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 Here is an image of the circuit

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.



1 Answer 1


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.


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