/*! 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} Q29P An 50.0 mH inductor is connect... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An 50.0mHinductor is connected, as in Figure to an ac generator with ∈m=30.0V.(a) What is the amplitude of the resulting alternating current if the frequency of the emf is 1.00kHz? and (b) What is the amplitude of the resulting alternating current if the frequency of the emf is 8.00kHz?

An inductor is connected across an alternating-current generator.

Short Answer

Expert verified
  1. The amplitude of the resulting alternating current if the frequency of the emf has a valuef=1.00kHzis 95.5mA.
  2. The amplitude of the resulting alternating current if the frequency of the emf has a value f=8.00kHzis 11.9mA.

Step by step solution

01

The given data

  1. Emf of the ac generator,∈m=30.0V
  2. Series inductance,L=5.0mH=5×10-3H
02

Understanding the concept of capacitive reactance and Ohm’s law

An inductor's obstruction in the current flow is known as inductive reactance. By determining the reactance due to the oscillation frequency within an inductor, we can get the current value by substituting this value in the voltage equation of Ohm's law.

The inductive reactance of a capacitor,

XL=2Ï€´ÚdL...... (i)

The voltage equation from Ohm’s law,

iL=∈mXL........ (ii)

Here, fdis the driving frequency of oscillations, Lis the inductance of the inductor, and ∈mis the emf applied across the circuit.

03

a) Calculation of the resulting current for frequency 1.00 kHz

Since the circuit contains only the inductor and sinusoidal generator, the potential difference across the inductor and sinusoidal generator are the same.

VL=∈m

∈mis the amplitude of the sinusoidal generator.

Thus, substituting equation (ii) in equation (i), we can get the resulting current through the inductor for frequencyf=1.00kHz as follows:

IL=∈m2Ï€´ÚdLIL=30V2Ï€1×103Hz50×10-3HIL=0.0955AIL=95.5mA

Hence, the value of current is 95.5mA.

04

b) Calculation of the resulting current for frequency 8.00 kHz

The amplitude of alternating current if the frequency of emf is8kHz.

In this case, the frequency is eight times larger than part a). Thus, the value of the resulting current is given as:

IL=18∈m2Ï€´ÚdLIL=95.5mA8IL=11.9mA

Hence, the value of the current is 11.9mA.

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!

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 Fig. 31-38, a three-phase generator G produces electrical power that is transmitted by means of three wires. The electric potentials (each relative to a common reference level) are V1=AsinÓ¬dtfor wire 1, V2=Asin(Ó¬dt-1200) for wire 2, and V3=Asin(Ó¬dt-2400)for wire 3. Some types of industrial equipment (for example, motors) have three terminals and are designed to be connected directly to these three wires. To use a more conventional two-terminal device (for example, a lightbulb), one connects it to any two of the three wires. Show that the potential difference between any two of the wires (a) oscillates sinusoidally with angular frequency Ó¬dand (b) has an amplitude ofA3.

A 0.50kgbody oscillates in SHM on a spring that, when extended2.0mmfrom its equilibrium position, has a8.0Nrestoring force. (a) What is the angular frequency of oscillation?(b) What is the period of oscillation (c) What is the capacitance of an LCcircuit with the same period if Lis5.0H?

In an RLC circuit such as that of Fig. 31-7 assume that R=5.0Ω,L=60.0mH,fd=60.0Hzand εm=30.0V. For what values of the capacitance would the average rate at which energy is dissipated in the resistance be (a) a maximum and (b) a minimum? What are (c) the maximum dissipation rate and the corresponding (d) phase angle and (e) power factor? What are (f) the minimum dissipation rate and the corresponding (g) phase angle and (h) power factor?

For a sinusoidally driven series RLCcircuit, show that over one complete cycle with period T (a) the energy stored in the capacitor does not change; (b) the energy stored in the inductor does not change; (c) the driving emf device supplies energy (12T)ϵm Icos ϕ and (d) the resistor dissipates energy.(12T)RI2(e) Show that the quantities found in (c) and (d) are equal.

An alternating source drives a series RLCcircuit with emf amplitude of, 6.00Vat a phase angle of+30.0°. When the potential difference across the capacitor reaches its maximum positive value of+5.00V, what is the potential difference across the inductor (sign included)?

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.