Q. A transmitter supplies $9\ kW$ to the aerial when unmodulated. The power radiated when modulated to 40% is
A) 5 $kW$
B) 9.72 $kW$
C) 10 $kW$
D) 12 $kW$
Explanation
(b) $P_{t} = P_{c}\left\lbrack 1 + \frac{m^{2}}{2} \right\rbrack = 9\left\lbrack 1 + \frac{(0.4)^{2}}{2} \right\rbrack$
$= 9\left\lbrack 1 + \frac{0.16}{2} \right\rbrack\ \ (\because m = 40\% = 0.4)$
$= 9(1.08) = 9.72\ kW$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.