Question

An ac series circuit contains a 20 ohm resistor, a capacitor of 0.75 micro farad, and an inductor of 120 mH. What frequency should be used to create a resonance condition

Answer 1

Answer:

530.5 Hz

Explanation:

Given :

C=0.75 micro

L=120 mH

Reactance capacitor can be determined by

=$\frac{1}{ wC}$

where c =capacitor

= $\frac{1}{w*0.75*10^-6}$

Reactance inductor can be determined by

$= w*L\\ = w*120*10^-3$

As they are in series so ,

$\frac{1}{w*0.75*10^-6} =w*120*10^-3$

On solving these value we get

$w=3333.33$

As we know that

f = $\frac{w}{2*3.14}$ =

530.5 Hz