# Instantaneous power

A sinusoidal voltage source v =10V sin(ωt) is connected across a 1k resistor.

1. Make a sketch of p(t), the instantaneous power supplied by the source.
2. Determine the average power supplied by the source.
3. Now, suppose that a square wave generator is used as the source. If the square wave signal has a peak-to-peak of 20 V and a zero average value, determine the average power supplied by the source.
4. Next, if the square wave signal has a peak-to-peak of 20 V and a 10 V average value, determine the average power supplied by the source.

This image has the picture of my partial solution: • Can you mark up $t_1$ and $T$ on your graph so we know what you're talking about. The image quality is not good. See if you can improve the contrast. – Transistor Apr 14 at 13:39
• @Transistor The image contains my solution to the average power (2), kindly help with the others. – Ademola Apr 14 at 16:43
• Draw the voltage and power graphs for Q3 and things should get quite clear for that question. Post a (better) photo into your question. – Transistor Apr 14 at 17:08
• @Transistor I've been trying to upload my new solution but it fails – Ademola Apr 14 at 18:19

## 1 Answer

Your answer to part 2 is correct (although I didn't check all of the intermediate steps). The plot of instantaneous power is a sine wave that has a minimum of zero and a maximum of $$\\frac{{10 V}^2}{1000 \Omega} = 0.1 W\$$. The average value of this waveform is half the peak, or 0.05 W.

The same analysis for the square waves is quite straightforward. In either case, there are only two values of instantaneous power that you need to consider.

• Thanks for verifying the answer for me. But I don't understand what to do to the 3rd and 4th question – Ademola Apr 14 at 18:59
• With a square wave, there are only two values of voltage that you need to consider over the entire cycle. Compute the instantaneous power for both voltages, then average them together. For example, for part 3, the two voltages are +10V and -10V. – Dave Tweed Apr 14 at 19:46
• after doing as you've said I got 0.1watt, is this correct? Better still you can put your explanation down in written form for better understanding. Thanks – Ademola Apr 14 at 19:56