Physics · Ch 13 — AC Circuits

Average Power Associated with Resistance

13.6.1

Average Power Associated with Resistance

For a pure resistor, current and voltage are exactly in phase: $e=e_0\sin\omega t$ and $i=i_0\sin\omega t$ . The instantaneous power is therefore $P=ei=e_0i_0\sin^2\omega t$ . Averaging $\sin^2\omega t$ over one complete cycle gives exactly $\dfrac{1}{2}$ (a standard result, since $\sin^2\omega t=\dfrac{1-\cos2\omega t}{2}$ and the $\cos2\omega t$ term itself averages to zero over a whole cycle), so the average power comes out to $P_{av}=\dfrac{1}{2}e_0i_0=\dfrac{e_0}{\sqrt2}\cdot\dfrac{i_0}{\sqrt2}=e_{rms}\,i_{rms}$ . …

Misc Ex.13.7RMS current and net power consumed by a resistor from its rms rating

Worked out. A 100 $\Omega$ resistor is connected to a 220 V (rms), 50 Hz supply. The rms current follows directly from Ohm's law applied to rms values: $i_{rms}=e_{rms}/R=220/100=2.2$ A. Since a pure resistor has zero phase difference between voltage and current (power factor $\cos\phi=1$ ), the net (average) power consumed over a full cycle is simply $P_{av}=e_{rms}\,i_{rms}=220\times2.2=484$ W -- this example fixes that for a PURELY resistive load, the $\cos\phi$ factor of the general power formula can be dropped entirely (it equals 1), so average power reduces to the same simple $VI$ f …

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