A) unchanged
B) doubled
C) increased by 50%
D) zero
Correct Answer: C
Solution :
[c] : The transmitted power in AM wave is\[{{P}_{t}}=\left( 1+\frac{{{\mu }^{2}}}{2} \right){{P}_{c}}\]where \[\mu \]is the modulation index and \[{{P}_{c}}\] is the power in carrier wave. When \[\mu =0,{{P}_{t}}={{P}_{c}}\]and when \[\mu =1,{{P}_{t}}=\left( 1+\frac{1}{2} \right){{P}_{c}}=\frac{3}{2}{{P}_{c}}\] \[\therefore \]The percentage increased in transmitted power \[=\frac{\frac{3}{2}{{P}_{c}}-{{P}_{c}}}{{{P}_{c}}}\times 100%=\frac{1}{2}\times 100%=50%\]
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