This is the question from a very important assignment I am working on: "A load is being driven by an AC source. At a particular time, the voltage across the load is 120cos(wt+50) and the current through the load is 20sin(wt-10). Calculate the average power in Watts dissipated by the load."
My working:
I first convert the current into cosine to give 20cos(wt-100)
Then I use the formula \$\ P=|Vrms||Irms|cos(\theta v -\theta i)\$
which gives me -1039.2 W
I assumed that the question asked for real power since it is the power dissipated in the resistor and thats the only equation the literature gave.
An online calculator gave me a complex number with the real part twice the value that I found. Is this correct? Negative power in a resistor? Where did I go wrong?
Best Answer
Be careful with online calculators and the inputs they expect. Many AC formulas expect RMS values.
Assuming the voltage and current given in your problem are not RMS, you've done it right.
$$ P_{avg} = \frac{1}{\sqrt{2}} V_o \cdot \frac{1}{\sqrt{2}} I_o \cos{150} \\ P_{avg} = \frac{120 \cdot 20}{2} * \hspace{2pt} \text{-}.866 \\ P_{avg} = - 1039.2 \hspace{2pt} \text{W} $$