Learn the basics of Electrical Engineering.

Norton Theorem

Norton theorem reduces the complex electrical networks to a single current source and a shunt resistor.

Norton theorem helps to reduce the complex circuit with many components to a single current source with a shunt resistor. The application of theorem is done in following steps:

  1. Remove the load from the circuit and mark the resulting terminals. Let's consider them 'a' and 'b' here.
  2. Calculate the equivalent resistance of circuit by removing all the current/voltage sources from the circuit. This equivalent resistance is the Norton resistance.
  3. Now short the terminals 'a' and 'b'
  4. Find the current passing through short circuit 'a' and 'b'. The current flowing through 'a' and 'b' is Norton current source.
  5. Shunting the overall current from step 4 as a current source to equivalent resistance of step 2
  6. Applying the current divider
Let's grasp our concepts by solving the example:

Norton's Theorem Example 1

Statement: Find the Norton equivalent for load resistor and then find the current across the load resistor. The value of the current source is 2A.

Step 1: Remove & Label

Remove the load resistor and label the remaining terminals:

Step 2: Equivalent resistance

The next step of the solution involves removing the current and voltage sources by open and short circuits respectively.

Step 3: Short circuiting the terminal

Revert back to step 1 and short the labeled terminals.

Step 4: Finding the current

Different techniques can be used for finding the current passing through the terminals. However, Superposition principle is the easiest technique which is feasible in the present scenario. Let's recap the steps involved in Superposition principle:
  1. Removing the current source by open circuit and finding current through terminals ab by applying Ohm's law
  2. Removing voltage source by short and applying the current divider
  3. Adding the calculated current from both steps.
And the pictorial form of our above steps:

Step 4a: 

= 1 A

Step 4b:

After replacing the voltage source by a short circuit, the short circuit ab comes in parallel to the 20-ohm resistor.
A short circuit cancels/burns all parallel resistors. All input current will now pass through ab.

Adding currents from step 4a and 4b.
Iequivalent = 2 A + 1 A = 3 A

Step 5:

Shunting the current source with equivalent resistance:

Step 6:

Connecting the load resistor and then applying the Current divider.