/*! 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 61 (II) A 36-k\(\Omega\) resistor i... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(II) A 36-k\(\Omega\) resistor is in series with a 55-mH inductor and an ac source. Calculate the impedance of the circuit if the source frequency is (a) 50 Hz, and (b) 3.0 \(\times\) 10\(^4\) Hz.

Short Answer

Expert verified
Impedance at 50 Hz is approximately 36000.00 Ω. Impedance at 30000 Hz is approximately 37367.47 Ω.

Step by step solution

01

Understanding Impedance in Series AC Circuits

Impedance in a series AC circuit consists of a resistor and an inductor is found using the formula \( Z = \sqrt{R^2 + (X_L)^2} \), where \( R \) is resistance and \( X_L \) is inductive reactance.
02

Calculating Inductive Reactance \(X_L\)

Inductive reactance \( X_L \) is calculated as \( X_L = 2\pi f L \), where \( f \) is the frequency in hertz and \( L \) is the inductance in henries.
03

Calculating \(X_L\) for 50 Hz Frequency

Substitute \( f = 50 \text{ Hz} \) and \( L = 55 \times 10^{-3} \text{ H} \) in the formula \( X_L = 2\pi f L \). \[ X_L = 2\pi \times 50 \times 55 \times 10^{-3} = 17.28 \Omega \].
04

Calculating Impedance \(Z\) for 50 Hz

Using \( R = 36 \times 10^3 \Omega \) and \( X_L = 17.28 \Omega \), calculate \( Z = \sqrt{(36 \times 10^3)^2 + (17.28)^2} \). \[ Z \approx 36000.00008 \Omega \].
05

Calculating \(X_L\) for 30000 Hz Frequency

Substitute \( f = 3.0 \times 10^4 \text{ Hz} \) and \( L = 55 \times 10^{-3} \text{ H} \) in the formula \( X_L = 2\pi f L \). \[ X_L = 2\pi \times 30000 \times 55 \times 10^{-3} = 10362.26 \Omega \].
06

Calculating Impedance \(Z\) for 30000 Hz

Using \( R = 36 \times 10^3 \Omega \) and \( X_L = 10362.26 \Omega \), calculate \( Z = \sqrt{(36 \times 10^3)^2 + (10362.26)^2} \). \[ Z \approx 37367.47 \Omega \].

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.

Impedance
Impedance is a crucial concept in AC circuits. It essentially measures the total opposition that a circuit presents to the flow of alternating current. Unlike simple resistance, impedance combines both resistive and reactive elements. This is because, in AC circuits, both resistors and components like inductors or capacitors influence the current.
In this exercise, the impedance of a circuit containing a resistor and an inductor in series is calculated. The key is using the impedance formula: \[ Z = \sqrt{R^2 + (X_L)^2} \]where:
  • R = resistance in ohms (\(\Omega\))
  • \(X_L\) = inductive reactance in ohms (\(\Omega\))
This formula helps determine how much the resistor and the inductor together impede the flow of AC. It's important to use the correct values of \( R \) and \( X_L \) specific to your circuit conditions.
Resistor
A resistor's role in any electrical circuit is to limit the current flowing through the circuit. In AC circuits, a resistor's operation remains straightforward; it provides a constant resistance (\( R \)) opposing the flow of current.
For this exercise, we deal with a 36-k\(\Omega\) resistor. Because resistors affect both AC and DC currents alike, they are often crucial in controlling circuit current levels.
When calculating impedance in AC circuits, the resistor contributes the resistive part of the impedance which is purely real and unaffected by frequency, unlike reactance.
Inductor
An inductor is a passive electrical component that stores energy in a magnetic field when electric current flows through it. It's commonly used to manage AC currents by affecting their phase and amplitude.
In this exercise, the inductor has an inductance \( L \) of 55 mH. The inductor induces a reactive component called inductive reactance in an AC circuit. This reactance depends on the frequency of the AC signal passing through the inductor.
Inductors oppose changes in current because of their nature to store and release energy, making their behavior frequency-dependent. This characteristic is quantified as inductive reactance, which we'll explore next.
Inductive Reactance
Inductive reactance, symbolized as \( X_L \), is a measure of an inductor's opposition to changes in current in an AC circuit. The faster the current alternates, the greater the inductor opposes it.
The reactance is calculated using:\[ X_L = 2\pi f L \]Here, \( f \) represents the frequency in hertz (Hz), and \( L \) is inductance in henries (H).
  • Lower frequencies mean smaller \( X_L \), allowing more current through.
  • Higher frequencies result in a larger \( X_L \), reducing current flow.
For example, at 50 Hz, the reactance is about 17.28 \(\Omega\), and at 30,000 Hz, it's 10,362.26 \(\Omega\). This shows that inductive reactance increases with frequency, influencing the overall impedance.

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

(I) A 3500-pF capacitor is connected in series to a 5.0-\(\mu\)H coil of resistance 4.00 \(\Omega\). What is the resonant frequency of this circuit?

Two resistanceless rails rest 32 cm apart on a 6.0\(^\circ\) ramp. They are joined at the bottom by a 0.60-\(\Omega\) resistor. At the top a copper bar of mass 0.040 kg (ignore its resistance) is laid across the rails. Assuming a vertical 0.45-T magnetic field, what is the terminal (steady) velocity of the bar as it slides frictionlessly down the rails?

A 35-turn 12.5-cm-diameter coil is placed between the pole pieces of an electromagnet. When the electromagnet is turned on, the flux through the coil changes, inducing an emf. At what rate (in T/s) must the magnetic field change if the emf is to be 120 V?

(III) A long thin solenoid of length \(\ell\) and cross-sectional area \(A\) contains \(N_1\) closely packed turns of wire. Wrapped tightly around it is an insulated coil of \(N_2\) turns, Fig. 21-63. Assume all the flux from coil 1 (the solenoid) passes through coil 2, and calculate the mutual inductance.

The \(\textbf{Q factor}\) of a resonant ac circuit (Section 21-15) can be defined as the ratio of the voltage across the capacitor (or inductor) to the voltage across the resistor, at resonance. The larger the \(Q\) factor, the sharper the resonance curve will be and the sharper the tuning. (\(a\)) Show that the \(Q\) factor is given by the equation \(Q = (1/R) \sqrt{L/C}\). (\(b\)) At a resonant frequency \(f_0 =\) 1.0 MHz, what must be the values of L and R to produce a Q factor of 650? Assume that \(C =\) 0.010 \(\mu\)F.

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.