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What is the total power radiated in Watts for the power density \(w_r=\frac{4sin\theta}{3r^2}a_r W/m^2\)?(a) 4π^2(b) 8π^2/3(c) 4π^2/3(d) 2π^2/3I had been asked this question in an online quiz.This key question is from Radiation Pattern in section Antenna Parameters of Antennas

Answer»

The correct answer is (C) 4π^2/3

The explanation: Total power radiated \(P_{rad}= ∯ w_r.a_n\overline{d}L \)

\(=\iint_{\THETA = 0}^{\pi} w_r.r^2 sin\theta d\theta d\EMPTYSET\)

\(=\iint_{\theta = 0}^π\frac{4sin\theta}{3r^2}r^2sin\theta d\theta d \emptyset\)

\(=\iint_{\theta = 0}^π \frac{4}{3}sin^2\theta d\theta d\emptyset=\iint_{\theta=0}^π\frac{4}{3}(\frac{1-cos2\theta}{2})d\theta d\emptyset\)

\(=\frac{4}{3}(\frac{1}{2})(π)(2π)=\frac{4}{3}\) π^2.


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