Explanation

(b) The power dissipated in any circuit is a function of the square of voltage across the circuit and the effective resistance of the circuit. Equation of AM wave reveals that it has three components of amplitude $E_{c},\ mE_{c}/2\ and\ mE_{c}/2.$ Clearly; power output must be distributed among these components.
Carrier power, $P_{c} = \frac{\left( E_{c}/\sqrt{2} \right)^{2}}{R}\ \text{=}\frac{E_{c}^{2}}{2R}$
Total power of side bands,
$P_{S} = \ \frac{\left( mE_{c}/2\sqrt{2} \right)^{2}}{R} + \frac{\left( mE_{c}/2\sqrt{2} \right)^{2}}{R} = \ \frac{m^{2}E_{c}^{2}}{4R}$
$\therefore\frac{P_{S}}{P_{C}} = \frac{1}{2}\ m^{2}$
and $P_{T} = P_{C} + P_{S} = P_{C}\ \left( 1 + \frac{m^{2}}{2} \right)$
$\therefore\frac{P_{T}}{P_{C}} = 1 + \frac{m^{2}}{2} \Rightarrow \left( \frac{I_{T}}{I_{C}} \right)^{2} = 1 + \frac{m^{2}}{2}$
Given that, $I_{T} = 8.93\ A,\ I_{C} = 8\ A,\ m = ?$
$\therefore\left( \frac{8.93}{8} \right)^{2} = 1 + \frac{m^{2}}{2}$
Or $1.246 = 1 + \frac{m^{2}}{2}\ or\ \frac{m^{2}}{2} = 0.246$
Or $m = \sqrt{2 \times 0.246} = 0.701 = 70.1\%$