AC, phasors and power

Choose AC and give an ω. Elements keep their own units — henries and farads, not reactances — and the solver does the conversion.

e,1,0,10
r,1,2,50
l,2,0,0.05

At ω = 1000 rad/s the inductor is 50 Ω of reactance, so the impedance is 50 + 50j:

You may also give reactances directly, as ohms, if that is how the problem is stated — a resistor-valued l is not a thing, so put the reactance in as an impedance and leave ω out of it.

Rectangular or polar is a display choice: tick Show AC answers as polar phasors in Settings and 0.1 - 0.1j reads 0.1414∠-45°. Nothing about the answer changes. The aa mini-tool does the same conversion for one value you type.

The four power answers

In AC an element reports more than in DC:

e,1,0,10
r,1,2,30
l,2,0,0.04

At ω = 1000 that is 30 + 40j — a 3-4-5 triangle — and the source reports complex power −0.6 − 0.8j VA. So P = 0.6 W, Q = 0.8 var, |S| = 1.0 VA and the power factor is 0.6, lagging.

An inductor reports no average power, only complex. That is not an omission: a pure reactance consumes none, and a zero printed every time would be noise.

The pf mini-tool takes a complex power or an impedance and returns the power factor with its lead/lag sense.

Three-phase

There is no three-phase mode, and none is needed. A three-phase circuit is an AC circuit with three sources whose values carry the phase — 120, 120*exp(-2j*pi/3), 120*exp(2j*pi/3) — and Y or Δ is just how you wire the nodes. Balanced and unbalanced are the same description with different values.

Line and phase quantities are then read off the elements directly: there is no separate answer for them, because each is some element's own current or voltage.