Discussion :: Exam Questions Papers - Exam Paper 1 (Q.No.4)
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4. | A rectangular waveguide, in dominant TE mode, has dimensions 10 cm x 15 cm. The cut off frequency is |
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Answer: Option B Explanation: For TE10 mode |
Vasundhara said: (Aug 16, 2014) | |
In general dimensions of rectangle are represented as l:b. So a = 10cm, but here they have considered a = 15cm. Can any one help me to understand this? |
Alok said: (Apr 13, 2015) | |
Answer will be 1.5 GHz. For TE mode we will consider a = 10 cm and f= (3*10^10) / (2*a (cm)). |
Santhosh said: (Apr 5, 2016) | |
Please explain clearly. I can't understand. |
Gopal said: (Sep 19, 2016) | |
@Alok. I agree with your answer. For Dominant mode TE10. Cutoff frequency f = c/2a. Put all the value and we get our answer. |
Gauri said: (Jan 20, 2018) | |
Yes, answer should be 1.5 GHz. |
Rajesh said: (Feb 24, 2018) | |
Dominant mode is TE10 only when dimensions are a>b but here in the question he gave a<b so dominant mode will be TE01 so Cut-off frequency is C/(2b). |
Navya Rekha said: (Mar 23, 2018) | |
@Gauri. Here a is wider band& b is narrow band means a=15,b=10. So,f=3*10^8 m/2*15 cm, 3*10^8/0.3=1ghz. |
Thenmozhi said: (Nov 20, 2018) | |
@Gopal. What is "c" here? |
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