QJEEM21Ph705e: JEE Main 2021 Physics - Class 12 JEEM21Ph705e Solution: The maximum and minimum amplitude of an amplitude modulated…

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QJEEM21Ph705e: JEE Main 2021 Physics - Class 12 JEEM21Ph705e Solution: The maximum and minimum amplitude of an amplitude modulated…

The maximum and minimum amplitude of an amplitude modulated wave is 16V and 8V respectively. The modulation index for this amplitude modulated wave is x

\times

10^{$-$ 2}. The value of x is __________.

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Accepted Answer

A_{m} = \frac{A_{max} - A_{min}}{2}

A_{c} = \frac{A_{max} + A_{min}}{2}

Modulation index (mi) =

\frac{A_{m}}{A_{c}} = \frac{\frac{A_{max} - A_{min}}{2}}{\frac{A_{max} + A_{min}}{2}} = \frac{A_{max} - A_{min}}{A_{max} + A_{min}}

mi =

\frac{16 - 8}{16 + 8} = \frac{8}{24} = \frac{1}{3} = 0.33

mi = 33

\times

10^{$-$ 2}

∴

x = 33

QJEEM21Ph705e: JEE Main 2021 Physics - Class 12 JEEM21Ph705e Solution: The maximum and minimum amplitude of an amplitude modulated…

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