Question

A transmitter supplies $9{\text{ kW}}$ to the aerial when unmodulated. The power radiated when modulated to $40\%$ is:
A. ${\text{ 5 kW}}$
B. ${\text{ 9}}{\text{.72 kW}}$
C. ${\text{ 10 kW}}$
D. ${\text{ 12 kW}}$

Answer 1

Here a transmitter is given which can radiate the power of $9{\text{ kW}}$ when the wave is unmodulated, we have to find the power radiated when this wave will be $40\%$ modulated. For finding this we will use the relation between the power transmitted, power carried and the modulated index.

Formula Used:
${{\text{P}}_t}{\text{ = }}{{\text{P}}_c}{\text{ }}\left( {1{\text{ + }}\dfrac{{{\mu ^2}}}{2}} \right)$
Here, ${{\text{P}}_t}$ is the power of modulated wave, ${{\text{P}}_c}$ is the power carried by unmodulated wave and $\mu$ be the modulated index.

Complete step by step answer:
Whenever a wave travels it carries some power with it. Sometimes this power of wave is also known as energy of wave. Thus when the wave is transmitted then it carries some power with it which helps it in travelling in the medium. According to the question , an unmodulated wave is transmitted which carries a power of $9{\text{ kW}}$. Now we will calculate the power of this radiated wave when it is modulated $40\%$.

Let us consider the power carried by a unmodulated be ${{\text{P}}_c}$, the modulated index of the modulate wave be $\mu$ then the power transmitted by this modulated wave will be given by using the relation as,
${{\text{P}}_t}{\text{ = }}{{\text{P}}_c}{\text{ }}\left( {1{\text{ + }}\dfrac{{{\mu ^2}}}{2}} \right)$
According to question,
${{\text{P}}_c}{\text{ = 9 kW}}$
\Rightarrow \mu {\text{ = 40% = 0}}{\text{.4}}
Now substituting the values we get,
${{\text{P}}_t}{\text{ = }}{{\text{P}}_c}{\text{ }}\left( {1{\text{ + }}\dfrac{{{\mu ^2}}}{2}} \right)$
$\Rightarrow {{\text{P}}_t}{\text{ = 9 }}\left( {1{\text{ + }}\dfrac{{{{\left( {0.4} \right)}^2}}}{2}} \right)$
$\Rightarrow {{\text{P}}_t}{\text{ = 9 }}\left( {1{\text{ + }}\dfrac{{0.16}}{2}} \right)$
$\Rightarrow {{\text{P}}_t}{\text{ = 9 }}\left( {1.08} \right)$
$\therefore {{\text{P}}_t}{\text{ = 9}}{\text{.72 kW}}$
Hence when the same is modulated $40\%$ then the power transmitted is increased to ${\text{9}}{\text{.72 kW}}$.

Therefore the correct option is B.

Note: Modulating percentage of wave must be converted into decimal value before putting it in the equation. Modulated waves can travel up to longer distances as the power carried by them is more than that of carried by unmodulated waves. Modulating index can be denoted by small $m$.