The total power content of an AM wave is 1500 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The total power of an $AM$ wave is given by the formula: $P_t = P_c (1 + \frac{m^2}{2})$,where $P_t$ is the total power,$P_c$ is the carrier power

Vedclass · Accepted Answer

$1000$

Vedclass · Answer

(D) The total power of an $AM$ wave is given by the formula: $P_t = P_c (1 + \frac{m^2}{2})$,where $P_t$ is the total power,$P_c$ is the carrier power,and $m$ is the modulation index.Given: $P_t = 1500 \, W$ and $m = 100 \% = 1$.Substituting the values into the formula: $1500 = P_c (1 + \frac{1^2}{2})$.$1500 = P_c (1 + 0.5) = P_c (1.5)$.$P_c = \frac{1500}{1.5} = 1000 \, W$.Therefore,the power transmitted by the carrier is $1000 \, W$.

The total power content of an $AM$ wave is $1500 \, W$. For $100 \%$ modulation,the power transmitted by the carrier is ........ $W$.

Similar Questions

For $100\%$ modulation $(AM)$,the useful part of the total power radiated is

The maximum amplitude for an amplitude modulated wave is found to be $12\, V$ while the minimum amplitude is found to be $3\, V$. The modulation index is $0.6\, x$ where $x$ is $....$

The frequency range suitable for beyond-the-horizon communication using sky waves is:

In the case of amplitude modulation,to avoid distortion,the modulation index $(\mu)$ should be:

What is the advantage of an optical fibre?

Vedclass Test Series

Exam Paper Generator

Online Exam Module