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:
23 comments Page 1 of 3.
Raey said:
4 years ago
Thanks for explaining @Jacb.
Ravi said:
4 years ago
Thanks @Jacb.
Charan said:
5 years ago
It's clear now, Thanks @Jacb.
Jacb said:
6 years ago
z0=√(R/G) , equate to G: G=R/(z0)^2.
atten const = √(R*G), subs. G :
atten const = √(R*(R/(z0)^2)).
atten const = √(0.1 * ( 0.1/ (50)^2)).
Hope that helps.
atten const = √(R*G), subs. G :
atten const = √(R*(R/(z0)^2)).
atten const = √(0.1 * ( 0.1/ (50)^2)).
Hope that helps.
(1)
NIDHEESH NELSON said:
6 years ago
Can any one explain this?
Mahesh said:
6 years ago
Can anyone explain me briefly?
Sachin kumar said:
6 years ago
How 0.1/2*50=0.1/2500 can anyone explain?
Shweta said:
6 years ago
How it is?
Zo= √ RG.
It is Zo= √( R/G).
Zo= √ RG.
It is Zo= √( R/G).
Ashwin said:
7 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:
7 years ago
R = α * z0. Compute by this formula.
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