for some context: I was self-studying and got confused about this question. This is a past paper question from the CAIE O Levels Board and its reference code is shown in the image attached below. I don't need help with the actual question, but rather the underlying concept.
My Thinking: As seen in part (b), the brightness of light shown on the LDR decreases, and so its resistance increases. Since resistance adds up in series, the total resistance of branch "A" increases and since current divides in parallel, branch "A" gets less current and branch "B" gets more current. Thus, the reading on ammeter 2 increases.
But apparently, my working is wrong and the correct answer is "The reading on ammeter 2 does not change because voltage across resistor N does not change". Now I get it, using the equation V = IR there would be same voltage across branch "B" and its resistance will not be affected, so its current will be the same.
But then does that mean my concept of current dividing across parallel branches was wrong?
Best Answer
I think your underlying misconception (a common one at that) is that you think of voltage sources as current sources.
A voltage source supplies a specific voltage. The current it supplies depends on the resistance of the supplied circuit. A current source supplies a specific current. The voltage it supplies depends on the resistance of the supplied circuit.
Real sources of electricity are neither, but in most cases they are much closer to a voltage source than a current source.
So now apply this to your problem. Assume the power supply is a battery. A battery does not supply a constant current which then has to be divided among the possible paths. It supplies a constant voltage, and additional parallel paths lead to larger currents from the battery. The total current supplied by the battery does have to be divided among the paths, but additional paths increase the total current instead of decreasing the current per path.