Electronics and Communication Engineering - Exam Questions Papers - Discussion
Discussion :: 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
[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.
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.
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,
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?
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.
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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