Four half-wave dipoles (broadside array), λ = 100 cm, spacing = 50 cm, each element current 0.5 A in-phase: compute total radiated power.

Difficulty: Medium

Correct Answer: 73 W

Explanation:


Introduction / Context:
This question examines array power estimation using the radiation resistance of a half-wave dipole and assumes in-phase elements with negligible mutual coupling for a first-order calculation. Such estimates are common in antenna design to size transmitters and ensure compliance with power limits.



Given Data / Assumptions:

  • Wavelength λ = 100 cm ⇒ frequency around 300 MHz (context only).
  • Four half-wave dipoles, spacing 0.5 λ = 50 cm.
  • Each element carries an RF current magnitude I = 0.5 A (assumed RMS for power calculation).
  • Standard radiation resistance of a half-wave dipole Rr ≈ 73 Ω.
  • Neglect mutual coupling and losses; total radiated power is sum of per-element radiated powers.


Concept / Approach:

Radiated power for one half-wave dipole: P_elem = I_rms^2 * Rr. For N identical elements (uncoupled approximation) and equal in-phase currents, P_total ≈ N * P_elem. This is a reasonable engineering estimate when pattern multiplication and coupling are not detailed.



Step-by-Step Solution:

Compute per-element power: P_elem = (0.5 A)^2 * 73 Ω = 0.25 * 73 = 18.25 W.Total for 4 elements: P_total = 4 * 18.25 W = 73 W.Therefore, radiated power ≈ 73 W.


Verification / Alternative check:

If the 0.5 A were peak rather than RMS, results would differ by a factor of 1/2. The problem's options (18.25 W and 73 W appearing) suggest the intended interpretation is RMS current per element.



Why Other Options Are Wrong:

  • 18.25 W: power for one element only, not the entire 4-element array.
  • 36.5 W and 9.125 W: inconsistent partial sums; likely from miscounting elements or squaring errors.
  • 196 W: implies incorrect radiation resistance or current usage.


Common Pitfalls:

  • Confusing RMS and peak currents.
  • Applying pattern gain directly to power without considering input power and efficiency.


Final Answer:

73 W

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