Home
Class 12
PHYSICS
The maximum peak-to-peak voltage of an A...

The maximum peak-to-peak voltage of an `AM` wave is `16 mV` and the minimum peak-to-peak voltage is `4 mV`. The modulation factor is equal to

A

`2/3`

B

`2/5`

C

`1/3`

D

`3/5`

Text Solution

Verified by Experts

The correct Answer is:
D
`E_"max"=16/2`=8 mV
`E_"min" = 4/2 =2 ` mV
`therefore mu=(E_"max"-E_"min")/(E_"max"+E_"min") =(8-2)/(8+2)=6/10=3/5`
Promotional Banner

Topper's Solved these Questions

  • COMMUNICATION SYSTEMS

    MARVEL PUBLICATION|Exercise TEST YOUR GRASP -20|10 Videos

  • CIRCULAR MOTION

    MARVEL PUBLICATION|Exercise TEST YOUR GRASP-20|1 Videos

  • CURRENT ELECTRICITY

    MARVEL PUBLICATION|Exercise MCQ|151 Videos

Similar Questions

Explore conceptually related problems

The maximum peak to peak voltage of an AM wire is 24 mV and the minimum peak to peak voltage is 8 m V. The modulation factor is

The maximum peak ro peak voltage of an AM wire is 24 mV and the minimum peak to peak voltage is 8mV . The modulation factor is

The maximum peak to peak voltage of an AM wave is 16 m V and the minimum peak to peak voltage is 8 m V. What is the modulation index ?

If minimum voltage in an AM wave was found to be 2V and maximum voltage 10V . The % modulation index

If minimum voltage in an AM wave was found to be 1.1 V and maximum voltage 10V . The % modulation index

The peak voltage in a 220 V AC source is

If the maximum and minimum voltages of an AM wave are V_(max.)andV_(min.) respectively then modulation factor -

if the maximum and minimum voltage of AM wave are V_(max) and V_(min) , respectively then modulation factor

The peak voltage of an ac supply is 440 V, then its rms voltage is