Question

A 100-cm long dipole is excited by a sinusoidally varying current with an amplitude i0=2 a. Determine the time average power radiated by the dipole if the oscillating frequency is 150 mhz

Answer 1

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

What is the time average power radiated by the dipole if the oscillating frequency is 150 mhz?

Generally, the equation for the   is mathematically given as

$\lambda =\frac{c}{f}$

Therefore

$\lambda=\frac{3\times 10^{8}}{10^{6}}$

lambda=300m

In conclusion, for the power

$P=40\pi^{2}(I_{0})^{2}(\frac{l}{\lambda})^{2}\\\\P=40* (3.14)^{2}\times6^{2} (\frac{1}{300})^{2}$

P=0.1577w

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Answer 2

The time-averaged power radiated by the dipole if the oscillating frequency is 150 MHz is 0.1577 w.

What is wavelength?

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,

$\lambda = \dfrac cf = \dfrac{3 \times 10^8}{10^6} = 300\rm\ m$

Now, the power can be written as,

$P=40 \pi^2 \times (I_o)^2 \times (\dfrac{l}{\lambda})^2\\\\P = 40 \times \pi ^2 \times 6^2 \times (\dfrac{1}{300})^2\\\\P = 0.1577\rm\ w$

Hence,  the time-average power radiated by the dipole if the oscillating frequency is 150 MHz is 0.1577 w.

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