Question

The cut off frequency of a waveguide with resonant cavity is given by (a) V√((m/a)^2 + (n/b)^2 + (p/d)^2)/2 (b) V√((m/a)^2 + (n/b)^2 + (p/d)^2) (c) 2V√((m/a)^2 + (n/b)^2 + (p/d)^2) (d) V√((m/a) + (n/b) + (p/d))/2 I have been asked this question in an interview. This intriguing question comes from Transients in division Waveguides of Electromagnetic Theory Select the correct answer from above options Electromagnetic Theory questions and answers, Electromagnetic Theory questions pdf, Electromagnetic Theory question bank, Electromagnetic Theory questions and answers pdf, mcq on Electromagnetic Theory pdf,electromagnetic theory book pdf,electromagnetic theory pdf notes,electromagnetic theory lecture notes,electromagnetic theory of light,maxwell electromagnetic theory,electromagnetic theory proposed by,electromagnetic theory engineering physics,electromagnetic theory nptel

Answer 1

Right option is (a) V√((m/a)^2 + (n/b)^2 + (p/d)^2)/2 The best explanation: The cut off frequency of the waveguide of dimensions a x b with the resonant cavity of dimension d is given by fc = √((m/a)^2 + (n/b)^2 + (p/d)^2)/2. Here m and n are the orders of the waveguide, p is the order of the cavity and v is the velocity.