Question

An AM wave has \(1800\,W\) of total power content. For \(100%\) modulation the carrier should have power content equal to

• A. 1000 W
• B. 1200 W
• C. 1500 W
• D. 1600 W

Hint:

AM total power: \(P_{total}=P_c\left(1+\dfrac{m^2}{2}\right)\). For \(m=1\), \(P_{total}=\dfrac{3}{2}P_c\).

Answer 1

The Correct Option is B

Step 1: Total power in AM wave.
\[ P_{total} = P_c\left(1+\frac{m^2}{2}\right) \] where \(m\) is modulation index.
Step 2: For 100% modulation.
\[ m = 1 \Rightarrow P_{total} = P_c\left(1+\frac{1}{2}\right) = \frac{3}{2}P_c \] Step 3: Solve for carrier power.
\[ 1800 = \frac{3}{2}P_c \Rightarrow P_c = 1800\times \frac{2}{3} = 1200\,W \] Final Answer: \[ \boxed{1200\ W} \]