Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 21: Problem 73

A capacitor to be used in a radio is to have a reactance of \(6.20 \Omega\) at a frequency of \(520 \mathrm{Hz}\). What is the capacitance?

Short Answer

Expert verified
Answer: The capacitance of the capacitor is 4.93 × 10^{-8} F.

Step by step solution

01

Write down the given values

We are given the reactance (\(X_C = 6.20 \Omega\)) and the frequency (\(f = 520 \mathrm{Hz}\)). Our goal is to find the capacitance (\(C\)).
02

Rearrange the formula for capacitance

Rearrange the formula for reactance in a capacitor to solve for capacitance: \(C = \frac{1}{2 \pi f X_C}\)
03

Plug the given values into the formula

Insert the given values for reactance and frequency into the rearranged formula: \(C = \frac{1}{2 \pi (520 \mathrm{Hz}) (6.20 \Omega)}\)
04

Solve for capacitance

Calculate the value of the capacitance: \(C = \frac{1}{2 \pi (520) (6.20)} = 4.93 \times 10^{-8} \mathrm{F}\) The capacitance of the capacitor is \(4.93 \times 10^{-8} \mathrm{F}\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 series \(R L C\) circuit has a \(0.20-\mathrm{mF}\) capacitor, a \(13-\mathrm{mH}\) inductor, and a \(10.0-\Omega\) resistor, and is connected to an ac source with amplitude \(9.0 \mathrm{V}\) and frequency \(60 \mathrm{Hz}\) (a) Calculate the voltage amplitudes \(V_{L}, V_{C}, V_{R},\) and the phase angle. (b) Draw the phasor diagram for the voltages of this circuit.
The circuit shown has a source voltage of \(440 \mathrm{V}\) ms, resistance \(R=250 \Omega\), inductance \(L=0.800 \mathrm{H},\) and capacitance $C=2.22 \mu \mathrm{F} .\( (a) Find the angular frequency \)\omega_{0}$ for resonance in this circuit. (b) Draw a phasor diagram for the circuit at resonance. (c) Find these rms voltages measured between various points in the circuit: \(V_{a b}, V_{b c}, V_{c d}, V_{b d},\) and \(V_{c d},\) (d) The resistor is replaced with one of \(R=125 \Omega\). Now what is the angular frequency for resonance? (e) What is the rms current in the circuit operated at resonance with the new resistor?
Transformers are often rated in terms of kilovolt-amps. A pole on a residential street has a transformer rated at $35 \mathrm{kV} \cdot \mathrm{A}$ to serve four homes on the strect. (a) If each home has a fuse that limits the incoming current to \(60 \mathrm{A}\) rms at \(220 \mathrm{V}\) rms, find the maximum load in \(\mathrm{kV} \cdot \mathrm{A}\) on the transformer. (b) Is the rating of the transformer adequate? (c) Explain why the transformer rating is given in kV.A rather than in kW.
An RLC series circuit is connected to a \(240-\mathrm{V}\) ms power supply at a frequency of \(2.50 \mathrm{kHz}\). The elements in the circuit have the following values: \(R=12.0 \Omega, C=\) \(0.26 \mu \mathrm{F},\) and $L=15.2 \mathrm{mH},$ (a) What is the impedance of the circuit? (b) What is the rms current? (c) What is the phase angle? (d) Does the current lead or lag the voltage? (e) What are the rms voltages across each circuit element?
The instantaneous sinusoidal emf from an ac generator with an mos emf of $4.0 \mathrm{V}$ oscillates between what values?
See all solutions

Recommended explanations on Physics Textbooks

Kinematics Physics

Read Explanation

Astrophysics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Atoms and Radioactivity

Read Explanation

Oscillations

Read Explanation

Mechanics and Materials

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.

Sign-up for free