This set of Electromagnetic Theory Problems focuses on “Transverse Electric Waves(TE)”.
1. In transverse electric waves, which of the following is true?
a) E is parallel to H
b) E is parallel to wave direction
c) E is transverse to wave direction
d) H is transverse to wave direction
View Answer
Explanation: In TE waves, the electric field strength will be transverse to the wave direction. Thus the TE waves are also called H waves.
2. The dominant mode in rectangular waveguide is
a) TE01
b) TE10
c) TM01
d) TM10
View Answer
Explanation: TE10 is the dominant mode in the rectangular waveguide. This is because it gives the minimum cut off frequency required for transmission.
3. The number of modes in a waveguide having a V number of 10 is
a) 10
b) 25
c) 100
d) 50
View Answer
Explanation: The number of modes is given by m = V2/2, where V is the v number. On substituting for V = 10, we get m = 100/2 = 50.
4. Does the waveguide with dimensions 3 cm x 5.5 cm exist?
a) Yes
b) No
View Answer
Explanation: For a waveguide, the dimension a should be greater than b. Here a = 3 and b = 5.5, thus such waveguide does not exist.
5. The mode which has the highest wavelength is called
a) Dominant mode
b) Evanescent mode
c) Generate mode
d) Degenerate mode
View Answer
Explanation: Dominant modes are the modes having least cut off frequency. This implies they have highest cut off wavelength.
6. The intrinsic impedance of a TE wave having a cut off frequency of 6 GHz at a frequency of 7.5 GHz in air is
a) 628.33
b) 338.62
c) 498.76
d) 342.24
View Answer
Explanation: The intrinsic impedance of a TE wave is given by ηTE = η/cos θ, where cos θ is given by √(1- (fc/f)2). On substituting for fc = 6 GHz, f = 7.5 GHz and η = 377, we get the intrinsic impedance as 628.33 units.
7. The cut off frequency of a rectangular waveguide of dimensions 3 cm x 1.5 cm is
a) 12 GHz
b) 6 GHz
c) 4 GHz
d) 5 GHz
View Answer
Explanation: The cut off frequency in dominant mode will be fc = mc/2a. On substituting for c = 3 x 108 and a = 0.03, we get the cut off frequency as 5 GHz.
8. The propagation constant for a lossless transmission line will be
a) Real
b) Complex
c) Real and equal to phase constant
d) Complex and equal to phase constant
View Answer
Explanation: The propagation constant is given by γ = α + jβ, where α and β are the attenuation and phase constants respectively. For a lossless line, the attenuation constant is zero. Thus γ = jβ. It is clear that γ is complex and equal to β.
9. The attenuation of a 50 ohm transmission line having a resistance of 100 ohm is
a) 0.01
b) 0.1
c) 1
d) 10
View Answer
Explanation: The attenuation of a wave is given by α = R/2Z0. On substituting for R = 100 and Z0 = 50, we get α = 100/(2 x 50) = 1 unit.
10. The cut off frequency of a TE wave with waveguide dimension of a= 3.5 cm in a medium of permittivity 2.2 is
a) 2.88 GHz
b) 3.32 GHz
c) 4.5 GHz
d) 2.12 GHz
View Answer
Explanation: The cut off frequency of a TE wave in any other medium is mc/2a√εr. On substituting for a = 0.035 and εr = 2.2, we get the cut off frequency as 2.88 GHz.
11. The phase constant and frequency are related by
a) Phase constant α ω
b) Phase constant α 1/ω
c) Phase constant α 1/2ω
d) Phase constant α ω/2
View Answer
Explanation: The phase constant is given by β = ω√LC. Thus the relation is β is directly proportional to ω.
12. Which of the following parameter is non zero for a lossless line?
a) Attenuation
b) Resistance
c) Conductance
d) Phase constant
View Answer
Explanation: The attenuation constant, resistance and conductance are zero for a lossless line. Only the phase constant is non zero.
Sanfoundry Global Education & Learning Series – Electromagnetic Theory.
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