# Calculating the average power dissipated in an AC circuit

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?

An online calculator gave me a complex number with the real part twice the value that I found.

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.

$$P_{avg} = \frac{1}{\sqrt{2}} V_o \cdot \frac{1}{\sqrt{2}} I_o \cos{150} \\ P_{avg} = \frac{120 \cdot 20}{2} * \hspace{2pt} \text{-}.866 \\ P_{avg} = - 1039.2 \hspace{2pt} \text{W}$$

Negative power in a resistor? Where did I go wrong?

The load is not a resistor. If it were a resistor, there would be no phase difference between the current and voltage waveforms.

In general, a negative power is possible --- it means the load has a power source within it and it is delivering power back to your source rather than taking power from it.

So I don't think you have approached the problem incorrectly.

• Then the average power would be zero? – Ian Nov 2 '17 at 20:44
• Just because it's not a resistor doesn't mean it's just a network or capacitors and inductors. It could be the mains. It could be some machinery. It could be another generator. – The Photon Nov 2 '17 at 20:49

Well, as you wrote the given information we have:

$$\text{P}=\frac{\hat{\text{V}}_{\space\text{load}}}{\sqrt{2}}\cdot\frac{\hat{\text{I}}_{\space\text{load}}}{\sqrt{2}}\cdot\cos\left(\varphi_{\space\text{load}}\right)\tag1$$

Where:

$$\varphi_{\space\text{load}}=\left|\arg\left(\underline{\text{V}}_{\space\space\text{load}}\right)-\arg\left(\underline{\text{I}}_{\space\space\text{load}}\right)\right|\tag2$$

Using the given values we get:

1. $$\hat{\text{V}}_{\space\text{load}}=120\tag3$$
2. $$\hat{\text{I}}_{\space\text{load}}=20\tag4$$
3. $$\varphi_{\space\text{load}}=\left|\arg\left(120e^{\frac{50\pi i}{180}}\right)-\arg\left(20e^{-\frac{5\pi i}{9}}\right)\right|=\frac{5\pi}{6}\tag5$$

So:

$$\text{P}=\frac{120}{\sqrt{2}}\cdot\frac{20}{\sqrt{2}}\cdot\cos\left(\frac{5\pi}{6}\right)=-600\sqrt{3}\approx-1039.23048\space\text{W}\tag6$$