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? Question

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    $\begingroup$ You explicitly expected something else than the correct answer. So you must be wrong. (Yes, the answer key is correct.) Your gut feeling is very reasonable and is a common beginner's misconception. $\endgroup$ Apr 25 at 7:05
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    $\begingroup$ You have assumed that the reading on ammeter 1 does not change, ie the effective resistance of the whole circuit has not changed. $\endgroup$
    – Farcher
    Apr 25 at 7:07
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    $\begingroup$ For ideal zero resistance ammeters: Vn = Vsource at all times SO In = V'R = Vsource/Rn at all times. || Im = V/R = + Vsource / (Rldr + Rm). $\endgroup$ Apr 25 at 10:16

3 Answers 3


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.


Current going through a branch in a circuit divides, with more going to branches of less resistance. This part of your reasoning is correct. What you failed to account for is that the current being divided isn't the same as the LDR's resistance changes. As the resistance increases, the current entering the branched top node (i.e. the current out of the source) decreases. At the same time, the way that current divides between the two branches changes, with the left branch taking a smaller fraction and the right branch taking more. In the right branch, the effects cancel out and the current stays constant.

You could follow my reasoning above with algebra: the combined resistance seen by the source is $$R=\frac{1}{\frac{1}{L+M}+\frac{1}{N}},$$ and the total current in the circuit is $I=V/R$ (where $L$ is the LDR's resistance). The fraction of the current going into the right branch (through ammeter 2) is $$\frac{I_2}{I}=\frac{1/N}{1/R}=\frac{R}{N}.$$ Again, you're perfectly correct that this ratio increases as $L$ increases. But the actual amount of current through the ammeter is still $$I_2=\frac{I_2}{I}\cdot I=\frac{R}{N}\cdot\frac{V}{R}=\frac{V}{N},$$ which doesn't change when $L$ changes.

But notice that you can also see $I_2=V/N$ directly from looking at the circuit diagram. This is much simpler (and faster!) than the above reasoning, where you have to balance two opposing facts about the current ("more of the current goes to the right" vs. "there is less current overall") and see that they cancel out.


Reduced to basics it should be clear. (For some values of should be :-) ).

Vsource = input voltage.
Vn = Voltage across N.
Vm = voltage across M
Rldr = resistance of LDR.

Ohms law says : I = V/R

So, for ideal zero resistance ammeters:

  • Vn = Vsource at all times SO
  • In = V'R = Vsource/Rn at all times.


  • Im = V/R = Vsource / (Rldr + Rm).


In is constant as Vn is constant.
Im varies as Rldr varies


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