# Microwave Engineering Questions and Answers – Circular Waveguide Cavity Resonators

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Circular Waveguide Cavity Resonators”.

1. A cylindrical cavity resonator can be constructed using a circular waveguide.
a) shorted at both the ends
b) open at both the ends
c) matched at both the ends
d) none of the mentioned

Explanation: A cylindrical cavity resonator is formed by shorting both the ends of the cylindrical cavity because open ends may result in radiation losses in the cavity.

2. The dominant mode in the cylindrical cavity resonator is TE101 mode.
a) true
b) false

Explanation: The dominant mode of propagation in a circular waveguide is TE111 mode. Hence, the dominant mode of resonance in a cylindrical cavity made of a circular waveguide is TE111 mode. In a cylindrical resonator, the mode of propagation depends on the length of the cavity.

3. Circular cavities are used for microwave frequency meters.
a) true
b) false

Explanation: Circular cavities are used for microwave frequency meters. The cavity is constructed with a movable top wall to allow the mechanical tuning of the resonant frequency.

4. The mode of the circular cavity resonator used in frequency meters is:
a) TE011 mode
b) TE101 mode
c) TE111 mode
d) TM111 mode

Explanation: Frequency resolution of a frequency meter is determined from its quality factor. Q factor of TE011 mode is much greater than the quality factor of the dominant mode of propagation.

5. The propagation constant of TEmn mode of propagation for a cylindrical cavity resonator is:
a) √ (k2-(pnm/a)2)
b) √ pnm/a
c) √ (k2+(pnm/a)2)
d) none of the mentioned

Explanation: The propagation constant for a circular cavity depends on the radius of the cavity, and the wave number. If the mode of propagation is known and the dimension of the cavity is known then the propagation constant can be found out.
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6. A circular cavity resonator is filled with a dielectric of 2.08 and is operating at 5GHz of frequency. Then the wave number is:
a) 181
b) 151
c) 161
d) 216

Explanation: Wave number for a circular cavity resonator is given by the expression 2πf011√∈r/C. substituting the given values in the above expression; the wave number of the cavity resonator is 151.

7. Given that the wave number of a circular cavity resonator is 151 (TE011 mode), and the length of the cavity is twice the radius of the cavity, the radius of the circular cavity operating at 5GHz frequency is:
a) 2.1 cm
b) 1.7 cm
c) 2.84 cm
d) insufficient data

Explanation: For a circular cavity resonator, wave number is given by √( (p01/a)2 + (π/d)2). P01 for the given mode of resonance is 3.832. Substituting the given values the radius of the cavity is 2.74 cm.

8. The loss tangent for a circular cavity resonator is 0.0004.Then the unloaded Q due to dielectric loss is:
a) 1350
b) 1560
c) 560
d) 2500

Explanation: Unloaded Q due to the dielectric loss in a circular cavity resonator is the reciprocal of the loss tangent. Hence, taking the reciprocal of the loss tangent, unloaded Q due to dielectric loss is 2500.

9. A circular cavity resonator has a wave number of 151, radius of 2.74 cm, and surface resistance of 0.0184Ω. If the cavity is filled with a dielectric of 2.01, then unloaded Q due to conductor loss is:
a) 25490
b) 21460
c) 29390
d) none of the mentioned

Explanation: Unloaded Q of a circular resonator due to conductor loss is given by ka/2Rs. is the intrinsic impedance of the medium. Substituting the given values in the equation for loaded Q, value is 29390.

10. If unloaded Q due to conductor loss and unloaded Q due to dielectric loss is 29390 and 2500 respectively, then the total unloaded Q of the circular cavity is:
a) 2500
b) 29390
c) 2300
d) 31890

Explanation: The total unloaded Q of a circular cavity resonator is given by the expression (Qc-1+ Qd-1)-1. Substituting the given values in the above expression, the total unloaded Q for the resonator is 2300.

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