I have a question about short circuits that I haven't been able to solve.
From this we know: The total voltage V is constant, and the total current I is inversely proportional to the total resistance R.
Due to this being a parallel circuit, the voltage of each branch is the same and is equal to the total voltage: $$ V_{AB}=V_{CD}=V(\text {constant}) $$
Or we can write this as: $$ I_{AB}R_{AB}=I_{CD}R_{CD}=V $$
Since current is inversely proportional to the total resistance, when RCD decreases, ICD will increase. At the same time, the total current I will also increase. $$I=I_{AB}+I_{CD}=\frac{V}{R}$$
But IAB will not change, otherwise the total voltage will change too, but the total voltage is constant.
Now comes my question: If this is so, let's suppose RCD approaches 0, which means there will be a short circuit. It is said that all the current will travel through the path of least resistance, which means no current passes through AB and thus the bulb won't light up.
However, in theory, there is still a current IAB flowing through the bulb notwithstanding the decrease in RCD which only affects ICD right? Shouldn't the bulb still light up then?
How do we square these two facts?
PS: I understand that if there is internal resistance, what I said no longer holds. But what if we imagine there was no internal resistance, does it mean that the bulb will light up?