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JEE Main & Advanced Physics Communication System / संचार तंत्र Question Bank Communication

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

    A)            10%                                          

    B)            20%

    C)            25%                                          

    D)            50%

    Correct Answer: D

    Solution :

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


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