A dipole with a length of 1.5 m operates at 100 MHz while the other has a length of 15 m and operates at 10 MHz. The dipoles are fed with same current. The power radiated by the two antennas will be

Concept:

P_rad = I_rms² R_rad

Where

I_rms = RMS current

R_rad = radiated resistance

P_rad = radiated power

So P_rad ∝ R_rad

and \({R_{rad}} = 80\;{\pi ^2}{\left( {\frac{l}{\lambda }} \right)^2}\)

l = length of antenna

λ = wavelength

Since \(\lambda = \frac{c}{f}\)

So R_rad ∝ l²f²

Solution:

Given:

 dioples are fed with same current (I₁ = I₂)

l₁ = 1.5 m, f₁ = 100 MHz = 100 ×  10⁶ Hz

l₂ = 15 m, f₂ = 10 MHz = 10 × 10⁶ Hz

So, \(\frac{{{P_{rad}}_1}}{{{P_{rad}}_2}} = \frac{{l_1^2f_1^2}}{{l_2^2f_2^2}}\)

\(\frac{{{P_{rad}}_1}}{{{P_{rad}}_2}} = {\left[ {\frac{{1.5 \;\times \;100\; \times \;{{10}^6}}}{{15 \;\times \;10\; \times \;{{10}^6}}}} \right]^2}\)

= \({\left[ {\frac{1}{1}} \right]^2}\)

So, P_rad1= P_rad2