Q69 / 90 · JEE Main 2022 (26 Jul Shift 2) · Physics

The maximum and minimum voltage of an amplitude modulated signal are $60 V$ and $20 V$ respectively. The percentage modulation index will be

A
$0.5 %$
Correct answer
B
$50 %$
C
$2 %$
D
$30 %$

Solution

If amplitude of carrier wave is $A_{c}$ and amplitude of message wave is $A_{m}$ ,
$A_{m a x} = A_{c} + A_{m}$ and $A_{m i n} = A_{c} - A_{m}$
Modulation index,
$M_{i} = \frac{A_{m}}{A_{c}} = \frac{\frac{A_{m a x} - A_{m i n}}{2}}{\frac{A_{m a x} + A_{m i n}}{2}} = \frac{60 - 20}{60 + 20} = \frac{1}{2} = 50 %$

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

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The maximum and minimum voltage of an amplitude modulated signal are 60 V and 20 V respectively. The percentage modulation index will be

Solution

Relevant Questions

The maximum and minimum voltage of an amplitude modulated signal are $60 V$ and $20 V$ respectively. The percentage modulation index will be