Class 12 Exam  >  Class 12 Questions  >  A message of frequency 5 kHz and peak voltage... Start Learning for Free
A message of frequency 5 kHz and peak voltage of 2 V is used to modulate a carrier wave of frequency 2 MHz and a peak voltage of 10 V. The modulation index is​
  • a)
    0.01
  • b)
    0.2
  • c)
    0.1
  • d)
    1
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
A message of frequency 5 kHz and peak voltage of 2 V is used to modula...
It's very simple.

Use formula, Em/ Ec.

where,
(Em-peak voltage of message signal.
Ec - Peak voltage of carrier wave.)

Am = 2/10
= 0.2
Free Test
Community Answer
A message of frequency 5 kHz and peak voltage of 2 V is used to modula...
To find the modulation index, we need to understand the concept of modulation and how it affects the carrier wave. Modulation is the process of modifying a high-frequency carrier wave with a lower-frequency message signal to transmit information. In this case, the message signal has a frequency of 5 kHz and a peak voltage of 2 V, while the carrier wave has a frequency of 2 MHz and a peak voltage of 10 V.

Modulation Index Formula
The modulation index (m) is calculated using the following formula:
m = (Vm / Vc),
where Vm is the peak voltage of the message signal and Vc is the peak voltage of the carrier wave.

Step 1: Convert Frequencies to Hz
Since the given carrier wave frequency is in MHz and the message signal frequency is in kHz, we need to convert both frequencies to Hz for accurate calculations.
Carrier wave frequency = 2 MHz = 2 × 10^6 Hz
Message signal frequency = 5 kHz = 5 × 10^3 Hz

Step 2: Calculate the Modulation Index
Using the formula, we can calculate the modulation index as follows:
Vm = 2 V (given)
Vc = 10 V (given)
m = (2 V / 10 V)
m = 0.2

Final Answer
Therefore, the modulation index is 0.2, which corresponds to option B.
Explore Courses for Class 12 exam

Similar Class 12 Doubts

Read the following text and answer the following questions on the basis of the same: Tuning a radio set: In essence the simplest tuned radio frequency receiver is a simple crystal set. Desired frequency is tuned by a tuned coil / capacitor combination, and then the signal is presented to a simple crystal or diode detector where the amplitude modulated signal, is demodulated. This is then passed straight to the headphones or speaker. In radio set there is an LC oscillator comprising of a variable capacitor (or sometimes a variable coupling coil), with a knob on the front panel to tune the receiver. Capacitor used in old radio sets is gang capacitor. It consists of two sets of parallel circular plates one of which can rotate manually by means of a knob. The rotation causes overlapping areas of plates to change, thus changing its capacitance. Air gap between plates acts as dielectric. The capacitor has to be tuned in tandem corresponding to the frequency of a station so that the LC combination of the radio set resonates at the frequency of the desired station.When capacitive reactance (XC) is equal to the inductive reactance (XL), then the resonance occurs and the resonant frequency is given by ω0 = 1/√LCcurrent amplitude becomes maximum at the resonant frequency. It is important to note that resonance phenomenon is exhibited by a circuit only if both L and C are present in the circuit. Only then do the voltages across L and C cancel each other (both being out of phas e) and the Current amplitude is Vm/R the total source voltage appearing across R. This means that we cannot have resonance in a RL or RC circuit.Resonance frequency is equal to

Read the following text and answer the following questions on the basis of the same:Tuning a radio set: In essence the simplest tuned radio frequency receiver is a simple crystal set. Desired frequency is tuned by a tuned coil / capacitor combination, and then the signal is presented to a simple crystal or diode detector where the amplitude modulated signal, is demodulated. This is then passed straight to the headphones or speaker. In radio set there is an LC oscillator comprising of a variable capacitor (or sometimes a variable coupling coil), with a knob on the front panel to tune the receiver. Capacitor used in old radio sets is gang capacitor. It consists of two sets of parallel circular plates one of which can rotate manually by means of a knob. The rotation causes overlapping areas of plates to change, thus changing its capacitance. Air gap between plates acts as dielectric. The capacitor has to be tuned in tandem corresponding to the frequency of a station so that the LC combination of the radio set resonates at the frequency of the desired station.When capacitive reactance (XC) is equal to the inductive reactance (XL), then the resonance occurs and the resonant frequency is given by ω0 = 1/√LCcurrent amplitude becomes maximum at the resonant frequency. It is important to note that resonance phenomenon is exhibited by a circuit only if both L and C are present in the circuit. Only then do the voltages across L and C cancel each other (both being out of phas e) and the Current amplitude is Vm/R the total source voltage appearing across R. This means that we cannot have resonance in a RL or RC circuit.Resonance may occur in

A message of frequency 5 kHz and peak voltage of 2 V is used to modulate a carrier wave of frequency 2 MHz and a peak voltage of 10 V. The modulation index is​a)0.01b)0.2c)0.1d)1Correct answer is option 'B'. Can you explain this answer?
Question Description
A message of frequency 5 kHz and peak voltage of 2 V is used to modulate a carrier wave of frequency 2 MHz and a peak voltage of 10 V. The modulation index is​a)0.01b)0.2c)0.1d)1Correct answer is option 'B'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A message of frequency 5 kHz and peak voltage of 2 V is used to modulate a carrier wave of frequency 2 MHz and a peak voltage of 10 V. The modulation index is​a)0.01b)0.2c)0.1d)1Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A message of frequency 5 kHz and peak voltage of 2 V is used to modulate a carrier wave of frequency 2 MHz and a peak voltage of 10 V. The modulation index is​a)0.01b)0.2c)0.1d)1Correct answer is option 'B'. Can you explain this answer?.
Solutions for A message of frequency 5 kHz and peak voltage of 2 V is used to modulate a carrier wave of frequency 2 MHz and a peak voltage of 10 V. The modulation index is​a)0.01b)0.2c)0.1d)1Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 12. Download more important topics, notes, lectures and mock test series for Class 12 Exam by signing up for free.
Here you can find the meaning of A message of frequency 5 kHz and peak voltage of 2 V is used to modulate a carrier wave of frequency 2 MHz and a peak voltage of 10 V. The modulation index is​a)0.01b)0.2c)0.1d)1Correct answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of A message of frequency 5 kHz and peak voltage of 2 V is used to modulate a carrier wave of frequency 2 MHz and a peak voltage of 10 V. The modulation index is​a)0.01b)0.2c)0.1d)1Correct answer is option 'B'. Can you explain this answer?, a detailed solution for A message of frequency 5 kHz and peak voltage of 2 V is used to modulate a carrier wave of frequency 2 MHz and a peak voltage of 10 V. The modulation index is​a)0.01b)0.2c)0.1d)1Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of A message of frequency 5 kHz and peak voltage of 2 V is used to modulate a carrier wave of frequency 2 MHz and a peak voltage of 10 V. The modulation index is​a)0.01b)0.2c)0.1d)1Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice A message of frequency 5 kHz and peak voltage of 2 V is used to modulate a carrier wave of frequency 2 MHz and a peak voltage of 10 V. The modulation index is​a)0.01b)0.2c)0.1d)1Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice Class 12 tests.
Explore Courses for Class 12 exam
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev