# 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

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.