My two cents:
First, we are talking about a single locomotive here.
If the power pack is "powerful enough" to run the loco while it is on track near the wiring terminals, then it is powerful enough to run the loco further away too.
If the power pack is "powerful enough" to run the loco on a small loop, then it is powerful enough to run the loco on a large multi-looped spaghetti bowl.
As 60103 mentions, the problem is the voltage drop which occurs as you get further away from the power pack. And if you have poor connections in the track, this voltage drop is going to occur whether you are using a $10 train set pack or a $200 fancy pack.
So, a high dollar "more powerful" pack is not a cure for poor connections which place high resistance in series with the locomotive. It is all Ohm's Law.
Certainly, a pack with more watts is necessary for running multiple locos. But we are only talking about a single loco.
Here is some math for anyone who is interested:
$10 cheapie pack vs $200 good pack
We have set the throttle at a position where each pack is producing 12 volts. Somewhat arbitrarily chosen, our locomotive motor has 35 ohms of resistance and it is sitting on track close to the pack, and the overall track connection resistance = 1 ohm.
In both cases, current = voltage divided by ohms. To calculate the ohms, we must add the track ohms to the loco ohms because this is a series circuit. The ohms = 35 for loco + 1 ohm for track = 36 total. So the current to the circuit is 12 volts divided by 36 ohms equals 1/3 amp. The actual voltage seen at the loco = current x ohms of the loco = 1/3 amp x 35 ohms = 11.67 volts. We lost 1/3 of a volt in the track. Not too bad, and the loco runs near full speed.
Now same deal, packs putting out 12 volts, but the loco is way over on the other side of the spaghetti bowl, and the track connection resistance is 25 ohms because of poor rail connections. So, current = 12 volts divided by resistance, resistance = 25 ohms in track + 35 ohms of loco = 60 ohms. Current = 12 divided by 60 = 1/5 amp flowing in circuit. Actual voltage at loco equals current x ohms of loco = 1/5 amp x 35 ohms = 7 volts. So, the loco on the other side of the track is only seeing 7 volts. We are losing 5 volts in the poor connections and the will run at barely over half speed versus when it is close to the power pack connections.
Notice all this has nothing to do with "power" which is watts, which is voltage x current. It all has to do with the voltage and the resistance of the path.
Now, I'll grant you that the nice power pack will give much better throttle response, and all kinds of other pluses, and may actually give out a purer DC versus the cheapie power pack, and this may help the above problem a little bit, but an expensive pack is in no way a fix for poor track connections.
Run as many feeder wires as you possibly can, to all portions of the track.
Oh... I have been in model trains for barely over 2 years, but am an electrician. If there is something I am missing here, please let me know!