Electronics and Communication Engineering - Electromagnetic Field Theory - Discussion
Discussion Forum : Electromagnetic Field Theory - Section 1 (Q.No. 10)
10.
A rectangular metal waveguide filled with a dielectric of relative permittivity εr = 4, has the inside dimensions 3 x 1.2 cm, the cut off frequency for the dominant mode is
Answer: Option
Explanation:
for dominant mode.
Discussion:
15 comments Page 1 of 2.
Kamlesh said:
1 decade ago
Option A is correct.
Phase velocity is c/2 hence f = c/4a.
Phase velocity is c/2 hence f = c/4a.
Vinay said:
1 decade ago
Yes option A is correct.
C' = C/2.
C' = C/2.
Allaa said:
1 decade ago
c = c0/sqrt(εr).
f = c/2a.
f = c/2a.
Ravi said:
9 years ago
Cut off frequency = c/(2a * sqrt (Er * Ur).
Anjli said:
9 years ago
As f = C'/2a.
And C' = C/2 (because Er = 4).
So f = c/4a.
And C' = C/2 (because Er = 4).
So f = c/4a.
Pk rai said:
9 years ago
A is correct.
Sunil Chavan said:
8 years ago
A is the answer.
Ronston said:
8 years ago
What does 'a' stand for? dimension?
Azar said:
8 years ago
fc = C/2a.
fc = Cut-off Frequency.
C = Speed of Light = 3 * 10^8,
a = Width = 3 cm.
fc = 3 * 10^8 / 2 *3 = 0.5 * 10^8 ==> 5 GHz.
fc = Cut-off Frequency.
C = Speed of Light = 3 * 10^8,
a = Width = 3 cm.
fc = 3 * 10^8 / 2 *3 = 0.5 * 10^8 ==> 5 GHz.
Vijay said:
8 years ago
Fc should be divided by 2.
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