For A 100-cm long dipole is excited by a sinusoidally varying current with an amplitude i0=2 , the time average power radiated is mathematically given as
P=0.1577w
Generally, the equation for the is mathematically given as
[tex]\lambda =\frac{c}{f}[/tex]
Therefore
[tex]\lambda=\frac{3\times 10^{8}}{10^{6}}[/tex]
lambda=300m
In conclusion, for the power
[tex]P=40\pi^{2}(I_{0})^{2}(\frac{l}{\lambda})^{2}\\\\P=40* (3.14)^{2}\times6^{2} (\frac{1}{300})^{2}[/tex]
P=0.1577w
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The time-averaged power radiated by the dipole if the oscillating frequency is 150 MHz is 0.1577 w.
The distance between identical points (adjacent crests) in adjacent cycles of a waveform signal carried in space or down a wire is defined as the wavelength.
Given that the current is 2 amp, while the frequency is 150 MHz. Therefore, the wavelength can be written as,
[tex]\lambda = \dfrac cf = \dfrac{3 \times 10^8}{10^6} = 300\rm\ m[/tex]
Now, the power can be written as,
[tex]P=40 \pi^2 \times (I_o)^2 \times (\dfrac{l}{\lambda})^2\\\\P = 40 \times \pi ^2 \times 6^2 \times (\dfrac{1}{300})^2\\\\P = 0.1577\rm\ w[/tex]
Hence, the time-average power radiated by the dipole if the oscillating frequency is 150 MHz is 0.1577 w.
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