Question:

For the following question, enter the correct numerical value upto TWO decimal places. If the numerical value has more than two decimal places, round-off the value to TWO decimal places. (For example: Numeric value 5 will be written as 5.00 and 2.346 will be written as 2.35) A 1 kW carrier is modulated to a depth of 80%. The total power in the modulated wave is ____ kW.

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For AM waves, total power increases due to sidebands and is given by \[ P_t = P_c \left(1 + \frac{m^2}{2}\right), \] where \(m\) is the modulation index.
Updated On: Jan 14, 2026
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Correct Answer: 1.32

Solution and Explanation

Step 1: Given: \[ P_c = 1 \text{ kW}, \quad m = 80% = 0.8 \]
Step 2: The total power of an amplitude modulated (AM) wave is given by: \[ P_t = P_c \left( 1 + \frac{m^2}{2} \right) \]
Step 3: Substitute the given values: \[ P_t = 1 \left( 1 + \frac{(0.8)^2}{2} \right) = 1 \left( 1 + \frac{0.64}{2} \right) \]
Step 4: \[ P_t = 1 (1 + 0.32) = 1.32 \text{ kW} \] Final Answer (up to two decimal places): \[ \boxed{1.32} \]
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