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Two identical rectangular waveguide are joined end to end where a = 2b. One guide is air filled and other is filled with a lossless dielectric of εr. it is found that up to a certain frequency single mode operation can be simultaneously ensured in both guide. For this frequency range, the maximum allowable value of εr is
  • a)
    4
  • b)
    2
  • c)
    1
  • d)
    6
Correct answer is option 'A'. Can you explain this answer?
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Two identical rectangular waveguide are joined end to end where a = 2b...
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Two identical rectangular waveguide are joined end to end where a = 2b...
To ensure single mode operation in a rectangular waveguide, the cutoff frequency of the dominant mode should be higher than the operating frequency. In this scenario, we have two identical rectangular waveguides joined end to end, where the cross-sectional dimensions of one waveguide are a and b, and the other waveguide is filled with a lossless dielectric of relative permittivity r.

1. Cutoff frequency of the waveguide:
The cutoff frequency of the dominant mode TE10 in a rectangular waveguide can be calculated using the formula:

fc = c / 2a

where fc is the cutoff frequency, c is the speed of light in vacuum, and a is the width of the waveguide.

2. Single mode operation:
For single mode operation, the operating frequency f should be lower than the cutoff frequency of the dominant mode in both waveguides. Since the waveguides are identical, the cutoff frequency for both waveguides will be the same.

3. Considering the air-filled waveguide:
The cutoff frequency for the air-filled waveguide is given by fc_air = c / 2a.

4. Considering the dielectric-filled waveguide:
The cutoff frequency for the dielectric-filled waveguide is given by fc_dielectric = c / (2a√r).

5. Equating the cutoff frequencies:
To ensure single mode operation in both waveguides, we need to equate the cutoff frequencies and solve for r:

c / 2a = c / (2a√r)

Simplifying the equation, we get:

1 = 1 / √r

Squaring both sides, we obtain:

1 = 1 / r

r = 1

6. Maximum allowable value of r:
The maximum allowable value of r is the one that satisfies the equation r = 1. Therefore, the maximum allowable value of r is 1.

Hence, the correct answer is option 'A' - 1.
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Two identical rectangular waveguide are joined end to end where a = 2b. One guide is air filled and other is filled with a lossless dielectric of εr. it is found that up to a certain frequency single mode operation can be simultaneously ensured in both guide. For this frequency range, the maximum allowable value of εrisa)4b)2c)1d)6Correct answer is option 'A'. Can you explain this answer?
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