Electronics and Communication Engineering - Networks Analysis and Synthesis - Discussion

Discussion Forum : Networks Analysis and Synthesis - Section 1 (Q.No. 13)
13.
Z(c) for the network shown in the figure is The value of C and R are, respectively
1/6 F and 4 Ω
2/9 F and 9/2 Ω
2/3 F and 1/2 Ω
1/2 F and 1 Ω
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
22 comments Page 1 of 3.

Purushothama said:   5 years ago
Can anyone explane the answer?
(1)

Monita said:   6 years ago
Z(0) = 8.
2 resistance sum = 4 So, 8-4 = 4(R value).
1 capacitor sum = 1/2 so, 1/(8-(1/1/2)) (C value).

Here for capacitance, we use 1/c.

I don't know whether it's correct or not.
(3)

Hephzibah said:   7 years ago
Please solve problem.

Saanvi said:   7 years ago
Can anyone explain this problem in detail?

Surya Dubey said:   7 years ago
Thank you @Satyam.

Roopa said:   8 years ago
Thank you @Satyam.

Ravi said:   8 years ago
Thanks @Satyam.

Arnab said:   8 years ago
After continued fraction expansion you get;

3+{1/(s/6)+{1/(4)+{1/(s/2)+2}}}
C=1/6, R=4.

Hari said:   8 years ago
@Gopal Agarwal.

I didn't understand that. Please make it clear.

Gopal agarwal said:   8 years ago
TRANSFER FUNCTION COMES OUT AS -: (2+2*R+s*R) + 3*[RC*s^2+s*(2*C+2*R*C+1)+2]
------------------------------------------------------------------------
RC*s^2+s*(2*C+2*R*C+1)+2.
From Denominator-:RC*s^2+s*(2*C+2*C*R+1)+2.
===== RC[ s^2 + s*(2/R+2+1/RC) + 2/RC].

And on comparing it from given equation(i.e. s^2+4*s+3 )
We get 3=2/RC ---> (1)
and from given option, only option (A) satisfy this condition.
that's why the answer is an option (A).
(1)


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