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_(tau)=P_(C)(1+(m^(2))/(2))` . . . .. . . . . .. (1)
where `P_(c)` is the power of carrier component and m is 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_(tau))/(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_(c)=6` Amp,m `=0.6`
`I_(m)=6[1+((0.6)^(2))/(2)]^(1//2)=6[1.086]=6.52` Amp.