This set of Antennas Multiple Choice Questions & Answers (MCQs) focuses on “Radiation – Hertzian Dipole”.
1. Hertzian dipole carries which type of current throughout its length while radiating?
a) Varying
b) Constant
c) Depends on type of polarization
d) Depends on radiation resistance
View Answer
Explanation: Hertzian dipole is a short linear antenna which carries a constant current throughout its length while radiating. It consists of two equal and opposite charges separated by a very short distance. It is infinitesimal current element.
2. Power radiated by a Hertzian dipole of length λ/30 and carrying a current 2A?
a) 0.87W
b) 3.51W
c) 2.51W
d) 8.77W
View Answer
Explanation: Power radiated by a Hertzian dipole Prad=Rrad I2
Rrad = \(80\pi^2(\frac{l}{\lambda})^2=80\pi^2(\frac{\frac{\lambda}{30}}{\lambda})^2=0.877\Omega\)
Prad = Rrad I2=0.877×2×2=3.51W
3. A Hertzian dipole consists of two _____ and ______ charges separated by a very short distance.
a) unequal, opposite
b) equal, same
c) equal, opposite
d) unequal, same
View Answer
Explanation: A Hertzian dipole consists of two equal and opposite charges separated by a very short distance. It is infinitesimal current element. It is a short linear antenna which carries a constant current throughout its length while radiating.
4. When Hertzian dipole is connected to a practical antenna, which of the following fields is observed to be absent when a uniform current flow is observed?
a) Radiation field
b) Induction field
c) Electrostatic field
d) Both radiation and Induction Field
View Answer
Explanation: Since a constant current flow and there is no any charge accumulation at the ends of the dipole, the term 1/r3disappears. Therefore, electrostatic field is absent.
5.Which of the following is the radiation resistance of the Hertzian dipole?
a) \(\frac{\eta_0 w^2 dl^2}{6\pi c^2}\)
b) \(\frac{\eta_0 wdl^2}{6\pi c^2}\)
c) \(\frac{\eta_0 w^2 dl^2}{3\pi c^2}\)
d) \(\frac{\eta_0 w^3 dl^2}{3\pi c^3}\)
View Answer
Explanation: Radiation resistance of a Hertzian dipole is \(R_{rad} = 80\pi^2(\frac{l}{\lambda})^2\)
By simplifying the options given above,
\(\frac{\eta_0 w^2 dl^2}{6\pi c^2} = \frac{120\pi(2\pi/\lambda)^2 dl^2}{6\pi} = 80\pi^2(\frac{l}{\lambda})^2\)
6. If the radiation resistance of a Hertzian dipole is 100Ω, then the radiation resistance of short dipole is ____Ω.
a) 25
b) 50
c) 73
d) 35.6
View Answer
Explanation: The radiation resistance of the short dipole is ¼ times the radiation resistance of a current element. So 100/4= 25Ω.
7. The radiation resistance of a monopole of height 1cm and operating at frequency 100MHz is ____ Ω.
a) 4.83m
b) 4.38k
c) 4.38m
d) 4.83k
View Answer
Explanation: The radiation resistance of a monopole is 1/8 times the current element.
\(R_{rad}=10\pi^2 (l/\lambda)^2\)
For a monopole height h= l/2 => l= 2h
\(R_{rad}=10\pi^2(\frac{2h}{\lambda})^2=40\pi^2 (\frac{hf}{c})^2=40\pi^2 (\frac{1×100×10^6}{3×10^{10} })^2=4.38m\Omega\)
8. The radiation resistance of a monopole is _____ times the current element.
a) 1/8
b) 1/4
c) 1/2
d) 1/16
View Answer
Explanation: The radiation resistance of monopole is ½ times the short dipole. But the radiation resistance of short dipole is ¼ time the current element.
\(R_{rad \,mono}=\frac{1}{2}×\frac{1}{4}×80\pi^2(\frac{l}{\lambda})^2=\frac{1}{8}×80\pi^2(\frac{l}{\lambda})^2=10\pi^2 (\frac{l}{\lambda})^2=\frac{R_{rad \,Herztian}}{8}\)
The radiation resistance of monopole is 1/8 times the current element.
9. Practically we don’t use Hertzian dipole.
a) True
b) False
View Answer
Explanation: Since the current distribution at the center is maximum and minimum at ends, there is no uniform distribution of current along length. But Hertzian dipole is derived by assuming a uniform current distribution along length and having infinitesimal length. So that is reason why we don’t use Hertzian dipole practically.
10. If the radiation resistance of a monopole is 18Ω, then the radiation resistance of a Hertzian dipole is _____________
a) 124Ω
b) 144Ω
c) 164Ω
d) 154Ω
View Answer
Explanation: The radiation resistance of monopole is 1/8 times the current element.
Rrad Herztian = 8×Rrad mono=8×18=144Ω
Sanfoundry Global Education & Learning Series – Antennas.
To practice all areas of Antennas, here is complete set of 1000+ Multiple Choice Questions and Answers.
