- #1

LagrangeEuler

- 717

- 20

[tex]i(t)=I_0\sin(\omega t+\varphi_0)[/tex]

period is ##T=\frac{2\pi}{\omega}##,

Power is defined as

[tex]p(t)=Ri^2(t)[/tex]. So period of power is not any more ##T=\frac{2\pi}{\omega}##. Why then average power is

[tex]P=\frac{1}{T}\int^T_0p(t)d t[/tex].

Why are we using the period of current and not of the power ##p(t)##?