Question

An air filled $5.0cm\times 2.5cm$ rectangular waveguide of 0.8 m long is to delivered 1.2 kW to a match load at 4.5 GHz. Assuming an atenuation constant of $0.05dB/m$, the total amount of power (in Watt) dissipated in the walls of the waveguide is _________- .

• A. 27.36
• B. 18.79
• C. 11.11
• D. 33.51

Answer 1

The correct option is C 11.11

$P_{lead}=P_{in}{e}^{-2\alpha l}\Rightarrow P_{in}=P_{lead}{e}^{2\alpha l}\dots \dots \dots \left(1\right)$

$\therefore P_{dissipated}=P_{in}-P_{load}$

$=P_{load}{e}^{2\alpha l}-P_{load}$from equation $\left(1\right)$

$P_{diss}=P_{load}|e^{2\alpha l}-1|\dots \dots \dots ...\left(2\right)$

From the given problem $P_{load}=1.2kW$

$a=0.05dB/m=\frac{0.05}{8.68}Np/m$

$\alpha =5.76\times {10}^{-3}Np/m$

and $\beta =0.8m$

Substituting all the values in equation $\left(2\right)$

$P_{diss}=1.2\times {10}^3[e^{2\times 5.76\times {10}^{-3}\times 0.8}-1]$

$P_{diss}=11.11\left(Watts\right)$