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When carrier wave of 2.5 MHz frequency i...

When carrier wave of 2.5 MHz frequency is amplitude modulated, the resulting AM wave has maximum of 15 V and minimum amplitude of 10 V. The modulation index is

A

0.3

B

0.2

C

0.1

D

0.4

Text Solution

Verified by Experts

The correct Answer is:
B

Molulation Index = `(V_(m))/(v_(c))= (((V_("max")-V_("min")))/(2))/( ((V_("max")+V_("min")))/(2))`
`= (15-10)/(15+10) xx 100%= 20%`
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