# When deriving the instantaneous power equation for an AC power system, I get different answers depending on how I model the V & I signals. Why?

When I define $$\v(t)\$$ & $$\i(t)\$$ like this:

$$\v(t)=V_m\cos(\omega t)\$$

$$\i(t)=I_m\cos(\omega t-b)\$$

where $$\V_m\$$=Voltage Amplitude, $$\I_m\$$=Current Amplitude, $$\\omega\$$=rotational frequency, $$\b\$$=phase shift between $$\v\$$ and $$\i\$$ relative to $$\v\$$.

I get: $$\p(t)=\frac{1}{2}V_mI_m[\cos(b)+\cos(2\omega t-b)]\$$

When I define $$\v(t)\$$ & $$\i(t)\$$ like this:

$$\v(t)=V_m\cos(\omega t-Av)\$$

$$\i(t)=I_m\cos(\omega t-Ai)\$$

I get: $$\p(t)=\frac{1}{2}V_mI_m[\cos(Av-Ai)+\cos(2\omega t-Av-Ai)]\$$, where $$\b=Av-Ai\$$

Can someone tell me why?

• they are the same – JonRB Jul 21 '18 at 20:57
• Can you show that? I can't seem to see it. I can't resolve Av+Ai with b unless Ai<=0 – Steve Jul 21 '18 at 21:05
• Why should you want to do this? And what is the EE content that is relevant? – Andy aka Jul 21 '18 at 21:05
• I am trying to understand if the sign convention imposed on reactive elements is arbitrary or necessarily enforced by this derivation – Steve Jul 21 '18 at 21:08
• The sign convention for reactive impedance is necessary to make the algebra work out correctly when using Ohm's law on reactive components in AC steady state analysis. – mkeith Jul 21 '18 at 22:20