# Lightbulbs in circuits. Why does the current go UP? [duplicate]

Possible Duplicate:
Why does the light bulb's brightness decrease?

So I have a circuit in the picture with three resistors.

If I were to remove/disconnect $R_2$ from the circuit, what would happen to the overall brightness of $R_3$ and $R_1$?

Now the answer I already have is that $R_1$'s brightness will go down and $R_3$ will go up.

I understand that $R_1$'s brightness goes down because by removing the $R_2$, the net resistance goes up (destroying a parallel combination)

But I don't understand why $R_3$ would go up? Conceptually this makes no sense to me. How do I know that the new current would always be larger than when the current splits at the junction?

I would like a conceptual physics answer and not equations comparing inequalities.

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## marked as duplicate by David Z♦May 18 '12 at 6:31

Think about the voltage drop across $R_{1}$ before and after removing $R_{2}$ – Jerry Schirmer May 18 '12 at 5:36
Jak, this is the same question, and has the same answer, as you asked previously: physics.stackexchange.com/a/28424/4066 – EnergyNumbers May 18 '12 at 6:28

The following answer is, I hope, at least plausible-sounding. It is by no means a proof, and ultimately you do need the equations to make a convincing argument, because not all plausible sounding arguments end up being correct when put to the test.

The fundamental idea is that R2 is limited in its ability to set the total current in the circuit because the current will always have to travel through R1.

The answer is easiest to understand when R2 is tiny compared to R3. Then R2 acts to leech away current from R3 because it provides a path of lesser resistance. Even though a small value of R2 will cause more current to flow through the entire circuit than if R2 were larger, the current can never get too large because it always has to travel through R1 first. On the other hand, the ability for R2 to leech current away from R3 is not really limited by anything. So it should seem reasonable that removing R2 in this case would increase the current flowing through R3.

The same argument basically applies for larger values of R2, but the amount by which the current in R3 increases when you remove R2 will get smaller and smaller if you consider larger and larger values of R2. In the end, you're always restoring more current to R3 that had been trickling through R2 than the amount of current you remove from the circuit by removing R2. This comes back to the idea that R2 is limited in its ability to set the total current in the circuit.

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I think this is easier to see if you instead imagine the situation in reverse: start with just R1 and R3 and look at what happens when you add R2. Initially you have some current I flowing through both R1 and R2, which causes a voltage drop V1=I*R1 across R1, so the voltage across R3 is V0-V1.

Now if you add R2, you know the total parallel resistance will be smaller and so the current flowing through R1 will increase by some amount. This means that the voltage drop V1 will also increase, and so the voltage across R3 will be decreased. Clearly this means that less current will now flow through R3.

Therefore, by removing R2, you will always increase the current flowing through R3, although this increase can be very small, depending on the values of all the parameters.

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In words, the total resistance to the current when all three resistances are present goes up when you remove R2. The reason is that adding R2 in parallel reduces the total resistance. Removing it increases the total resistance , lowering the current and thus R1 loses brightness but R3 gets a higher current than it would get while sharing the voltage drop with R2, so it brightens.