-1 Ω

125

02:10:35

Change the V/A of the current-controlled voltage source to change the resistance (which is multiplied by -1)

published 3 years ago

592azy2circuitdude

3 years ago

This circuit is very intriguing. I can't figure out how it works. Would you like to please explain it more?

guillianj

3 years ago

For each Amp going through that current-controlled voltage source you have 1 Volt in the opposite direction. Ohms are basically Volts per Amp, so if the voltage is in the opposite direction, it’s negative resistance. All the other stuff around it is just to play around with. If you use that switch on the bottom it adds 1 Ohm in parallel with the -1 Ohm and no current flows through (R = 1*(-1)/(1-1) = -1/0 = ∞) and if you use the switch at the end it adds 1 Ohm in series and it becomes a short (R = (-1) + 1 = 0).

guillianj

3 years ago

Woops actually it’s 2 Ohms at the end, so R = (-1) + 2 = 1. You can see the polarity of the sine wave flip. If you change it to 1 Ohm you will have 1000000 Amps flowing through since there’s only 1 microOhm in the start to avoid crashing the simulation x)

592azy2circuitdude

3 years ago

I think I get it now. I didn't realize that a ccvs could act like a resistor by generating volts per amount of current flow, but it makes sense. Now I'm led to another question: Using the bottom switch to add 1ohm in parallel causes current to flow in a loop through the ccvs. What on earth causes that??

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