# Antenna Parameters Questions and Answers – Power Radiation from Half Wave Dipole

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This set of Antenna Parameters Interview Questions and Answers for Experienced people focuses on “Power Radiation from Half Wave Dipole”.

1. If the current input to the antenna is 100mA, then find the average power radiated from the half-wave dipole antenna?
a) 365mW
b) 0.356mW
c) 0.365mW
d) 356mW

Explanation: Average Power radiated from the half-wave dipole Pavg=$$I_{rms}^2 R=(\frac{I_m}{\sqrt 2})^2 R$$
Radiation resistance of a half-wave dipole is 73Ω.
Given Im=100mA => Pavg=$$(\frac{100×10^{-3}}{\sqrt 2})^2×73=365mW.$$

2. The average radiated power of half-wave dipole is given by ______
a) $$73I_{rms}^2$$
b) $$36.5I_{rms}^2$$
c) $$13.25I_{rms}^2$$
d) $$146I_{rms}^2$$

Explanation: Radiation resistance of a half-wave dipole is 73Ω.
Average Power radiated from the half-wave dipole $$P_{avg}=I_{rms}^2 R=73I_{rms}^2$$
Radiation resistance of a quarter-wave monopole is 36.5Ω.

3. If the power radiated by a quarter-wave monopole is 100mW then power radiated by a half wave dipole under same current input is _____
a) 100W
b) 100mW
c) 200W
d) 200mW

Explanation: Average Power radiated from the half-wave dipole $$P_{avg-hlf}=I_{rms}^2 R_{hlf}$$
⇨ $$\frac{P_{avg-hlf}}{P_{avg\, mono}} = \frac{R_{hlf}}{R_{mono}} = \frac{73}{36.5}=2$$ (under same current)
⇨ Pavg-hlf = 2Pavg mono = 2*100mW=200mW

4. Power radiated by a half wave dipole is how many times the power radiated by a quarter wave monopole under same input current to antennas?
a) 2
b) 3
c) 4
d) 1

Explanation: Average Power radiated from the half-wave dipole $$P_{avg-hlf}=I_{rms}^2 R_{hlf}$$
⇨ $$\frac{P_{avg-hlf}}{P_{avg\, mono}} = \frac{R_{hlf}}{R_{mono}} = \frac{73}{36.5}=2$$ (under same current)

5. Find the magnetic field if the electric field radiated by the half-wave dipole is 60mV/m?
a) 159μA/m
b) 195μA/m
c) 159mA/m
d) 195mA/m

Explanation: η=$$\frac{E}{H}$$
⇨ 120π=60m/H
⇨ H = 159μA/m

6. In which of the following the power is radiated through a complete spherical surface?
a) Half-wave dipole
b) Quarter-wave Monopole
c) Both Half-wave dipole & Quarter-wave Monopole
d) Neither Half-wave dipole nor Quarter-wave Monopole

Explanation: In a half-wave dipole the power is radiated in the entire spherical surface and in quarter wave monopole the power is radiated only through a hemispherical surface. Hence its radiation resistance is also twice that of the quarter wave monopole.

7. Find the power radiated from the half wave dipole at 2km away with magnetic field at point $$\theta=\frac{\pi}{2}$$ is 10μA/m ?
a) 0.576mW
b) 0.576W
c) 0.756W
d) 0.675W

Explanation: Magnetic field strength $$H=\frac{I_m}{2\pi r}(\frac{cos⁡(\frac{\pi}{2}cos\theta)}{sin\theta})$$
⇨ 10×10-6=$$\frac{I_m}{2\pi×2×10^3}(\frac{cos⁡(\frac{\pi}{2}cos\frac{\pi}{2})}{sin\frac{\pi}{2}})$$
⇨ Im=0.125A
Now Average power radiated $$P_{avg}=I_{rms}^2 R=(\frac{I_m}{\sqrt 2})^2 R=(\frac{0.125}{\sqrt 2})^2 ×73×=0.576W$$

8. For the same current, the power radiated by half-wave dipole is four times that of the radiation by quarter wave monopole.
a) True
b) False

Explanation: The radiation resistance of a half wave dipole is 73Ω and that of a quarter wave monopole is 36.5Ω. So the power radiated by half-wave dipole is two times that of the radiation by quarter wave monopole.

9. If the power radiated by a half wave dipole is 100mW then power radiated by a quarter wave monopole under same current input is _____
a) 50mW
b) 200mW
c) 100mW
d) 50W

Explanation: Average Power radiated from the half-wave dipole $$P_{avg-hlf}=I_{rms}^2 R_{hlf}$$
⇨ $$\frac{P_{avg-hlf}}{P_{avg\, mono}} = \frac{R_{hlf}}{R_{mono}} = \frac{73}{36.5}=2$$ (under same current)
$$P_{avg \,mono}=\frac{P_{avg-hlf}}{2}=\frac{100mW}{2}=50mW.$$

10. If the power radiated by a quarter wave monopole is 100mW, then power radiated by a half wave dipole with doubled current is ______
a) 800mW
b) 400mW
c) 200mW
d) 100mW

Explanation: Average Power radiated from the half-wave dipole $$P_{avg-hlf}=I_{rms}^2 R_{hlf}=4I_{rms}^2 R_{hlf}$$
⇨$$\frac{P_{avg-hlf}}{P_{avg\, mono}}=\frac{4I_{rms}^2 R_{hlf}}{I_{rms}^2 R_{mono}}=4*\frac{73}{36.5}=8$$ (under same current)
Pavg-hlf=8(Pavg mono)=800mW

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