all the power is being effectively used, whereas values less than 1 point towards inefficiency.
In the given RCL circuit, with a resistance \(R\) of \(47.0 \, \Omega\) and impedance \(Z\) of \(56.51 \, \Omega\), the power factor is calculated to be approximately \(0.832\).
Thus, this value indicates a fairly efficient circuit where about 83.2% of the power is being used effectively.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

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 A. As the generator frequency is increased, at what frequency will the fuse burn out?

A series \(\mathrm{RCL}\) circuit contains only a capacitor \((C=6.60 \mu \mathrm{F})\) an inductor \((L=7.20 \mathrm{mH}),\) and a generator (peak voltage \(=32.0 \mathrm{V}\), frequency \(=1.50 \times 10^{3} \mathrm{Hz}\) ). When \(t=0 \mathrm{s}\), the instantaneous value of the voltage is zero, and it rises to a maximum one-quarter of a period later. (a) Find the instantaneous value of the voltage across the capacitor/ inductor combination when \(t=1.20 \times 10^{-4} \mathrm{s} .\) (b) What is the instantaneous value of the current when \(t=1.20 \times 10^{-4} \mathrm{s} ?\) (Hint: The instantaneous values of the voltage and current are, respectively, the vertical components of the voltage and current phasors.)

In the absence of a nearby metal object, the two inductances \(\left(L_{\mathrm{A}}\right.\) and \(\left.L_{\mathrm{B}}\right)\) in a heterodyne metal detector are the same, and the resonant frequencies of the two oscillator circuits have the same value of \(630.0 \mathrm{kHz} .\) When the search coil (inductor \(\mathrm{B}\) ) is brought near a buried metal object, a beat frequency of \(7.30 \mathrm{kHz}\) is heard. By what percentage does the buried object increase the inductance of the search coil?

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, \(f_{0 A}=1 /\left[2 \pi\left(L_{A} C\right)^{1 / 2}\right]\) and \(f_{0 B}=1 /\left[2 \pi\left(L_{B} C\right)^{1 / 2}\right] .\) Any difference between these frequencies is detected through earphones as a beat frequency \(\left|f_{0 B}-f_{0 A}\right| \cdot\) 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 \(f_{0 \mathrm{B}}\) 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 \(8.2-\mathrm{mH}\) inductor is connected to an ac generator \((10.0 \mathrm{V} \mathrm{rms},\), \(620 \mathrm{Hz}\) ). Determine the peak value of the current supplied by the generator.
See all solutions

Recommended explanations on Physics Textbooks

Kinematics Physics

Read Explanation

Waves Physics

Read Explanation

Medical Physics

Read Explanation

Work Energy and Power

Read Explanation

Modern Physics

Read Explanation

Fundamentals of Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.