An amplitude modulated waves is represented by expression \(v_m=5(1+0.6\cos6280t)\sin(211×10^4t) \text{ volts.} \) The minimum and maximum amplitudes of the amplitude modulated wave are, respectively:
(1) 5V, 8V
(2) 5/2V, 8V
(3) 3V, 5V
(4) 3/2V, 5V
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1 Answer
Correct option according to NTA is \({5\over 2}V , 8V \)
Solution ::
\(V_m=5(5(1+0.6 \cos 6280t)\sin(2π×10^4t)\)
\(V_m=(5+3 \cos 6280t)\sin(2π×10^4t)\)
\(V_{max}= 5+3=8\)
\(V_{min}= 5-3=2\)
Correct answer can be (2V,8V)
But JEE main is a exam where sometimes correct answer not given instead likely to be correct nearby accurate answer so go according as 2.5 is nearby 2