91Ó°ÊÓ

Given that$V\left(t\right)=V_0\sin \left(2\mathrm{Ï€´Ú³Ù}\right)$ and$P\left(t\right)=\frac{{\left[V\right(t\left)\right]}^2}{R}$

$$\begin{gathered} P\left(t\right)=\frac{{\left[V\right(t\left)\right]}^2}{R} \\ =\frac{\left[V_0\sin {\left(2\mathrm{Ï€´Ú³Ù}\right)}^2\right]}{R} \\ =\frac{{V_0}^2}{R}{\sin }^2\left(2\mathrm{Ï€´Ú³Ù}\right) \end{gathered}$$

Hence,it is proved thatrole="math" localid="1646469531336" $P\left(t\right)=\frac{{V_0}^2}{R}{\sin }^2\left(2\mathrm{Ï€´Ú³Ù}\right)$

In the given graph, there is a vertical shift up by $\frac{{V_0}^2}{2R}$units. If we shift the graph down by $\frac{{V_0}^2}{2R}$unit, the given graph has characteristics of a sine function as the graph will pass through origin. So, we view the equation as a sine function $y=A\sin \left(\mathrm{Ó\neg }x\right)$with $\left|A\right|=\frac{{V_0}^2}{2R}$as the curve lies between $-\frac{{V_0}^2}{2R}$and $\frac{{V_0}^2}{2R}$ on y-axis, and period $T=\frac{1}{t}$as one cycle in the graph begins at $t=0$and ends at$t=\frac{1}{f}$. The given sine function will not be negative as power in an ac generator is always positive.
Now,

$$\begin{gathered} \mathrm{Ó\neg }=\frac{2\mathrm{Ï€}}{T} \\ =\frac{2\mathrm{Ï€}}{{\displaystyle \frac{1}{f}}} \\ =2\mathrm{Ï€´Ú} \end{gathered}$$

After adding the vertical shift value, equation will be of the form $y=A\sin \left(\mathrm{Ó\neg }x\right)+c$where is the vertical shift. Hence, the sine function whose graph is given is: localid="1646475784402" $P\left(t\right)=\frac{{V_0}^2}{2R}\sin \left(2\mathrm{Ï€´Ú³Ù}\right)+\frac{{{\mathrm{V}}_0}^2}{2\mathrm{R}}$