This set of Electromagnetic Theory Multiple Choice Questions & Answers (MCQs) focuses on “Cut off Frequency and Wavelength”.
1. The real part of the propagation constant is the
a) Attenuation constant
b) Phase constant
Explanation: The propagation constant is given by γ = α + jβ. Here the real part is the attenuation constant and the imaginary part is the phase constant.
2. The phase constant of a wave is given by
Explanation: The phase constant of a wave in a transmission line is given by β = ω√(LC), where L and C are the specifications of the line.
3. The cut off frequency of the dominant mode in a TE wave in the line having a and b as 2.5 cm and 1 cm respectively is
a) 4.5 GHz
b) 5 GHz
c) 5.5 GHz
d) 6 GHz
Explanation: The dominant mode in TE is TE10. The cut off frequency will be mc/2a, where m = 1 and a = 0.025 are given. On substituting, we get the frequency as 1 x 3 x 108/2 x 0.025 = 6 GHz.
4. The cut off frequency of the TE01 mode will be
Explanation: The cut off frequency consists of modes m and n. For m = 0, the dimension b will be considered. Thus the frequency is nc/2b, where c is the speed of the light.
5. The condition which will satisfy the dimensions of the waveguide is
a) a = b
b) a > b
c) a < b
d) ab = 0
Explanation: The dimensions a and b represent the broad wall and the side wall dimensions respectively. The broad wall will be greater than the side wall. Thus the condition a>b is true.
6. The cut off wavelength of the TE10 mode having a broad wall dimension of 5cm is
Explanation: The cut off wavelength of the waveguide is given by λc = 2a/m. on substituting for a = 0.05 and m = 1, we get λc = 2 x 0.05/1 = 0.1 units.
7. The broad wall dimension of a waveguide having a cut off frequency of 7.5 GHz is
Explanation: The cut off frequency and the broad wall dimension are related by fc = mc/2a. On substituting for m = 1 and fc = 7.5 GHz, we get a = 0.02 or 2 cm.
8. The sin θ in the waveguide refers to the ratio of the
a) Frequency to wavelength
b) Wavelength to frequency
c) Cut off frequency to frequency
d) Frequency to cut off frequency
Explanation: The ratio of the cut off frequency to the frequency at any point gives the sin θ in a waveguide.
9. Is the transmission of a frequency 5 GHz possible in waveguides?
Explanation: The cut off frequency for waveguide operation is 6 GHz. Thus a wave of 5 GHz is not possible for transmission in a waveguide.
10. The dimension for a waveguide in dominant mode with a cut off wavelength of 2 units is
Explanation: The cut off wavelength of a waveguide is given by λc = 2a/m. For the dominant mode, m = 1. Given that λc = 2, thus we get a = 4 units.
11. The waveguides are used in a transmission line for
a) Increasing transmission coefficient
b) Increasing reflection coefficient
c) Decreasing transmission coefficient
d) Decreasing reflection coefficient
Explanation: The waveguides are used to increase the transmission efficiency of the waves travelling through it.
12. The attenuation coefficient of the wave having a resistance of 15 ohm in a 50 ohm line is
Explanation: The attenuation coefficient of a wave with a resistance of R in a line of characteristic impedance Zo is α = R/2Zo. On substituting for R = 15 and Zo = 50, we get α = 15/(2 x 50) = 0.15 units.
Sanfoundry Global Education & Learning Series – Electromagnetic Theory.
To practice all areas of Electromagnetic Theory, here is complete set of 1000+ Multiple Choice Questions and Answers.