Spintronics Community Forum

Finding hard to replicate Kirchhoff's law

Theory
#1

Hi, this is my first post so no offence if my question has been answered in the past. So it goes as follows:

To create a circuit with three resistors in series and measure the voltage drop I have come up with this circuit: Spintronics Simulator

However, once you remove any of the three capacitors/voltmeters the results are spoiled. I can see that this is related to the nature of the component that is essentially a capacitor but shouldn’t somehow have a component that will act as a pure voltmeter? In the circuit in question, you would expect 2V drop to be consistent and not related to how many caps/v-meters we have in the cirquit.

Kind Regards and thank you
Yiannis

#2

Hey @Janagn!
Great question! There are a couple of things that I think are causing the confusion. Firstly, as designed in the simulator, the resistors are not exactly in series. Here’s a line diagram on the circuit you have:
image

The next thing is about how you’re “disconnecting” the capacitors in your system. The way the junctions work, if there’s nothing connected to one of the sprockets, it will act as a “zero” resistance path. So if, in your circuit, you just remove the chain connecting the capacitor to the junction, you will have the following circuit:
image

As you can see, the path through the circuit, will effectively by-pass the resistor at that split where the capacitor is removed (the middle one) resulting in that resistance effectively being removed from the circuit, resulting in a different voltage drop. I think this circuit here is more along the lines of what you were interested in demonstrating:

Let me know if that clarifies it for you!

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#3

Hey Aaron,

this was a brilliant answer. So in essence I need to “block” with a switch the sproket of the junction to essentially simulate that the circuit is “open” via this path and thus, if you want to use only ony v-meter, you need to have the switches in each other resistor in the off position. Makes absolute sense.

Kind Regards
Yiannis

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#4

Exactly!

It’s one of the things about Spintronics that isn’t super intuitive to someone who already has experience with electronics. In electronics, we open a circuit by simply removing a wire, where in Spintronics you open a circuit by stopping the movement of the chain.

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