Why does all the current flow through a short circuit if its voltage drop is considered zero? Path of least resistance vs. short circuit
I know the path of least reisistance has been clarified already However, to derive the equations you need to assume that the voltage of each parallel component is equal. But we know for a short circuit current the voltage is zero (By Ohm's law!) So there is a glaring contradiction!
 A: In analyzing circuits one must always consider the possibility that things you've ignored in some situations cannot be ignored in others.
For instance, there is always the possibility of an internal series resistance in a voltage supply or a parallel resistance in a current supply.  Usually we ignore those if the external resistance is much higher than the internal series resistance  (or lower than the internal parallel resistance).
In the case of the short circuit in the external wiring, the voltage of the EMF will be dropped across the internal resistance, causing the supply to get very warm or fail.
For the circuit on the left, the total current out of the supply will be $V_s/25.1 \, \Omega$ so the total voltage across the parallel circuit will be
$$V_\text{parallel} = (V_s / 25.1 \, \Omega) 25\, \Omega = 0.996V_s \, .$$
The internal resistance might be reasonably ignored, depending on the precision required in the application. The power consumed by the internal resistor will be $0.00016 \, V_s^2 / \Omega$.
For the circuit on the right, assuming an ideal short ($R_\text{short}=0$), the current supplied will be $V_s/0.1 \, \Omega$, the external voltage will be zero and the power consumed by the internal resistor will be $10V_s^2 / \Omega,$ more than ten thousand times larger.
Even if the short circuit is $0.01\,\Omega$, the current will be $V_s / 0.11\,\Omega$, the external voltage drop $0.091V_s$, and the power consumed internally will be $8.3V_s^2/\Omega$.
In the circuit on the right, the internal resistance can't be ignored.
A: yes, there is $\epsilon$ Volts in the 2 branches, but one is $\epsilon'$ Ohms and the other a lot more. So I = U/R really hugely prefers the very smaller denominator. By they inverse ratio of resistance.
$\epsilon$, $\epsilon'$ : because there is no such things than zero and infinity in real (classical) physical world.
