Modulation index of a amplitude modulated wave is 75%. If its maximum…

The correct answer is Option (1) → \( 8\,\mathrm{V} \)

The modulation index of an amplitude modulated wave is the ratio of the amplitude of the modulating signal to that of the carrier, \( \mu = A_m/A_c \). The envelope of the AM wave reaches its maximum when the signal adds to the carrier.

    \( A_{\max} = A_c + A_m = A_c(1 + \mu) \)

    \( \displaystyle A_c = \frac{A_{\max}}{1 + \mu} = \frac{14\,\mathrm{V}}{1 + 0.75} = \frac{14\,\mathrm{V}}{1.75} = 8\,\mathrm{V} \)

Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), and it also appears to be out of the CUET 2026 syllabus.

NCERT (Pre-rationalised edition, 2022-23):
Amplitude Modulation – “… Let \( c(t) = A_c \sin\omega_c t \) represent carrier wave and \( m(t) = A_m \sin\omega_m t \) represent the message or the modulating signal … The modulated signal \( c_m(t) \) can be written as

    \( \displaystyle c_m(t) = (A_c + A_m \sin\omega_m t)\sin\omega_c t = A_c\left(1 + \frac{A_m}{A_c}\sin\omega_m t\right)\sin\omega_c t \qquad (15.3) \)

… Here \( \mu = A_m/A_c \) is the modulation index; in practice, \( \mu \) is kept \( \le 1 \) to avoid distortion.”