
For an AM-system the total power of modulated signal is 600W and that of carrier is 400W, the modulation index is
(A) 0.25
(B) 0.36
(C) 0.54
(D) 1
Answer
216.6k+ views
Hint The modulation index for the total power of the modulated signal and carrier signal is related to total power and carrier power is given by ${P_T} = {P_C}\left( {1 + \dfrac{{{m_a}^2}}{2}} \right)$ . Substitute the power values and simplify to get the modulation index.
Complete step-by-step answer
The process of changing the amplitude of a carrier wave in accordance with the amplitude of the audio frequency signal is known as amplitude modulation (AM).

Power modulation index is given by,
${P_T} = {P_C}\left( {1 + \dfrac{{{m_a}^2}}{2}} \right)$
Where, ${P_T}$ is the total power modulated, ${P_C}$ is the carrier power, and ${m_a}$ is the modulation index.
It is given in the question that,
Total power of modulated signal$ = 600W$
Carrier power$ = 400W$
Substitute the given data in the above expression.
$600 = 400\left( {1 + \dfrac{{m_a^2}}{2}} \right)$
$\dfrac{3}{2} = 1 + \dfrac{{m_a^2}}{2}$
$\dfrac{{m_a^2}}{2} = \dfrac{1}{2}$
${m_a} = 1$
Hence, the modulation index is 1 and the correct option is D.
Note Maximum power in the AM without distortion will occur when ${m_a} = 1$
If ${I_C}$is unmodulated current and${I_T}$ is total modulated current then,
$\dfrac{{{P_T}}}{{{P_C}}} = {\left( {\dfrac{{{I_T}}}{{{I_C}}}} \right)^2}$
The ratio of change of amplitude of carrier wave to the amplitude of original carrier wave is called modulation factor or degree of modulation to modulation index. It is also called modulation depth.
Complete step-by-step answer
The process of changing the amplitude of a carrier wave in accordance with the amplitude of the audio frequency signal is known as amplitude modulation (AM).

Power modulation index is given by,
${P_T} = {P_C}\left( {1 + \dfrac{{{m_a}^2}}{2}} \right)$
Where, ${P_T}$ is the total power modulated, ${P_C}$ is the carrier power, and ${m_a}$ is the modulation index.
It is given in the question that,
Total power of modulated signal$ = 600W$
Carrier power$ = 400W$
Substitute the given data in the above expression.
$600 = 400\left( {1 + \dfrac{{m_a^2}}{2}} \right)$
$\dfrac{3}{2} = 1 + \dfrac{{m_a^2}}{2}$
$\dfrac{{m_a^2}}{2} = \dfrac{1}{2}$
${m_a} = 1$
Hence, the modulation index is 1 and the correct option is D.
Note Maximum power in the AM without distortion will occur when ${m_a} = 1$
If ${I_C}$is unmodulated current and${I_T}$ is total modulated current then,
$\dfrac{{{P_T}}}{{{P_C}}} = {\left( {\dfrac{{{I_T}}}{{{I_C}}}} \right)^2}$
The ratio of change of amplitude of carrier wave to the amplitude of original carrier wave is called modulation factor or degree of modulation to modulation index. It is also called modulation depth.
Recently Updated Pages
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Electricity and Magnetism Explained: Key Concepts & Applications

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

NCERT Solutions For Class 11 Physics Chapter 8 Mechanical Properties Of Solids

Motion in a Straight Line Class 11 Physics Chapter 2 CBSE Notes - 2025-26

NCERT Solutions for Class 11 Physics Chapter 7 Gravitation 2025-26

Understanding Atomic Structure for Beginners

