
The total power content of an AM wave is $1500W$ . For $100\% $ modulation, the power transmitted by the carrier is
$\left( A \right){\text{ 500W}}$
$\left( B \right){\text{ 700W}}$
$\left( C \right){\text{ 750W}}$
$\left( D \right){\text{ 1000W}}$
Answer
563.1k+ views
Hint: So in this question, we can see that the modulation index is $100\% $ and if we convert it into the value we can say $\mu = 1$ , By using the formula, $\dfrac{{{P_t}}}{{{P_c}}} = 1 + \dfrac{{{\mu ^2}}}{2}$ . And from this, we will calculate the power transmitted by the carrier.
Formula used
The ratio of total power by the transmitted power is given by,
$\dfrac{{{P_t}}}{{{P_c}}} = 1 + \dfrac{{{\mu ^2}}}{2}$
Here, ${P_t}$ , will be the total power content, ${P_c}$ , will be the total power transmitted by the charge carrier, $\mu $ , will be the modulation index.
Complete step by step answer:
So from the question we have the values are given, where ${P_t} = 1500W$ and $\mu = 1$ . So by using the formula,
$ \Rightarrow \dfrac{{{P_t}}}{{{P_c}}} = 1 + \dfrac{{{\mu ^2}}}{2}$
Or the above formula can also be written as
$ \Rightarrow {P_t} = {P_c}\left( {1 + \dfrac{{{\mu ^2}}}{2}} \right)$
So now on substituting the values, we will get the equation as
$ \Rightarrow 1500 = {P_c}\left( {1 + \dfrac{{{1^2}}}{2}} \right)$
Now on solving thee braces first, so we will get the equation as
$ \Rightarrow 1500 = {P_c}\left( {\dfrac{3}{2}} \right)$
And from here solving for the value of transmitted power by the carrier, we get the equation as
$ \Rightarrow {P_c} = \left( {\dfrac{{2 \times 1500}}{3}} \right)W$
And on solving the above equation we will get the values as
$ \Rightarrow {P_c} = 1000W$
Therefore, the power transmitted by the carrier is $1000W$ .
Correct option is (D).
Note:
Here we can see that for just making the students confused the percentage value of modulation is given. So we should keep in mind that if the percentage value is given then first we have to get the normal value of it. And then substituting the values, also to memorize the formula I would recommend memorizing the formula for the ratio of power transmitted and the power carrier as it will help to get any value associated with the formula easily.
Formula used
The ratio of total power by the transmitted power is given by,
$\dfrac{{{P_t}}}{{{P_c}}} = 1 + \dfrac{{{\mu ^2}}}{2}$
Here, ${P_t}$ , will be the total power content, ${P_c}$ , will be the total power transmitted by the charge carrier, $\mu $ , will be the modulation index.
Complete step by step answer:
So from the question we have the values are given, where ${P_t} = 1500W$ and $\mu = 1$ . So by using the formula,
$ \Rightarrow \dfrac{{{P_t}}}{{{P_c}}} = 1 + \dfrac{{{\mu ^2}}}{2}$
Or the above formula can also be written as
$ \Rightarrow {P_t} = {P_c}\left( {1 + \dfrac{{{\mu ^2}}}{2}} \right)$
So now on substituting the values, we will get the equation as
$ \Rightarrow 1500 = {P_c}\left( {1 + \dfrac{{{1^2}}}{2}} \right)$
Now on solving thee braces first, so we will get the equation as
$ \Rightarrow 1500 = {P_c}\left( {\dfrac{3}{2}} \right)$
And from here solving for the value of transmitted power by the carrier, we get the equation as
$ \Rightarrow {P_c} = \left( {\dfrac{{2 \times 1500}}{3}} \right)W$
And on solving the above equation we will get the values as
$ \Rightarrow {P_c} = 1000W$
Therefore, the power transmitted by the carrier is $1000W$ .
Correct option is (D).
Note:
Here we can see that for just making the students confused the percentage value of modulation is given. So we should keep in mind that if the percentage value is given then first we have to get the normal value of it. And then substituting the values, also to memorize the formula I would recommend memorizing the formula for the ratio of power transmitted and the power carrier as it will help to get any value associated with the formula easily.
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