- #1
AngusBurger
- 5
- 0
This is an enquiry into whether something is possible, so the numbers might be rubbish to work with.
If I apply a voltage of v=25sin(wt+0.2rads) to a circuit, with a resultant current of i=15sin(wt-0.5rads), can I work out all three sides of a power triangle? I've been working on it for about an hour but with no such luck.
This is what I have done so far:
VRMS*IRMS = 25*15 = 375 Volt-Amps
So long as that is correct, my guess would be that I have to find the True Power, perhaps using the formula P=VIcosϕ, and then I can use Pythagorus theorem to find the volt-amps reactive (var), which would be with the formula var2=VA2-W2. If that's not right, I assume I would have to work backwards in order to calculate an impedance?
Any help with this would be much appreciated. I think that it is possible using only a source voltage and resultant current, but am obviously to new to the subject to properly hack it out.
If I apply a voltage of v=25sin(wt+0.2rads) to a circuit, with a resultant current of i=15sin(wt-0.5rads), can I work out all three sides of a power triangle? I've been working on it for about an hour but with no such luck.
This is what I have done so far:
VRMS*IRMS = 25*15 = 375 Volt-Amps
So long as that is correct, my guess would be that I have to find the True Power, perhaps using the formula P=VIcosϕ, and then I can use Pythagorus theorem to find the volt-amps reactive (var), which would be with the formula var2=VA2-W2. If that's not right, I assume I would have to work backwards in order to calculate an impedance?
Any help with this would be much appreciated. I think that it is possible using only a source voltage and resultant current, but am obviously to new to the subject to properly hack it out.