JEE Main 12 Jan 2019 Morning Question 26
Question: A 100 V carrier wave is made to vary between 160 V and 40 V by a modulating signal. What is the modulation index? [JEE Main Online Paper Held On 12-Jan-2019 Morning]
Options:
A) 0.4
B) 0.5
C) 0.6
D) 0.3
Show Answer
Answer:
Correct Answer: C
Solution:
- Modulation Index $ \mu =\frac{{A _{\max }}-{A _{\min }}}{{A _{\max }}+{A _{\min }}}=\frac{160-40}{160+40}=\frac{120}{200}=0.6 $ f
