Find the modulation index of an AM wave having 8 V variation where maximum amplitude of the AM wave is 9 V.

We are given that the variation (peak-to-peak change) in the amplitude of the AM wave is 8 V and the maximum amplitude of the AM wave is $$A_{\max} = 9$$ V.

We know that for an amplitude-modulated wave, the maximum and minimum amplitudes are related to the carrier amplitude $$A_c$$ and the modulating signal amplitude $$A_m$$ by $$A_{\max} = A_c + A_m$$ and $$A_{\min} = A_c - A_m$$. The "8 V variation" means the total swing from minimum to maximum, so $$A_{\max} - A_{\min} = 8$$ V. This gives us $$A_{\max} - A_{\min} = (A_c + A_m) - (A_c - A_m) = 2A_m = 8$$, hence $$A_m = 4$$ V.

Now, since $$A_{\max} = A_c + A_m = 9$$, we get $$A_c = 9 - 4 = 5$$ V.

The modulation index is defined as $$\mu = \dfrac{A_m}{A_c} = \dfrac{4}{5} = 0.8$$.

Hence, the correct answer is Option A.