Take the wrench/car battery story and lets say that the wrench has a 1 ohm resistance including contact resistance. And the battery has a cold cranking amp rating of 400 and delivers 12 volts. The internal resistance of the battery is ~0.030 ohms.
Power = I^2*R and if I assume a large contact resistance of ~1 ohm and neglecting the 0.030 ohms, 160,000 Watts could be produced in a very short amount of time.
Whilst I am fully in agreement with the spirit of your post, the electrical theory and mathematics leave a lot to be desired.
If the wrench has a resistance of 1 ohm and is connected across a 12v car battery, the current will be
I = V/R = 12/1 = 12 amps.
the power dissipated in the wrench will be
W = I²R = 12² x 1 =144 watts
However, a more likely scenario is that the wrench resistance is much less than 1ohm, maybe 0.01 ohm.
Also, the battery, if it has a CCA rating of 400 amps, its internal resistance will be much less than 0.03 ohm.
You cant just calculate R = V/I = 12/400 = 0.03, where is the voltage for the starter motor?
When the 400amps are flowing there will be 9 or 10 volts across the load (the starter motor).
JimB