Electronics and Communication Engineering - Exam Questions Papers - Discussion
Discussion Forum : Exam Questions Papers - Exam Paper 1 (Q.No. 2)
2.
A transmission line has a characteristic impedance of 50 Ω and a resistance of 0.1 Ω/m, if the line is distortion less, the attenuation constant (in Np/m) is
Answer: Option
Explanation:
Distortion less 
a = RG and Z0 = RG → 
.
Discussion:
24 comments Page 2 of 3.
Shweta said:
8 years ago
How it is?
Zo= √ RG.
It is Zo= √( R/G).
Zo= √ RG.
It is Zo= √( R/G).
Ashwin said:
9 years ago
We know that in distortion less medium R/L=G/C, so alpha= R.sqrt(C/L) and Zo= sqrt(L/C), so alpha= R/Zo.
Sonal said:
9 years ago
R = α * z0. Compute by this formula.
Shweta said:
1 decade ago
Can anyone explain it further please?
Gopal said:
9 years ago
Hi,
Propogation constant = attenuation constant + i(phase constant).
Attenuation constant = R/2zo.
= 0.1/(2 * 50).
Propogation constant = attenuation constant + i(phase constant).
Attenuation constant = R/2zo.
= 0.1/(2 * 50).
Jitendra said:
9 years ago
Anyone, please explain this.
Rajesh said:
9 years ago
I understood the explanation.
Sowmya said:
9 years ago
Can any 1 explain how 0.1/2500 came?
Jai2x said:
10 years ago
The real formula is Zo = sqrt(R/G) if you analyze the solution for number 2.
SAI SURESH said:
1 decade ago
I can't understand can any one explain this.
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