Electronic – Circuit Analysis and Voltage

circuit analysishomework

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:

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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?

Best Answer

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.