Venkat, you bring up a very good point.
Two or more branches in parallel, connected at both extremes, will have the same voltage. The braches can't be mere lines, as you said, since these lines represent bare conductors which can be applied as short circuits. Branches have some sort of resistive or reactive elements in them. This problem uses resistor loads, so insert resistive elements into the branches you mention and yes, even with different resistance values in each branch, both parallel branches have the same identical voltage across them. Here is a simple KVL equation that supports this [ see attached PDF ].
Here is my argument:
In the picture you provided, the current supplied by the 5V battery will seek the path of least resistance. The entire circuit is a resistive network; except for the short directly in front of the battery. The mesh enclosing the battery and this shorting wire is the path of least resistance. All or most of the battery's current will travel through this path. In doing so, the zero resistance of the short (assuming ideal conductors) will produce a zero votage (use Ohm's law with 0 resistance). This last result with Ohm's law can serve as a first indicator that the voltage, as read across the battery terminals being shorted together, should be zero. Unfortunately, using KVL here does not work for me either. 5 - 0 = 0 is False. I think this is a limitation of KVL, with respect to short-circuits. Just realize that the inequality (5 = 0) is telling you something. Knowing something is up, I resort to Ohm's law, and it tells me that there is zero volts across the short. Or, I recall that for lumped circuit analysis, the potentials at the ends of a perfect conductor are the same (no matter what is connected to these ends), therefore their difference is zero.
Here was my approach to his part of the circuit:
Every Copper conductor has a cross sectional area or width, such as a wire and its AWG or guage, or a PCB trace. The problem did not specify the short's, or any other conductor's, wire gauge or thickness, so I have to assume it conducts whatever current is available. Every Copper conductor also has some resistivity. Nothing of the sort is mentioned in the problem statement, so I have to assume somewhat ideal conductors (zero or near zero resistance) everywhere. This assumption, and either Ohm's law applied to the wire (with 0 resistance), or note that the difference in the potentials at the ends of this wire is zero, results in a potential difference across the wire of zero.
Another point of view:
Looking at it realistically, the battery will supply its rated current, the conductor - causing the short - will heat up (a real conductor with some resistivity), and one of the two (or both) will fail or stop working after some amount of time. It bears mentioning that a real circuit with a power supply would have included some type of over-current protection (like a fuse or circuit breaker), as required by safety standards and codes (NEC/NFPA 70), to stop the threat of over-current which results in overheating situations and potential fire. Nothing of the type is present in this circuit. Continuing on with the analysis, I'm therefore forced to again assume one of the following: the battery fails first (either completely discharges or fails internally such that it stops supplying the 5 volts, and therefore does not drive any current) or, the conductor fails first (it opens like a fuse), or both fail at the same time. Being it is a DC circuit with relatively low levels, I will go with battery failure first (i.e., discharging all the way down) and the conductor surviving. So only the conductor (perhaps with little or no insulation left since it likely melted) placed across the (dead) battery terminals survives. A dead battery supplying a potential of 0 Volts will drive zero current out of its terminals. The short-circuit is still in place. These results suggest an "inert" portion of the circuit, i.e. supplies no power from this end. The grand result is that the remaining short merely supplies the connections that result in a node.
Hopes this helps Venkat...
