Question

A rectangular waveguide 2.29 cm \(\times\) 1.02 cm operates at a frequency of 11 GHz and a cut-off frequency of 6.55 GHz in TE10 mode. If the maximum potential gradient of the signal is 5 kV/cm, then the maximum power handling capacity of the waveguide will be

• A. 31.11 mW
• B. 31.11 W
• C. 31.11 kW
• D. 3.111 mW

Hint:

The power handling of a waveguide is limited by the breakdown voltage of the dielectric (usually air). The formula connects this physical limit (\(E_{max}\)) to the waveguide dimensions and operating frequencies through the wave impedance. Always ensure all units are in the base SI system (meters, volts/meter, Hz, ohms) before calculating.

Answer 1

The Correct Option is C

Step 1: Define the formula for maximum power in TE10 mode.The maximum power capacity (\(P_{max}\)) of a rectangular waveguide is limited by the maximum electric field strength (\(E_{max}\)) the dielectric can withstand before breakdown. For the TE10 mode, the maximum power is given by:\[ P_{max} = \frac{a \cdot b}{4 \eta_g} E_{max}^2 \]where \(a\) and \(b\) are the waveguide's width and height, respectively, and \(\eta_g\) represents the guide impedance.
Step 2: Calculate the guide impedance \(\eta_g\).The guide impedance is calculated as:\[ \eta_g = \frac{\eta_0}{\sqrt{1 - (f_c/f)^2}} \]where \(\eta_0\) is the free space impedance (\(\approx 377 \, \Omega\)), \(f\) is the operating frequency, and \(f_c\) is the cutoff frequency. Given \(f = 11\) GHz and \(f_c = 6.55\) GHz:\[ \eta_g = \frac{377}{\sqrt{1 - (6.55/11)^2}} = \frac{377}{\sqrt{1 - (0.595)^2}} = \frac{377}{\sqrt{1 - 0.354}} = \frac{377}{\sqrt{0.646}} \approx \frac{377}{0.804} \approx 469 \, \Omega \]
Step 3: Convert dimensions and electric field to SI units.\( a = 2.29 \text{ cm} = 0.0229 \text{ m} \)\( b = 1.02 \text{ cm} = 0.0102 \text{ m} \)\( E_{max} = 5 \text{ kV/cm} = 5 \times 10^3 \text{ V/cm} = 5 \times 10^5 \text{ V/m} \)
Step 4: Compute the maximum power.\[ P_{max} = \frac{(0.0229)(0.0102)}{4 \times 469} (5 \times 10^5)^2 \]\[ P_{max} = \frac{2.3358 \times 10^{-4}}{1876} (25 \times 10^{10}) \]\[ P_{max} \approx (1.245 \times 10^{-7}) \times (25 \times 10^{10}) = 3.1125 \times 10^4 \text{ W} \]\[ P_{max} \approx 31.125 \text{ kW} \]This result corresponds to option (C).