Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 21: Problem 84

A certain circuit has a \(25-\Omega\) resistor and one other component in series with a \(12-\mathrm{V}\) (rms) sinusoidal ac source. The rms current in the circuit is 0.317 A when the frequency is \(150 \mathrm{Hz}\) and increases by \(25.0 \%\) when the frequency increases to \(250 \mathrm{Hz}\). (a) What is the second component in the circuit? (b) What is the current at $250 \mathrm{Hz} ?$ (c) What is the numerical value of the second component?

Short Answer

Expert verified
Answer: The second component in the series circuit is a capacitor with a value of 37.79 μF.

Step by step solution

01

Find Impedance at Given Frequencies

We can calculate the impedance of the circuit at two given frequencies using Ohm's law. At 150 Hz, the rms current is 0.317 A, and the rms voltage is 12 V. The resistance R is given as 25 Ω. The impedance Z can be found using Ohm's law: \(Z = \frac{V}{I} = \frac{12 \mathrm{V}}{0.317 \mathrm{A}} \approx 37.87 \Omega\) At 250 Hz, the current increases by 25%, so the new current I' is: \(I'= 0.317 \mathrm{A} \times 1.25 = 0.39625\, \mathrm{A}\) Now let's find the new impedance Z': \(Z' = \frac{V'}{I'} = \frac{12 \mathrm{V}}{0.39625 \mathrm{A}} \approx 30.28 \Omega\)
02

Determine the Type of the Second Component

As the frequency increases, the impedance decreases, meaning that the unknown component is capacitive. The capacitive reactance \(X_C\) can be calculated as: \(X_C = Z^2 - R^2 \Rightarrow X_C = \sqrt{Z^2 - R^2}\) And, \(X'_C = Z'^2 - R^2 \Rightarrow X'_C = \sqrt{Z'^2 - R^2}\) Now, let's calculate \(X_C\) and \(X'_C\): \(X_C = \sqrt{37.87^2 - 25^2} \approx 28.11 \Omega\) \(X'_C = \sqrt{30.28^2 - 25^2} \approx 18.39 \Omega\)
03

Calculate Current at 250 Hz and Numerical Value of the Second Component

(b) We already found the current at 250 Hz in Step 1, which equals to \(I' = 0.39625 \, \mathrm{A}\). (c) Now, we will determine the capacitance of the second component. The capacitive reactance is given by: \(X_C = \frac{1}{2 \pi f C}\) Solving for C: \(C = \frac{1}{2 \pi f X_C}\) By plugging in the values at 150 Hz: \(C = \frac{1}{2 \pi \times 150 \mathrm{Hz} \times 28.11 \Omega} \approx 37.79 \, \mu\mathrm{F}\) Thus, the second component is a capacitor with a value of 37.79 μF.

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

In an RLC series circuit, these three elements are connected in scries: a resistor of \(60.0 \Omega,\) a 40.0 -mH inductor, and a \(0.0500-\mathrm{F}\) capacitor. The series elements are connected across the terminals of an ac oscillator with an rms voltage of \(10.0 \mathrm{V}\). Find the resonant frequency for the circuit.
A large coil used as an electromagnet has a resistance of \(R=450 \Omega\) and an inductance of \(L=2.47 \mathrm{H} .\) The coil is connected to an ac source with a voltage amplitude of \(2.0 \mathrm{kV}\) and a frequency of $9.55 \mathrm{Hz}$. (a) What is the power factor? (b) What is the impedance of the circuit? (c) What is the peak current in the circuit? (d) What is the average power delivered to the electromagnet by the source?
An \(R L C\) series circuit is connected to an ac power supply with a \(12-\mathrm{V}\) amplitude and a frequency of \(2.5 \mathrm{kHz}\) If $R=220 \Omega, C=8.0 \mu \mathrm{F},\( and \)L=0.15 \mathrm{mH},$ what is the average power dissipated?
A 40.0 -mH inductor, with internal resistance of \(30.0 \Omega\) is connected to an ac source $$\varepsilon(t)=(286 \mathrm{V}) \sin [(390 \mathrm{rad} / \mathrm{s}) t]$$ (a) What is the impedance of the inductor in the circuit? (b) What are the peak and rms voltages across the inductor (including the internal resistance)? (c) What is the peak current in the circuit? (d) What is the average power dissipated in the circuit? (e) Write an expression for the current through the inductor as a function of time.
The FM radio band is broadcast between \(88 \mathrm{MHz}\) and $108 \mathrm{MHz}$. What range of capacitors must be used to tune in these signals if an inductor of \(3.00 \mu \mathrm{H}\) is used?
See all solutions

Recommended explanations on Physics Textbooks

Electrostatics

Read Explanation

Mechanics and Materials

Read Explanation

Work Energy and Power

Read Explanation

Electricity and Magnetism

Read Explanation

Astrophysics

Read Explanation

Further Mechanics and Thermal 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.

Sign-up for free