In the previous article, we discussed one of the network\circuit analysis theorem the Thevenin theorem which is very similar to the Norton Theorem.

As You can see that the above figure is very similar to Thevenin’s network similarly both Thevenin theorem and Norton Theorem are similar. But the difference is in Norton’s equivalent circuit consists of a current source and the equivalent resistance of the circuit are connected in parallel to the current source.

Table of Contents

**Norton theorem statement**

**Norton theorem:** It states that any linear, bilinear circuit can be resolved into Norton equivalent circuit. Where several active and passive elements are converted into equivalent current sources in parallel with Norton equivalent resistance and load resistance. It is known as Norton’s equivalent circuit.

That means a circuit consists of various active and passive elements that can be simplified into an equivalent circuit only consisting of short circuit current of the circuit connected in series with an equivalent resistance of the circuit with load resistor or resistor across which current have to find.

**Main steps to solve :**

- Remove the resistance across which you have to obtain current.
- Now obtain the equivalent resistance by converting the voltage source into a short-circuit wire and current source into an open terminal (if any)
- Obtain the equivalent current by making terminal short-circuited wire
- At last, connect equivalent Norton voltage in parallel with equivalent resistance with load resistance and find current across the load

Now, Let’s solve the above steps with help of an example.

**Q. Find the current in resistance R _{L} having 25 ohms of the network shown below.**

**Step 1.** First, remove the resistance across which you have to find current or voltage and make it short-circuit, and mark the open terminals ‘a and b’.

Current across short circuit terminal a-b :

I = Total voltage / total resistance

= 100/22

I = 4.54 A

**Step 2**. Then, if the circuit has any current or voltage source, convert the voltage source into short-circuit and the current source into an open-circuit. Now find the equivalent resistance of the circuit by looking from terminal ‘a-b’.

R_{N} = (10+20)||20 + 10

R_{N} = (30*20/30+20) + 10

= (600/50) + 10

= 22

**Step 3.** Then, find the short-circuit current across terminal ‘a & b’.

I_{sc} = current across 10 ohm resistor

I_{sc }= I*10/(10+20) (by current divider rule) =45.4/30

= 1.51 A

**Step 4**. At last, now obtain the current across load resistor by connecting Norton equivalent across terminal ‘a-b’ in parallel with equivalent resistance with the load resistance.

Current across 25-ohm resistor

I_{L }= 1.51 *22/25+22

= 33.22/47

= 0.706 A

Therefore, the current flowing across the 25-ohm resistor is 0.706 A. So, the above question gives you the concept to solve the question. The process of solving questions is very similar to the Thevenin theorems method.