Math Problem Statement

At a frequency 50% below its resonance frequency, the input impedance of a series RLC circuit is given by

[A]. R−j1.5LC−−√L/C

[B]. R+j1.5LC−−√L/C

[C]. R+j2.25CL−−√C/L

[D]. R−j2.25CL−−√C/L

A series RLC resonator has a unloaded Q of 900 at a resonance frequency of 100 MHz. If the resistance is 120 Ω, the value of the inductor is

[A]. 0.5πµ H

[B]. 54π mH

[C]. 540πµ H

[D]. 0.5π mH

A series RLC resonator has an unloaded Q of 800. When a load resistance RL is connected to the resonator, the loaded Q is 200. If the value ofRL is 400 Ω , the value of R is

[A]. 1003Ω

[B]. 2003Ω

[C]. 100Ω

[D]. 4003Ω

The magnitude of input impedance of a series RLC circuit attains its minimum value at the resonance frequency [A]. True [B]. False

A parallel RLC resonant circuit has an unloaded Q of 1000 and R=150 Ω . When a load of 75 Ω is connected to the resonator, the loaded Q of the resonator is

[A]. 1003

[B]. 225

[C]. 10003

[D]. 450

A short-circuited λ/4 long section of a low loss transmission line behaves as a series RLC resonator

[A]. True [B]. False

For an air-filled cavity resonator of inner cross section 3 cm X 1.5 cm and inner length of 4 cm, the dominant resonant mode has a frequency of [A]. 4.75 GHz [B]. 5.30 GHz [C]. 6.25 GHz [D]. 6.75 GHz

A short-circuited λ/2 transmission line has its first resonance at a frequency of 100 MHz. The attenuation constant of the line is π1000 Np/m and the velocity of wave propagation is 2.5×108 m/s. The unloaded Q of this resonator at its first resonance is [A] 200 [B] 300 [C] 400 [D] 500 gimme as of these mcqs correctly

Solution

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Series RLC Circuits
Resonance Frequency
Q Factor
Impedance

Formulas

Input Impedance of Series RLC Circuit
Q Factor Calculation
Resonance Frequency Calculation

Theorems

Behavior of RLC Resonators

Suitable Grade Level

Advanced Engineering