Why a pure resistive AC circuit absorbs power at all times according to this equation

Here is the equation of average power in a pure resistive AC circuit:

"Fundamentals of electric circuits - Alexander Sadiku"

How this equation imply that the circuit absorbs power all times, I understand that resistor only absorb power and there is no reactive elements in the circuit like capacitors and inductors, but I want to know how it was concluded from the equation above.

• Only if V is not zero.... – Solar Mike Mar 9 at 21:12
• Capacitors and inductors are reactive elements, not active ones. – Hearth Mar 9 at 21:14
• @Hearth Yeah you are right, I edited it.Thanks. – Bishoy Essam Mar 9 at 21:30
• @BishoyEssam The portion you quoted makes a circular argument. They transform one expression into another by assuming their result. They should have instead proven that only the resistive portion of an impedance dissipates energy and that, for any one given frequency, time drops out (due to the fact that $1=\operatorname{sin}^2+\operatorname{cos}^2$.) Once you see that time disappears in this case, the claim is made. – jonk Mar 9 at 22:01

I suspect that what the author is suggesting is that if the equations demonstrate that the maximum possible power is being extracted from the AC then the circuit is purely resistive. Since $$\ V_{rms} = \frac {1}{\sqrt 2} V_{max} \$$ and $$\ I_{rms} = \frac {1}{\sqrt 2} I_{max} \$$ then $$\ P = V_{rms}I_{rms} = \frac {1}{\sqrt 2} V_{max} \frac {1}{\sqrt 2} I_{max} = \frac {1}{2} V_{max}I_{max} \$$. A reactive circuit would always give a lesser value for power.