# Calculating instantaneous and average power

sorry that this is such a basic question but I need to calculate the average and instantaneous power for the following circuit:

A voltage supply in series with a resistor: V1 = 150sin(wt)V and R1 = 25ohms

I understand that t = time, w = 2*pi*f but what is confusing is that I am not given either time or frequency do I just assume that sin(wt) = 0 and V = 150V?

If this is the case then is the instantaneous power just P = VI (or (V^2)/R)?

How do i calculate average power if Pavg = VIcos(phi)?

Any help to these questions would be amazing. This is a practice paper for an exam not homework.

thanks

## 1 Answer

The instantaneous power is, as you surmised, $$\ P= V^2/R=900 \sin^2(\omega t)\,\rm W \$$.

The average value of $$\ \sin^2(\omega t) \$$ over any whole number of cycles is $$\ 1/2 \$$, so the average power is $$\ 450\,\rm W \$$.

To address your other formula, $$\P=VI\cos\phi\$$, observe that the peak voltage is $$\150\,\rm V\$$, so the RMS voltage is $$\150/\sqrt2\,\rm V\$$, and the RMS current is $$\6/\sqrt 2\,\rm A\$$. Since the load is resistive, the phase angle $$\\phi\$$ is zero, and the (average) power is $$\900/2 = 450\,\rm W\$$ as before.