Alternating current and Average power

[LagrangeEuler]

For a current
$$i(t)=I_0\sin(\omega t+\varphi_0)$$
period is $T=\frac{2\pi}{\omega}$,
Power is defined as
$$p(t)=Ri^2(t)$$. So period of power is not any more $T=\frac{2\pi}{\omega}$. Why then average power is
$$P=\frac{1}{T}\int^T_0p(t)d t$$.
Why are we using the period of current and not of the power $p(t)$?

 

[BvU]

Counter example: a postive sawtooth ($\ 0...A\$) in T: Period doesn't change.

 

[LagrangeEuler]

Ok. But if the period changes why we used the period of the current and not of the power?

 

[Dale]

Ok. But if the period changes why we used the period of the current and not of the power?

For consistency. The fundamental period of the current is always a period of the power, even if it is not the fundamental period of the power.

 

[kuruman]

For a current
$$i(t)=I_0\sin(\omega t+\varphi_0)$$
period is $T=\frac{2\pi}{\omega}$,
Power is defined as
$$p(t)=Ri^2(t)$$. So period of power is not any more $T=\frac{2\pi}{\omega}$. Why then average power is
$$P=\frac{1}{T}\int^T_0p(t)d t$$.
Why are we using the period of current and not of the power $p(t)$?

Who says we are not using the period of the power? Using simple trigonometry,
$$P(t)=I_0^2R\sin^2(\omega t)=\frac{1}{2}I_0^2R[1-\cos(\Omega t)]~~~~~(\Omega \equiv 2\omega)$$Thus, the period of the power $\frac{2\pi}{\Omega}$ is half the period of the current. The average power is $$ \langle P \rangle=\frac{\frac{1}{2}I_0^2R \int_0^{\frac{2\pi}{\Omega}}[1-\cos(\Omega t)]dt}{\int_0^{\frac{2\pi}{\Omega}}dt}=\frac{1}{2}I_0^2R.$$Same result.