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: Write the formula for maximum power in a waveguide for TE10 mode. The maximum power (\(P_{max}\)) that a rectangular waveguide can handle is determined by the maximum electric field (\(E_{max}\), or potential gradient) the dielectric inside can withstand before breakdown. For the TE10 mode, the formula is: \[ P_{max} = \frac{a \cdot b}{4 \eta_g} E_{max}^2 \] where \(a\) is the wide dimension, \(b\) is the narrow dimension, and \(\eta_g\) is the wave impedance in the guide.
Step 2: Calculate the wave impedance \(\eta_g\). The wave impedance is given by: \[ \eta_g = \frac{\eta_0}{\sqrt{1 - (f_c/f)^2}} \] where \(\eta_0\) is the impedance of free space (\(\approx 377 \, \Omega\)), \(f\) is the operating frequency, and \(f_c\) is the cutoff frequency. Given: \(f = 11\) GHz, \(f_c = 6.55\) 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 given dimensions and fields 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: Calculate 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 matches option (C).