JEE Main & Advanced Physics Communication System / संचार तंत्र Question Bank Self Evaluation Test - Communication System

  • question_answer
    The maximum peak to peak voltage of an AM wire is \[24\text{ }mV\] and the minimum peak to peak voltage is \[8\text{ }mV\]. The modulation factor in percentage is

    A) \[10%\]

    B) \[20%\]

    C) \[25%\]

    D) \[50%\]

    Correct Answer: D

    Solution :

    [d] Here \[{{V}_{\max }}=\frac{24}{2}=12\,mV\] and  \[{{V}_{\min }}=\frac{8}{2}=4\,mV\] Now, \[m=\frac{{{V}_{\max }}-{{V}_{\min }}}{{{V}_{\max }}+{{V}_{\min }}}=\frac{12-4}{12+4}=\frac{8}{16}=\frac{1}{2}=0.5=50%\]


You need to login to perform this action.
You will be redirected in 3 sec spinner