An electrical impedance can be expressed as a voltage/current ratio and in an AC scenario its value falls within the complex plane for which we have a real and an imaginary component (resistance and reactance). I was wondering if the magnitude of this complex value has any physical meaning?
4
-
1\$\begingroup\$ real part causes losses from current, reactive part stores energy \$\endgroup\$– D.A.S.Commented Sep 14, 2017 at 14:10
-
\$\begingroup\$ Magnitude is directly decomposing to real and the imaginary components while each of them has a physical meaning as above. \$\endgroup\$– Eugene Sh.Commented Sep 14, 2017 at 14:11
-
1\$\begingroup\$ Magnitude ratio determines current = V/|Z| while phase defines shift \$\endgroup\$– D.A.S.Commented Sep 14, 2017 at 14:16
-
\$\begingroup\$ Yes, it has physical meaning. \$\endgroup\$– user57037Commented Sep 14, 2017 at 14:27
Add a comment
|
2 Answers
\$\begingroup\$
\$\endgroup\$
If you apply a sinusoidal voltage v(t) = \$ V_{RMS}\sqrt{2}\cdot\sin(\omega t)\$ to a complex impedance Z and measure the current with an ammeter you will read the scalar value \$I_{RMS} = V_{RMS}/|Z|\$.
answered Sep 14, 2017 at 14:48
\$\begingroup\$
\$\endgroup\$
In a simple scenario, take an X-rated capacitor and close a 120VRMS AC circuit with it; you now have a capacitor as the only load. You know the complex impedance of a capacitor, ideally, has no real part, so it doesn't "waste" any energy, all of it fluctuates back and forth. This would give you a power factor of 0, because all of the power is reactive power.
So back to your question, if you have a complex impedance with non-zero real and imaginary parts, with the magnitude of the impedance you could account for the energy actually "consumed" (Real power, in Watts) AND the energy that moves back and forth (expressed as Reactive Power in Volt-Amper Reactive). This means, is the energy provided to you expressed in Volt-Amper, even if you don't "use" it all (power factor less than 1).
