Answer:
"192.3 watt" is the right answer.
Explanation:
Given:
Efficient amplifier,
= 65%
or,
= 0.65
Power,
[tex]P_c=250 \ watt[/tex]
As we know,
⇒ [tex]P_t=P_c(1+\frac{\mu^2}{2} )[/tex]
By putting the values, we get
[tex]=P_c(1+\frac{1}{2} )[/tex]
[tex]=1.5 \ P_c[/tex]
Now,
⇒ [tex]P_i=(P_t-P_c)[/tex]
[tex]=1.5 \ P_c-P_c[/tex]
[tex]=\frac{P_c}{2}[/tex]
DC input (0.65) will be equal to "[tex](\frac{P_c}{2} )[/tex]".
hence,
The DC input power will be:
= [tex]\frac{250}{2}\times \frac{1}{0.65}[/tex]
= [tex]\frac{125}{0.65}[/tex]
= [tex]192.3 \ watt[/tex]
