Electronics and Communication Engineering - Exam Questions Papers - Discussion

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

[A]. 500
[B]. 5
[C]. 0.014
[D]. 0.002

Answer: Option D

Explanation:

Distortion less

a = RG and Z0 = RG

.


Shweta said: (May 9, 2014)  
Can anyone explain it further please?

Shweta said: (May 9, 2014)  
Got it by myself.

Correct formula for distortionless transmission line is:

Characteristic impedance = Square root of (resistance/Conductance).

Aditya said: (Dec 21, 2014)  
Z0= sqt of RG.

Neelu said: (Feb 13, 2015)  
Zo = sqrt(R/G).

Dilip N said: (Feb 16, 2015)  
G/C = R/L.
Zo = sqrt(L/C).
Zo = sqrt(R/G).

Pavi said: (Aug 10, 2015)  
Z0 = sqrt of RG.

Sai Suresh said: (Aug 18, 2015)  
I can't understand can any one explain this.

Jai2X said: (Oct 16, 2015)  
The real formula is Zo = sqrt(R/G) if you analyze the solution for number 2.

Sowmya said: (Jun 26, 2016)  
Can any 1 explain how 0.1/2500 came?

Rajesh said: (Aug 11, 2016)  
I understood the explanation.

Jitendra said: (Sep 9, 2016)  
Anyone, please explain this.

Gopal said: (Sep 19, 2016)  
Hi,

Propogation constant = attenuation constant + i(phase constant).
Attenuation constant = R/2zo.
= 0.1/(2 * 50).

Fadhil said: (Dec 5, 2016)  
Now I undestand it thanks.

Sonal said: (Dec 16, 2016)  
R = α * z0. Compute by this formula.

Ashwin said: (Feb 1, 2017)  
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.

Shweta said: (Jul 4, 2017)  
How it is?

Zo= √ RG.
It is Zo= √( R/G).

Sachin Kumar said: (Aug 26, 2017)  
How 0.1/2*50=0.1/2500 can anyone explain?

Mahesh said: (Oct 2, 2017)  
Can anyone explain me briefly?

Nidheesh Nelson said: (Feb 13, 2018)  
Can any one explain this?

Jacb said: (Mar 31, 2018)  
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.

Charan said: (Sep 16, 2018)  
It's clear now, Thanks @Jacb.

Ravi said: (Jul 28, 2019)  
Thanks @Jacb.

Raey said: (Aug 21, 2019)  
Thanks for explaining @Jacb.

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