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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:

Correct Answer: 29390
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.

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:

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 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:

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

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

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:

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

A coaxial cable is air filled with air as dielectric with inner and outer radius equal to 1 mm and 4 mm. If the surface resistivity is 1.84*10-2Ω,then the attenuation due to conductor loss is:

Find the wave number of a rectangular cavity resonator filled with a dielectric of 2.25 and designed to operate at a frequency of 5 GHz.