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## Summary:

- KCL got declined while trying to find Inorton of an op amp circuit

Hi.

Let's say we have a circuit like that, and we want to find the Thevenin/Norton circuit from a and b points.

So for V

So in the Norton circuit, the wire would connect

The positive pin of the amplifier is connected to ground and as long as it's an ideal amplifier, the voltage of the negative pin is also considered as 0v due to the positive pin, which means the both ends of the 6K ohm resistor is 0v and no current should pass the 6K ohm resistor.

Now let's write KCL at the negative pin node. The voltage of this node is 0 due to the positive pin, also no current enters the amplifier as it's an ideal one. We have one current from the 3K ohm resistor (one end V

So KCL says: -V

Why KCL got declined here? What's wrong with my solution?

Let's say we have a circuit like that, and we want to find the Thevenin/Norton circuit from a and b points.

So for V

_{th}, we calculate the V_{oc}, in which we remove the 2K ohm resistor and calculate the V_{oc}. I calculated it correctly. But I have problem in calculating the Norton current. So in the Norton circuit, we should calculate I_{sc}, where we remove the 2K ohm resistor and put a wire instead of it.So in the Norton circuit, the wire would connect

**ground**to**V**of the amplifier, so the voltage of V_{output}_{output}would be zero. So as long as we have no potential difference at the two ends of the resistors, no current would pass the 12K and 15K ohm resistors. Also the voltage of the right end of the 6K ohm resistor is also zero (because V_{output}= 0v/gnd).The positive pin of the amplifier is connected to ground and as long as it's an ideal amplifier, the voltage of the negative pin is also considered as 0v due to the positive pin, which means the both ends of the 6K ohm resistor is 0v and no current should pass the 6K ohm resistor.

Now let's write KCL at the negative pin node. The voltage of this node is 0 due to the positive pin, also no current enters the amplifier as it's an ideal one. We have one current from the 3K ohm resistor (one end V

_{s}and the other end gnd, so we have a current entering the node). We have 3 branches at this node, one has current entering the node, one goes to the amplifier and doesn't have current as it's an ideal op amp and one leads to the 6K ohm resistor, but the voltage of both ends of the 6K ohm resistor is zero and the current doesn't flow in that breach either.So KCL says: -V

_{s}/3K = 0, which declines the law, because we have an actual known voltage from the voltage source and a number never would be equal to zero.Why KCL got declined here? What's wrong with my solution?

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