Electronis Discussion
0 like 0 dislike
18 views

The steady state output of the circuit shown in the figure is given by $y(t)=A(\omega) \sin (\omega t + \phi ( \omega))$. If the amplitude $\mid A (\omega ) \mid =0.25$, then the frequency $\omega$ is

  1. $\frac{1}{\sqrt{3} \: R \: C}$
  2. $\frac{2}{\sqrt{3} \: R \: C}$
  3. $\frac{1}{R \: C}$
  4. $\frac{2}{R \: C}$
in Network Solution Methods by (15.8k points)
edited by | 18 views

Please log in or register to answer this question.

Answer:
Welcome to GO Electronics, where you can ask questions and receive answers from other members of the community.
1,174 questions
78 answers
11 comments
43,898 users