This still gives Z = 0.0058 + j0.0072 and a voltage drop of 3.7. If you don't use the effective Z option and apply the power factor to the current, i.e. (
[email protected]º)*(0.00725 +j0.012), then you get a voltage drop of 5.6.
I still can't figure out why myself why it doesn't work the other way...