The dominant mode in the cylindrical cavity resonator is TE101 mode.

1.	The dominant mode in the cylindrical cavity resonator is TE₁₀₁ mode.
A.	true
B.	false
Answer» C.

Discussion

No Comment Found

Related MCQs

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 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:
The loss tangent for a circular cavity resonator is 0.0004.Then the unloaded Q due to dielectric loss is:
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:
The propagation constant of TEmn mode of propagation for a cylindrical cavity resonator is:a) √ (k2-(pnm/a)2)b) √ pnm/ac) √ (k2+(pnm/
The dominant mode in the cylindrical cavity resonator is TE101 mode.
A_CIRCULAR_CAVITY_RESONATOR_HAS_A_WAVE_NUMBER_OF_151,_RADIUS_OF_2.74_CM,_AND_SURFACE_RESISTANCE_OF_0.0184‚ÂÀ√≠¬¨¬©._IF_THE_CAVITY_IS_FILLED_WITH_A_DIELECTRIC_OF_2.01,_THEN_UNLOADED_Q_DUE_TO_CONDUCTOR_LOSS_IS:?$#
If unloaded Q due to conductor loss and unloaded Q due to dielectric loss is 29390 and 2500 respectively, then the total unloaded Q of the circular cavity is:$
The loss tangent for a circular cavity resonator is 0.0004.Then the unloaded Q due to dielectric loss is?
Given that the wave number of a circular cavity resonator is 151 (TE<sub>011</sub> 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:

The dominant mode in the cylindrical cavity resonator is TE101 mode.

Discussion

No Comment Found

Related MCQs

Reply to Comment

The dominant mode in the cylindrical cavity resonator is TE₁₀₁ mode.