# Why is carrier wave affected by percentage modulation?

In amplitude modulation, Power of a carrier wave is represented by : $$P_c = \frac{A_c ^2}{2R}$$

Which is independent of modulation index! Question: At what factor is the power of carrier wave dependent on!?

A conventional AM signal can be written as

$$x_{AM}(t)=A_c\big(1+\alpha\; m(t)\big)\cos(\omega_ct)\tag{1}$$

where $$\alpha$$ is the modulation index and $$m(t)$$ is the message signal. In Eq. $$(1)$$ it is assumed that the message signal $$m(t)$$ is normalized, i.e., $$\max_t|m(t)|= 1$$. The average power of $$x_{AM}(t)$$ is

$$\overline{x^2_{AM}(t)}=\frac{A_c^2}{2}\left(1+\alpha^2\overline{m^2(t)}\right)\tag{2}$$

If you assume now that the power $$(2)$$ is fixed to some value $$P$$, then the total available power must divided between the carrier term and the message-bearing term in $$(2)$$. Equating $$(2)$$ with the available power $$P$$, and expressing the carrier term $$A_c^2/2$$ in terms of the modulation index and the average power of the message signal gives

$$\frac{A_c^2}{2}=\frac{P}{1+\alpha^2\overline{m^2(t)}}\tag{3}$$

In the figure in your question it looks like they assumed $$\overline{m^2(t)}=\frac12$$, which for a maximum modulation index $$\alpha=1$$ results in

$$\frac{A_c^2}{2}=\frac23 P\tag{4}$$

This value corresponds to the right-most point of the curve labeled "carrier".