Small Dipole with No Heat Loss
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For a small uniform current element the available output power is equal to the induced voltage squared divided by four times the radiation resistance. Thus
![{\displaystyle P_{r}={\frac {E^{2}a^{2}}{4R_{rad}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/052b326b135a1ca9f56d4d6a19f0076c9ac36419)
where
E=effective value of the electric field of the wave.
a=length of the current element.
Rrad. = radiation resistance of the current element
Since the power flow per unit area is equal to the electric field squared divided by the impedance of free space,
i.e., , we have
![{\displaystyle A_{dip}={\frac {P_{r}}{P_{0}}}={\frac {3\lambda ^{2}}{8\pi }}=0.1193\lambda ^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3473d5e60d52138f87e0bb143c42b5c4ce5339ff)
The effective area of a half-wavelength dipole with no heat loss is only 9.4 per cent, 0.39 decibels,2 larger than the effective area of the small dipole. Therefore
The area of a rectangle with one-half wavelength and one-quarter wavelength sides is and it is, therefore, a good approximation for the effective areas of small dipoles and half-wavelength dipoles.
![{\displaystyle A_{0.5}=0.1305\lambda ^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b0880c0975860b2f2f6c723ad172fad6f2aa6b30)
![{\displaystyle 0.1305\lambda ^{2}=\pi r^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/15f877a6c4131204d9c5c3b401f3c9412f204e59)
![{\displaystyle {\frac {0.1305}{\pi }}\lambda ^{2}=r^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1432db3685b5a52925ce0cdf079cc90c65c31522)
![{\displaystyle 0.04154\lambda ^{2}=r^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/428de49295b70db5e6bc776ff3782a02ac25c111)
![{\displaystyle {\sqrt {0.04152\lambda ^{2}}}={\sqrt {r^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/91637e6fa15e21cea046874106725d194e507fe1)
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