A transmitter supplies 9 KW to the aerial when unmoderated. The power radiated when modulated to 60 % is:

Question

A transmitter supplies 9 KW to the aerial when unmoderated. The power radiated when modulated to 60 % is: (A) 10.10 KW (B) 10.62 KW (C) 9.96 KW (D) 9.16 KW

Tardigrade Admin · Accepted Answer

We know the relation (pt/pc)=[1+((m)2/2)] Given, m=60 %=(60/100)=0.6 (pt/9)=(1+((0.6)2/2)) ⇒ pt=9[1+((0.36)/2)] pt=9(1+0.18)=9(1.18) pt=10.62 k W

Q. A transmitter supplies $9 K W$ to the aerial when unmoderated. The power radiated when modulated to $60%$ is:

10.10 KW

10.62 KW

9.96 KW

9.16 KW

We know the relation
$\frac{p _{t}}{p _{c}} = [1 + \frac{( m ) ^{2}}{2}]$
Given, $m = 60% = \frac{60}{100} = 0.6$
$\frac{p _{t}}{9} = (1 + \frac{( 0.6 ) ^{2}}{2})$
$\Rightarrow p_{t} = 9 [1 + \frac{( 0.36 )}{2}]$
$p_{t} = 9 (1 + 0.18) = 9 (1.18)$
$p_{t} = 10.62 kW$

Q. A transmitter supplies 9KW to the aerial when unmoderated. The power radiated when modulated to 60% is: