$\eta=90 \%=0.9$
Efficiency $\eta=0.9=\frac{P_{s}}{P_{p}} \Rightarrow P_{s}=0.9 P_{p}$
$\mathrm{V}_{\mathrm{s}} \mathrm{I}_{\mathrm{s}}=0.9 \times \mathrm{V}_{\mathrm{P}} \mathrm{I}_{\mathrm{P}} \quad(\because \mathrm{P}=\mathrm{VI})$
${I_s} = \frac{{0.9 \times 2300 \times 5}}{{230}} = 45{\mkern 1mu} {\text{A}}$
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${V}_{{m}}({t})=10 \sin \left(2 \pi \times 10^{5} {t}\right)$ $volts$ and
Carrier signal
${V}_{{c}}({t})=20 \sin \left(2 \pi \times 10^{7} {t}\right)$ $volts$
The modulated signal now contains the message signal with lower side band and upper side band frequency, therefore the bandwidth of modulated signal is $\alpha\, {kHz}$. The value of $\alpha$ is :
