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The loss tangent for a circular cavity resonator is 0.0004.Then the unloaded Q due to dielectric loss is:

Correct Answer: 2500
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.

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:

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:

If a dielectric resonator has a dielectric constant of 49, then the reflection coefficient of the dielectric resonator is:

The approximate loaded Q due to dielectric loss for a dielectric resonator given the loss tangent is 0.0001 is:

The required length of the cavity resonator for l=1 mode (m=1, n=0) given that the wave number of the cavity resonator is 157.01 and the broader dimension of the waveguide is 4.755 cm:

If the attenuation due to dielectric loss and attenuation due to conductor loss in a microstrip transmission line is 0.024Np/m and 0.0724 Np/m, then the unloaded quality factor if the propagation constant is 151 is:

A circular cavity resonator is filled with a dielectric of 2.08 and is operating at 5GHz of frequency. Then the wave number is:

Unloaded Q of a rectangular waveguide cavity resonator:

A cavity in the shape of a frustum of a cone has diameter 30 cm and 60 cm and the height is 80 cm. If the cavity is maintained at temperature of 800 K, determine the heat loss from the cavity when the smaller diameter is at the bottom