Q160 / 160 · TS EAMCET 2024 (10 May Shift 2) · Physics

For an amplitude modulated wave, the maximum and minimum amplitudes are found to be 10 V and 2 V respectively. Then the modulation index is

  • A
    $\frac{1}{3}$
  • B
    $\frac{3}{4}$
  • C
    $\frac{1}{5}$
  • Correct answer
    D
    $\frac{2}{3}$

Solution

For amplitude modulated wave, $A_{\max }=10 \mathrm{~V}, \mathrm{~A}_{\min }=2 \mathrm{~V}$ $\therefore$ Modulation index, $\mu=\frac{\mathrm{A}_{\max }-\mathrm{A}_{\min }}{\mathrm{A}_{\max }+\mathrm{A}_{\min }}=\frac{10-2}{10+2}=\frac{2}{3}$

Hence, the correct answer is option: 'D'.

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