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I'm trying to find the Norton Equivalent circuit using Norton's Theorem for the circuit below. Since I need to find the Norton Resistance, I removed all the sources from the circuit. The problem is, I can't find the equivalent resistance because I don't know what to do with the 10kΩ resistance. I would appreciate it if I can get an explanation on what I should do. Also, what should come to mind when trying to find equivalent resistance of complicated circuits? Thanks.

Circuit for Finding Norton Equivalent

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Circuit for Finding Norton Resistance

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  • \$\begingroup\$ Remove 10kohm resistance as it is shorted ,and for complicated circuits generally we apply a voltage or current source (here between ab) and measure current through it or voltage across it (source) and then V/I gives your equivalent resistance \$\endgroup\$ – user215805 Nov 29 '20 at 11:00
  • \$\begingroup\$ I just realized, would a source conversion for the 10mA current source solve the problem? But, if I do that we will get the same circuit for finding the equivalent resistance. So, I should just remove it. \$\endgroup\$ – Kalamakra Nov 29 '20 at 11:02
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    \$\begingroup\$ The 10k is shorted so the equivalent resistance is 5k||15k \$\endgroup\$ – Dennis Ernst Nov 29 '20 at 11:17
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Norton Resistance

You already did the essential steps, e.g. remove load, replace voltage and current sources with short and open circuits respectively.

$$R_{norton}=5k\Omega||15k\Omega=3.75k\Omega$$

Norton Current

Simplify the circuit by converting current to voltage sources:

schematic

simulate this circuit – Schematic created using CircuitLab

Calculate the voltage at the load:

$$V_{a,b}=30V-\dfrac{45V}{20k\Omega}\cdot15k\Omega = -3.75V$$

Calculating the norton current:

$$I_{norton}=\dfrac{-3.75V}{3.75k\Omega}=-1mA$$

Redrawing:

schematic

simulate this circuit

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