Why does a folded dipole present about four times the impedance of a plain one?
Answer: C
The current divides between the two conductors, so for the same radiated power the current at the terminals is halved and, since power goes as current squared times resistance, the impedance rises by four. The radiation is that of a single dipole carrying the total current. The wires carry current in the same sense rather than in antiphase, which is why the arrangement radiates at all, and its resonant frequency is essentially that of the equivalent dipole.
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