# How do you add the resistors in these circuits? In the first one, can I simply add R3 and R4 in series, R1 and R2 in series, and then R34 and R12 in parallel? I'm confused because of the voltage drop, I'm unsure what it means or how it effects my calculations In the second, I know I can add R1 and R2 in series, but can I also add R3 and R4 in series?

• Eventually, a load is added to the circuit. But in this case, I'm disregarding it and focusing on measuring the voltage of the circuit without the load. However, in order to do that, I first have to calculate Rtotal, right? So that I can apply Ohm's law V=Iq*Rtotal. But my problem is with adding the resistors. – CaptainAwesom E Jan 20 '18 at 0:53
• Sure, in that case you can just add the resistors in series, use that to compute the current, and then use that current in the original circuit to figure out $V_A$ and $V_B$. – Chris Jan 20 '18 at 1:01
• Your two circuits are identical - if I assume that $U_{AB}$ is not a source, but a measurement. The only difference is that $U_{AB}$ is measured between different nodes. So yes, you can add $R_3$ and $R_4$. However, if the eventual load is connected between A and B, then the two circuits will be different. – hdhondt Jan 20 '18 at 1:12

As drawn your first diagram shows a voltage drop between nodes A and B but no conductor between A and B.

This means that no current flows directly between A and B so the currents through the left hand resistors is the same which means that they can be taken to be resistors in series.
The same is true for the right hand resistors.

Electrically your two diagrams are identical.

You can only add resistors in series if the same current is going through them. If you apply a voltage between $A$ and $B$, then some current will travel from $A$ to $B$, which breaks the balance. $I_1\ne I_2$ and $I_3\ne I_4$, so you cannot add them in series. If you are not loading the circuit like that, then you can add them in series. But keep in mind that this equivalent solution only works when there is no load.

If you want to find the equivalent resistance, you cannot do in such conventional way.Rather, you should apply KVL and KCL along the circuit...

Apply KVL at three closed loops,from which you can find the total current, alongside the current through the branches...

The you simply use Ohm's law to find equivalent resistance...