All these bulbs have equal resistance and the equivalent current is 6A. I want to know what will happen if bulb 3 is fused. All the sources I have came across say that the bulb 1 and bulb 2 will glow the same, because voltage is same.
I know I am going wrong but I don't know where I am going wrong. But the equivalent current is sum of all current, so if these are just two parallels, wont it get divided (3A and 3A), and the bulbs will get more current and glow more?
Please help.
You should get some parts, build it, and see what happens. That's the best way to really understand it.
The abstract way to capture what's going on depends on whether the 6A is a constant or a variable. If the power source is delivers constant current, then your analysis is correct: the bulbs get brighter. On the other hand, if your power source is constant voltage, the 6A drops to 4A, and then the bulbs glow the same.
To understand this consider that R=3 and V=6, V being the battery source voltage. With all the bulbs intact, the net resistance is 1 and so current drawn is 6A fitting with your question. Now when one bulb is fused, that part of the circuit may be ignored so that our circuit modifies to a net resistance of 1.5 with net current drawn equal to 4A (6/1.5) which will be divided equally between the remaining bulbs giving them the exact same current as before, i.e. 2A. Thus the bulbs glow the same.
