# Antenna Parameters Questions and Answers – Power Gain

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This set of Antennas Multiple Choice Questions & Answers (MCQs) focuses on “Power Gain”.

1. The ratio of power radiated in a particular direction to the total input power of antenna is called as _____
a) Directive gain
b) Power gain
c) Directivity
d) Partial directivity

Explanation: The ratio of power radiated in a particular direction to the actual power input to antenna is called Power gain. $$G_p=\frac{P_{d(\theta,\emptyset)}}{P_{in}}$$. Directive gain is the ratio of power radiated in desired direction to the average power radiated from the antenna. Partial directivity is the part of radiation intensity in a particular polarization to radiation intensity in all directions.

2. What is the maximum power gain of antenna with radiation efficiency 98% and directive gain 1?
a) 0.98
b) 1.02
c) 1.98
d) 1

Explanation: Gpmaxr Gdmax where ηr is radiation efficiency
∴ Gpmax=0.98×1=0.98.

3. Which of the following expression is correct for radiation efficiency?
a) $$\eta_r = \frac{P_r}{P_l}$$
b) $$\eta_r = \frac{P_r}{P_r-P_l}$$
c) $$\eta_r = \frac{P_r}{P_r+P_l}$$
d) $$\eta_r = \frac{P_l}{P_r+P_l}$$

Explanation: Radiation efficiency is defined as the ratio of power radiated to the total input power to the antenna. Total input power is the sum of the radiated power Pr and the ohmic losses Pl.
$$\eta_r = \frac{P_r}{P_{in}} = \frac{P_r}{P_r+P_l}$$

4. Which of the following represents the relation between maximum power gain and maximum directivity gain of the antenna?
a) Gpmax = ηrGdmax
b) Gpmax = ηr/Gdmax
c) ηr = $$\sqrt{(G_{pmax} G_{dmax})}$$
d) ηr = $$\frac{G_{dmax}+G_{pmax}}{G_{dmax}-G_{pmax}}$$

Explanation: Maximum power gain is obtained when there are no ohmic losses. Gpmax=$$\frac{U_{max}}{P_{in}/4π}$$
Maximum directive gain Gdmax=$$\frac{U_{max}}{P_r/4π}\, and\, \eta_r=\frac{P_r}{P_{in}}$$
∴Gpmaxr Gdmax

5. What is the maximum power gain when the radiation resistance is 72Ω, loss resistance is 8Ω and the maximum directive gain is 1.5?
a) 1.15
b) 1.35
c) 1.25
d) 1.53

Explanation: Maximum power gain Gpmaxr Gdmax
Radiation efficiency ηr=$$\frac{R_r}{R_r+R_l}=\frac{72}{72+8}=\frac{72}{80}=0.9$$
Now, Gpmaxr Gdmax=0.9×1.5=1.35

6. The radiation efficiency value is ______
a) 0
b) 1<η<∞
c) 0≤η≤1
d) ∞

Explanation: Radiation efficiency ηr=$$\frac{R_r}{R_r+R_l}$$
Rr+Rl>Rr So $$\frac{R_r}{R_r+R_l}$$ < 1 and If Rl=0 then ηr=1
Therefore, the value of efficiency lies in the range 0 to 1.

7. The value of maximum power gain is always greater than or equal to the maximum directive gain.
a) True
b) False

Explanation: Since GpmaxrGdmax and the value of radiation efficiency lies in range 0 to 1, The maximum power gain will be always less than or equal to the maximum directive gain of antenna. 