# A doubt on potential difference

This is a schematic diagram of the circuit.

Is it possible to take the voltage difference between AB as E? I have learned that the potential difference in parallel circuits are the same, Therefore I did assume the potential difference across AB as E. But I want to know whether I am correct or wrong?

Because I hope this concept is a crucial factor when solving circuits, since I am beginner I would like to know it.

• As explained by curd, they aint in parallel. For 2 or more things to be parallel, they must have both terminals common. For your case, all the vertical resistors as well as battery share only one terminal common ie Ground. Other terminals dont end up at a common point. May 12, 2015 at 6:23
• What are the values of the resistors? You can look at this as two potential dividers. If for example all resistors are the same then the voltage at the junction of the top two resistors will be $\frac{2}{5} \cdot E$ and the output voltage $\frac{1}{5} \cdot E$ May 12, 2015 at 9:03

As @Curd Mentioned A-B are not in parallel. An easy way to tell if any 2 elements are in parallel is to check if they share the same nodes. This is how you solve the circuit.

Note how i assume the 2 resistors to be in series, this can only be done because there is no load connected across A and B. After that all you have to do is apply 2 voltage dividers.

Edit 1, The Values in the diagram below have nothing to do with the solved axample above

The Nodes in the above circuit are represented as the following (Ignore the values):

simulate this circuit – Schematic created using CircuitLab

In simple terms why is AB not in parallel with E? It is because E shares the color Red and black, when AB shares the colors yellow and black. 2 Elements may be considered as being in parallel if they share the same "colors" (node). In the diagram each nodes has been colored in a distinct color.

So how does parallel look like? It looks like the following:

simulate this circuit

As you can see R2 and R3 share the same color (node) hence they are in parallel.

• Thanks a lot Sada93 for the answer,I just start to read your answer,so I have found the term "node" could you explain it, I mean the term "node" May 12, 2015 at 10:37
• A node is a point where circuit elements meet. In the circuit above there are 4 nodes. For 2 resistors to be in parallel they need to share the same nodes. For 2 resistors to be in series the same current needs to flow through them with no current devision. I'm finding it difficult to give a clearer explanation without drawing it so here is a link. en.wikipedia.org/wiki/Node_%28circuits%29 May 12, 2015 at 10:43
• In the above picture R and 2R share the nodes C and D hence they are in parallel. May 12, 2015 at 10:44
• Could you please explain bit more about finding nodes in a circuit May 12, 2015 at 10:59
• Any how ,thank you so much for clearing my doubt . May 12, 2015 at 11:29

A-B is not parallel to E.
Don't be deceived by the geometrical positions of A and B. What counts is the topology, i.e. which node is connected to what other nodes, no matter how the nodes are arranged geometrically.

You can see that in order to get from top of E to A you have to follow two resistors in series. Therefore A-B can't be parallel to E.

• Then if that two resistors in series are not there what would be the result? May 12, 2015 at 6:52
• Yes, if the top two resistors were replaced by connections A-B would be parallel to E.
– Curd
May 12, 2015 at 7:07
• Then to be parallel what is the requirement , should the both A and B be connected to either negative or positive terminal of the battery directly? May 12, 2015 at 7:16
• Yes, or as Plutonium smuggler said: to be parallel they must share both terminals: E has the two terminal E_top and E_bottom. E_bottom already is connected to B. Only if E_top is connected directly (without other components inbetween) to A then AB is parallel to E.
– Curd
May 12, 2015 at 8:21
• It means both A and B are should directly connect to cell May 12, 2015 at 8:24

You have two voltage dividers. Taking the left-most 2 resistors, (horizontal and vertical) and assuming they are each of value "R" (the actual value is irrelevant provided they are equal), then the potential at the junction between those is E/2 when referenced to point B.

The next two resistors that have a junction at point A are also a voltage divider, and once more if they are each the same value, they form a 2:1 voltage divider.

From that (and ignoring loading by a measurement instrument), the potential between points A and B is E/4.

• Are you sure? The first potential divider does not give a $\frac{E}{2}$ since the effective bottom resistor is not the same as the top one because there are are two series resistors in parallel with it. May 12, 2015 at 9:06

HERE'S ANOTHER PROOF :

If we make the following change by removing two resistors:

simulate this circuit – Schematic created using CircuitLab

then we have $V_{\text AB} = E$. The resistors do not matter because the circuit is open, and so no current flows. Since no current flows, the resistors do not create a voltage drop. $V = IR$, but $I = 0$.

When we add a parallel resistor, we cause current to flow:

simulate this circuit

Now, current is flowing through R1, and so there is a voltage drop across R1. Consequently, $V_{\text AB} \gt E$.

If we add another resistor across AB, to create a schematic similar to yours, it is more of the same; current then also flows through R2, which further drops the voltage.

I remember doing lots of analysis like this at university! Ah, they where the days. There are only two times a voltage is the same at two points is if 1)there is a straight line connecting the two points together and no components are in that line. 2) Where a resistor is in the line and no current is drawn. For example an unconnected battery which comprises a voltage source and series resistance (the internal resistance of the battery). The voltage on both sides of the resistor will be the same as the voltage of the voltage source.

The rest of the analysis, is simply understanding how to combine resistors in parallel and series. Combine the resistors together, simplify the circuit (but you have to do it right!), redraw the circuit with new effective resistances then you can analyse further until you get to the point where you can calculate currents.

1) Research potential (voltage) divders, there's only a couple of simple equations for these and previous posters have done a good job at pointing them out.

Two other further key rules: (Kirchoff's Laws) 1) Sum of potential differences around a loop must some to zero. (sum of potential differences across loads = sum of potential differences of voltage sources).

2) Sum of currents into a junction (or node) equals the sum of currents flowing out of that node. That is, you can't just lose current, if it goes into a junction on a circuit diagram, it all must come out.
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