Total transmitted power of an AM signal at 50 percent modulation

Step-by-step derivation

1. A tone-modulated AM signal is $$s(t) = A_c\big[1 + \mu\cos(2\pi f_m t)\big]\cos(2\pi f_c t),$$ which expands into a carrier plus two sidebands at $f_c \pm f_m$, each of amplitude $\mu A_c/2$.
2. Power is proportional to the square of amplitude. Taking the carrier power as $P_c = A_c^2/2$, each sideband carries $$P_{SB} = \frac{(\mu A_c/2)^2}{2} = \frac{\mu^2}{4}\cdot\frac{A_c^2}{2} = \frac{\mu^2}{4}P_c.$$
3. There are two sidebands, so the total is $$P_t = P_c + 2 \cdot \frac{\mu^2}{4}P_c = P_c\left(1 + \frac{\mu^2}{2}\right).$$
4. Substitute $P_c = 100$ W and $\mu = 0.5$: $$P_t = 100\left(1 + \frac{0.25}{2}\right) = 100\,(1 + 0.125) = \mathbf{112.5\ W}.$$
5. The efficiency point. Only the sidebands carry information: $$\eta = \frac{P_t - P_c}{P_t} = \frac{12.5}{112.5} = 11.1\%.$$ Even at full modulation ($\mu = 1$) efficiency only reaches $33.3\%$ - two thirds of a broadcast AM transmitter's power is spent on a carrier that conveys nothing. That is the entire motivation for DSB-SC and SSB.