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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?

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  • 1
    \$\begingroup\$ they are the same \$\endgroup\$ – JonRB Jul 21 '18 at 20:57
  • \$\begingroup\$ Can you show that? I can't seem to see it. I can't resolve Av+Ai with b unless Ai<=0 \$\endgroup\$ – Steve Jul 21 '18 at 21:05
  • \$\begingroup\$ Why should you want to do this? And what is the EE content that is relevant? \$\endgroup\$ – Andy aka Jul 21 '18 at 21:05
  • \$\begingroup\$ I am trying to understand if the sign convention imposed on reactive elements is arbitrary or necessarily enforced by this derivation \$\endgroup\$ – Steve Jul 21 '18 at 21:08
  • \$\begingroup\$ 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. \$\endgroup\$ – mkeith Jul 21 '18 at 22:20

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