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An AM wave is expressed as $e = 10 (1 + 0.6 \cos 2000 π t) \cos 2 x 10^{8} π t$ volts, the min and max values of modulated carrier wave are

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10 V, 20 V

8 V, 20 V

4 V, 8 V

16 V , 4 V

answer is C.

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Detailed Solution

$E_{\max} = (1 + μ) E_{c} = (1 + 0.6) 10 = 16 V E_{m i n} = (1 - μ) E_{c} = (1 - 0 .6) 10 = 4 V$

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