JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Semiconductors

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

x \times 1 0^{- 2}

. The value of

x

is ____

Answer: 33.33

Solution

The modulation index is given by

μ = \frac{A _{ma x} - A _{min}}{A _{ma x} + A _{min}}

, where

A_{ma x} = 16

V and

A_{min} = 8

V. Substituting:

μ = \frac{16 - 8}{16 + 8} = \frac{8}{24} = \frac{1}{3} = 0.3333

. Since the answer is in the form

x \times 1 0^{- 2}

, we have

0.3333 = x \times 1 0^{- 2}

, giving

x = 33.33

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Numerical answer