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

enter image description here

Circuit for Finding Norton Resistance

enter image description here

  • \$\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, 2020 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, 2020 at 11:02
  • 1
    \$\begingroup\$ The 10k is shorted so the equivalent resistance is 5k||15k \$\endgroup\$ Nov 29, 2020 at 11:17

1 Answer 1


Norton Resistance

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


Norton Current

Simplify the circuit by converting current to voltage sources:


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:




simulate this circuit


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.