# Circuit Analysis and Voltage

I'm trying to get a better understanding of the node method for circuit analysis. I understand the process, but I'm trying to verify that each step in the method is consistent with my overall understanding of what is going on in the circuit physically. I realize that the node method is a way that we can approach circuit analysis abstractly, but I feel uncomfortable using it without understanding the physics behind it.

I'm mainly confused about the voltages (potential differences) throughout the circuit. I understand that the electric field is a conservative vector field so the total potential difference of a closed loop circuit is 0, the potential difference between any two points in the electric field does not depend on the path taken, etc. I'm confused about why the potential difference is constant through ideal wires.

For example: I understand that the potential difference from H to A is 12 volts since its given and I can see that that the electric field does positive work when pushing a positive charge from H to A assuming that the H side of the voltage source is the low potential (-) and the A is the high potential (+), so it makes sense that the voltage they gave across the voltage source is positive. I don't understand why the potential difference across AB is 0 i.e. the potential at A = potential at B. I thought that since electric potential = $$\ \int_{r_A}^{r_B} \vec{E} \cdot d \vec{r} \$$ where $$\ r_A \$$ and $$\ r_B \$$ are the magnitudes of the position vectors pointing towards A and B respectively from the origin which I'll let be point H and since $$\|\vec{r_A}| - |\vec{r_B}|\$$ is not equal to zero then $$\ \int_{r_A}^{r_B} \vec{E} \cdot d \vec{r} \$$ is not equal to zero. I understand that the total potential difference of the circuit is 0 since $$\ \Delta \vec{r}_{total} = \vec{r_H} - \vec{r_H} = \vec{0} \$$. I just assumed that the potential difference is different at every point and since some potential differences across certain segments of the circuit are negative i.e. when charges are moving with the field (instead of against it), we can still have it that the total potential difference is 0.

I realize that the path from A to B is an ideal wire with no resistance and I understand that ohms law supports that the voltage (potential difference) is 0 from A to B since $$\ V = IR = I(0) = 0 \$$ which is why every example shows the potential at A is equal to the potential at B , but wouldn't that go against the definition of potential difference?

• Suppose that each of the resistors in your problem is 100 ohms. And each wire is 10 milliohms. How much does it change the final results (node voltages or branch currents) to approximate the wire resistance as 0 instead of 10 mOhms? – The Photon Jun 28 '20 at 0:56
• And you could imagine that each wire was a superconductor, which just slightly extends what The Photon said in their comment. – Ed V Jun 28 '20 at 1:04
• Does this answer your question? Flow of current, isn't make potential difference? – The Photon Jun 28 '20 at 1:08
• By the way, very similar questions get asked at least a couple times a month. Usually from the perspective of "how can current flow in the ideal wire if the voltage drop across it is zero?" But if you think a bit you can see that this is equivalent to your question. Probably someone else can find a more clearly asked example of a duplicate than the one I linked to, this was just the first one I found in my old answers. – The Photon Jun 28 '20 at 1:10
• Here's a related question on Physics.SE: Electric Field Topology in a Wire – The Photon Jun 28 '20 at 1:18

## 4 Answers

You clearly have to differentiate between schematic and real circuits. In a schematic you, most of the times, just represent the actual components you have in your circuit. In your example this would be your 3 resistors.

As you already stated, the wires between these components are ideal, this means they have 0 resistance and therefore 0 loss. As a consequence, node A and node B of your ideal schematic have the same potential so there's no potential difference.

When you switch to reality, of course there are no ideal wires, this means every wire that carries a current will automatically suffer from a voltage drop so when you build up this circuit, node A and B will not have exactly the same potential - for most cases this is negligible.

Wires and they're parasitic characteristics start to play a role at very high frequencies or high currents where you either have to keep your voltage drop in mind or control the effects of loop inductance at high current transients.

But to start, when analysing a schematic as yours, wires are ideal and therefore have no voltage drop.

• I do understand that we are taking ideal wires to have zero resistance and ideal wires don't really exist. My question is in regards to why ohms law and my understanding of the definition of potential difference seem to be inconsistent here. The only way that ∫𝐸⃗ ⋅𝑑𝑟 = 0 from A to B is if E is zero or if A = B (closed loop). I don't think E is zero here along the wires even if they are ideal. Is that not the case? – ktmath Jun 28 '20 at 15:53
• It is per definition, schematic wires are ideal so they connect points of identical potential, if you want to take the parasitic characteristic into account you have to add the parasitic wire components. – po.pe Jun 28 '20 at 16:36

The potential difference between points A and B is not zero. It is equal to the product of the the wire's resistance and the current flowing through the circuit. This is if you assume the wire to have a non-zero resistance.

If you assume the wire to have zero resistance, then there is no electric field inside the wire. So the integral $$\int_A^B E \,dr$$ also equals zero.

So the Kirchhoff's laws and Maxwell's equations are still in agreement.

• That was my initial thought too that perhaps there was no electric field in the wire and thus the voltage from A to B would then be zero.I can see that the electric field lines are the most dense in the voltage source, but they do exist outside of the voltage source. I was first thinking that perhaps the electric field outside of the source was pretty much negligible, so we take it to be zero, but I'm not sure that is the case.How can we have a potential difference across different components of a circuit, if we take the electric field to be 0 (or close to it)? – ktmath Jun 28 '20 at 15:44

You are right that potential difference between two points is: $$\int \vec{E}.\vec{dr}$$ And indeed $$\\vec{dr}\$$ is not zero from A to B in your diagram. But that is just half the story. You did not consider what is the electric field in the wire.
The microscopic form of Ohm's Law states that: $$\vec{E} = \rho \vec{j}$$ Here $$\\rho\$$ is the resistivity and j is the current density through the wire. I guess you know that the ideal wire has zero resistivity. Well, then guess what so is the electric field. It's zero. And no electric field means no potential difference.
Real wires do have a non-zero but very small resistivity and consequently some voltage drop across them. But this drop is negligible compared to the drop across the resistors and hence is neglected.

It does not violate potential difference definition because electric field between A and B is zero. In an ideal schematic, potential differences exist between two ends of a component, not between two ends of the wire. Taking the classic analogy, we can view the electric potential as gravitational potential. The region between A and B are equivalent to a flat, friction-less surface. There is zero electric-field between A and B, in the sense that a hockey puck placed in the middle will not accelerate toward either place. Each electrical component, however, is either a ski lift or a down-ward slope. A voltage source is a ski-lift, while a resistor is a down-ward slope that decreases potential. There is non-zero electric field WITHIN the voltage source and WITHIN the resistor.