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An amplitude modulated wave is represented by the equation $V_m=5(1+0.6 \cos 6280 t) \sin \left(211 \times 10^4 t\right) V$. The minimum and maximum amplitudes of the amplitude modulated wave are, respectively :
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Verified Answer
The correct answer is:
$2 V, 8 V$
$A_{\max }=A_c(1+\mu)$
The minimum amplitude is given by,
$A_{\text {min }}=A_c(1-\mu)$
From the given expression,
$V_m=5(1+0.6 \cos 6280 t) \sin \left(211 \times 10^4 t\right)$
Modulation index, $\mu=0.6$
$\begin{aligned} & A_c=5 \mathrm{~V} \\ & \Rightarrow A_{\max }=5(1+0.6)=8 \mathrm{~V} \\ & \Rightarrow A_{\min }=5(1-0.6)=2 \mathrm{~V}\end{aligned}$
Hence, option (2) is correct.
The minimum amplitude is given by,
$A_{\text {min }}=A_c(1-\mu)$
From the given expression,
$V_m=5(1+0.6 \cos 6280 t) \sin \left(211 \times 10^4 t\right)$
Modulation index, $\mu=0.6$
$\begin{aligned} & A_c=5 \mathrm{~V} \\ & \Rightarrow A_{\max }=5(1+0.6)=8 \mathrm{~V} \\ & \Rightarrow A_{\min }=5(1-0.6)=2 \mathrm{~V}\end{aligned}$
Hence, option (2) is correct.
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