Fraction of total power carried by side bands is given by:

\frac{P_{S}}{P_{T}} = m^{2}

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\frac{P_{S}}{P_{T}} = \frac{1}{m^{2}}

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\frac{P_{S}}{P_{T}} = \frac{2 + m^{2}}{m^{2}}

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\frac{P_{S}}{P_{T}} = \frac{m^{2}}{2 + m^{2}}

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Solution

The correct option is C $\frac{P_{S}}{P_{T}} = \frac{m^{2}}{2 + m^{2}}$
Total power of AM wave is

$P_{T} = \frac{m^{2} E_{c}^{2}}{4 R} + \frac{E_{c}^{2}}{2 R}$

and Side band power is given by

$P_{s} = \frac{m^{2} E_{c}^{2}}{4 R}$

So $\frac{P_{s}}{P_{T}} = \frac{\frac{m^{2} E_{c}^{2}}{4 R}}{\frac{m^{2} E_{c}^{2}}{4 R} + \frac{E_{c}^{2}}{2 R}}$

on simplifying $\frac{P_{s}}{P_{T}} = \frac{m^{2}}{2 + m^{2}}$

Hence option D is correct.

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Similar questions

(P_{S} / P_{T})

P t (s) | H_{2} (g, 1 b a r) | H^{+} (a q, 1 M) | | M^{4 +} (a q), M^{2 +} (a q) | P t (s)

E_{c e l l} = 0.092

\frac{[M^{2 +} (a q)]}{[M^{4 +} (a q)]} = 10^{x}

: E_{M^{4 +} / M^{2 +}}^{0} = 0.151 V; 2.303 \frac{R T}{F} = 0.059 V

∠ T P S = ∠ Q P R

{PR}^{2} = {PS}^{2} + QR \times ST + {(\frac{QR}{2})}^{2}

{PQ}^{2} = {PS}^{2} - QR \times ST + {(\frac{QR}{2})}^{2}

p_{s} = 12 sin ϕ m C / m^{2}

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