The load current in the transmitting antenna of an unmodulated AM transmitter is 6 Amp. What will be the antenna current when modulation is `60%`.

Explanation

Total power carried by AM wave
`P_(T)= P_(C)(1+ (m^(2))/(2))`……….(1)
where `P_(c)` is the power of carrier component and mis the modulation factor. If R is the resistance, `I_(m)` the antenna load current when modulation is 60% and `I_(c)` is the antenna load current when un modulated, then
`(P_(T))/(P_(C))= (I_(m)^(2)R)/(I_(c)^(2)R) :. 1+ (m_(2))/(2)= (I_(m)^(2))/(I_(c)^(2))` using (1)
or `I_(m) = I_(c)sqrt({(1+ (m^(2))/(2))}`
Given `I_(m)= 6Amp, m=0.6`
`I_(m)= 6[1+ ((0.6)^(2))/(2)]^(1//2)= 6[1.086]= 6.52 Amp`