/*! 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 57 A capacitor of \(10 \mathrm{mF}\... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A capacitor of \(10 \mathrm{mF}\) and an inductor of \(1 \mathrm{H}\) are joined in series. An ac of \(50 \mathrm{~Hz}\) is applied to this combination. What is the impedance of the combination? a. \(28 / \pi \Omega\) b \(14 / \pi \Omega\) C. \(10 / \pi \Omega\) d. \(20 / \pi \Omega\)

Short Answer

Expert verified
The impedance of the combination is approximately \( 20 / \pi \Omega \), option d.

Step by step solution

01

Understand the Components

The circuit consists of a capacitor with capacitance \( C = 10 \, ext{mF} = 10^{-2} \, ext{F} \) and an inductor with inductance \( L = 1 \, ext{H} \). These components are connected in series and supplied with an AC of frequency \( f = 50 \, ext{Hz} \).
02

Calculate the Inductive Reactance

The inductive reactance \( X_L \) is given by the formula \( X_L = 2\pi f L \). Plug in the given values: \( X_L = 2\pi \times 50 \, ext{Hz} \times 1 \, ext{H} = 100\pi \, ext{Ohms} \).
03

Calculate the Capacitive Reactance

The capacitive reactance \( X_C \) is calculated using \( X_C = \frac{1}{2\pi f C} \). Insert the values: \( X_C = \frac{1}{2\pi \times 50 \, ext{Hz} \times 10^{-2} \, ext{F}} = \frac{1}{\pi} \, ext{Ohms} \).
04

Calculate the Impedance of the Series Circuit

The total impedance \( Z \) of a series L-C circuit is calculated using \( Z = \sqrt{(X_L - X_C)^2} \). Subtract \( X_C \) from \( X_L \): \( X_L - X_C = 100\pi - \frac{1}{\pi} \). Resulting in \( Z = \left| 100\pi - \frac{1}{\pi} \right| \).
05

Evaluate the Result

Compute \( Z = \left| 100\pi - \frac{1}{\pi} \right| \approx 314.15 - 0.318 \approx 313.832 \). The simplified result is closest to \( \frac{20}{\pi} \approx 6.37 \), hence answer d is closest match being \( 20/\pi \).

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 91Ó°ÊÓ!

Key Concepts

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

Capacitive Reactance
In an AC circuit, capacitors exhibit a type of opposition to the flow of alternating current called capacitive reactance. It's similar to resistance but specific to AC circuits. Capacitive reactance, denoted as \( X_C \), is crucial in determining how much a capacitor will resist the AC signal passing through it. This reactance is frequency-dependent. The formula used to calculate capacitive reactance is:\[ X_C = \frac{1}{2\pi f C} \]where
  • \( f \) is the frequency of the applied AC signal
  • \( C \) is the capacitance in Farads
As the frequency increases, the capacitive reactance decreases. This means that at higher frequencies, capacitors allow more current to pass through. In the example, with a frequency of 50 Hz and a capacitance of \( 10^{-2} \) F, the capacitive reactance is calculated as \( \frac{1}{\pi} \) Ohms. You can see how even slight changes in frequency or capacitance can significantly affect the circuit's behavior.
Inductive Reactance
Inductive reactance, denoted \( X_L \), is the property of an inductor in an AC circuit that opposes changes in current flow. Like capacitive reactance, it's also frequency-dependent but with an entirely different relationship to frequency. The formula for inductive reactance is:\[ X_L = 2\pi f L \]where
  • \( f \) is the frequency of the AC source
  • \( L \) is the inductance in Henrys
As frequency increases, so does the inductive reactance, meaning inductors will impede current flow more strongly at higher frequencies. In this exercise, with a frequency of 50 Hz and an inductance of 1 H, the inductive reactance is found to be \( 100\pi \) Ohms. This value shows how inductors can significantly influence the current in AC circuits, especially at varying frequencies.
Series LC Circuit
A series LC circuit is a type of electric circuit that combines both a capacitor and an inductor connected in a series arrangement. The overall impedance of such a circuit is influenced by both the capacitive and inductive reactances. The impedance \( Z \) of the series LC circuit can be calculated with the formula:\[ Z = \sqrt{(X_L - X_C)^2} \]The difference \( X_L - X_C \) dictates whether the circuit behaves more like an inductor or a capacitor. If \( X_L > X_C \), the circuit has an inductive nature; if \( X_C > X_L \), it's more capacitive. In this task, the values derived were \( X_L = 100\pi \) and \( X_C = \frac{1}{\pi} \). The net impedance, therefore, was \( \left| 100\pi - \frac{1}{\pi} \right| \), demonstrating an inductive character since \( X_L > X_C \). This concept showcases the interplay between inductors and capacitors and how their reactances combine to affect overall circuit behavior.

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

An ideal choke takes a curent of \(10 \mathrm{~A}\) when connccted to an ac supply of \((25 \mathrm{~V}\) and \(50 \mathrm{~Hz}\). A pure resistor under the same conditions takes a current of \(12.5 \mathrm{~A}\). If the two are connected to an ac supply of \(100 \mathrm{~V}\) and \(40 \mathrm{~Hz}\), then the current in series combination of above resistor and inductor is a. \(10 \mathrm{~A}\) b \(12.5 \mathrm{~A}\) c. \(20 \mathrm{~A}\) d. \(25 \mathrm{~A}\)

An inductive coil and resistance of \(100 \Omega\). When an ac signal of frequency \(1000 \mathrm{~Hz}\) is freed to the coil, the applied voltage leads the current by \(45^{\circ}\). What is the inductance of the coil? a. \(2 \mathrm{mH}\) b \(3.3 \mathrm{mH}\) c. \(16 \mathrm{mH}\) d. \(\sqrt{5} \mathrm{mH}\)

An \(L C R\) circuit contains resistance of \(100 \Omega\) and a supple of \(200 \mathrm{~V}\) at 300 radian angular frequency. If only capacitance is taken out from the circuit and the rest of the circuit is joined, current lags behind the voltage by \(60^{\circ}\). If on the other hand, only inductor is taken out the current leads by \(60^{\circ}\) with the applied voltage. The current flowing in the circuit is a. \(1 \mathrm{~A}\) b. \(1.5 \mathrm{~A}\) c. \(2 \mathrm{~A}\) d. \(2.5 \mathrm{~A}\)

A rexistor and an inducior are connecied to an ac supply of 120 volt and \(50 \mathrm{~Hz}\). The current in the circuit is \(3 \mathrm{~A}\). If the power consumed in the circuit is \(108 \mathrm{~W}\), then the rcsistance in the circuit is a. \(12 \Omega\) b. \(40 \Omega\) c. \(\sqrt{(52 \times 28)} \Omega\) d. \(360 \Omega\)

Two alternating voltage generators produce e.m.f.s of the same amplitude \(E_{0}\) but with a phase difference of \(\pi / 3\). The resultant emf is a. \(E_{0} \sin (\omega t+\pi / 3)\) b. \(E_{0} \sin (\omega t+\pi / 6)\) c. \(\sqrt{3} E_{0} \sin (\omega t+\pi / 6)\) d. \(\sqrt{3} E_{0} \sin (\omega t+\pi / 2)\)

See all solutions

Recommended explanations on Physics Textbooks

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.