I emailed my teacher a sample of our question...here's my email to him:
The microphone preamp is powered by 6 volts dc and has a current draw of 500mAh (with phantom power off). The preamp also provides 48 volt phantom power to the microphone. The draw of the microphone is 4mAh.
What is the TOTAL draw of the preamp when suppling 48 volt phantom power to the microphone?
and here is his reply:
I would suggest that we take the power consumed by the preamp. 6 Volts X .5 Amperes = 3 Watts and add the power consumed by the mike, 48 Volts X .004 Amperes = .192 Watts and add the powers together which gives 3.192 Watts. Now divide the 3.192 Watts by 6 Volts and you will have .532 Amps or 532 MA.
Now since this calculation assumes 100% voltage conversion efficiency, which doesn't exist, I would expect to see slightly more total current. Probably something like 550 MA or so which would represent a 64% conversation efficiency which would seem likely.
So it seems that we can't just add the mics in...you're right mr skalinder!