# Norton’s Theorem. Easy Step by Step Procedure with Example (Pictorial Views)

## Norton’s Theorem

Easy Step by Step Procedure with Example (Pictorial Views)

This is another useful theorem to analyze electric circuits like Thevenin’s Theorem, which reduces linear, active circuits and complex networks into a simple equivalent circuit. The main difference between Thevenin’s theorem and Norton’s theorem is that, Thevenin’s theorem provides an equivalent voltage source and an equivalent series resistance, while Norton’s theorem provides an equivalent Current source and an equivalent parallel resistance.
Norton’s Theorem may be stated under:
Any Linear Electric Network or complex circuit with Current and Voltage sources can be replaced by an equivalent circuit containing of a single independent Current Source IN and a Parallel Resistance RN.
Simple Steps to Analyze Electric Circuit through Norton’s Theorem
2. Calculate / measure the Short Circuit Current. This is the Norton Current (IN)
3. Open Current Sources, Short Voltage Sources and Open Load Resistor.
4. Calculate /measure the Open Circuit Resistance. This is the Norton Resistance (RN)
5. Now, Redraw the circuit with measured short circuit Current (IN) in Step (2) as current Source and measured open circuit resistance (RN) in step (4) as a parallel resistance and connect the load resistor which we had removed in Step (3). This is the Equivalent Norton Circuit of that Linear Electric Network or Complex circuit which had to be simplified and analyzed. You have done.
6. Now find the Load current flowing through and Load Voltage across Load Resistor by using the Current divider rule. IL = IN / (RN / (RN+ RL)) ((For better understanding…check the solved example)
Example:

Find RN, IN, the current flowing through and Load Voltage across the load resistor in fig (1) by using Norton’s Theorem.

Click image to enlarge

Solution:-
Step 1.
Short the 1.5Ω load resistor as shown in (Fig 2).
Step 2.
Calculate / measure the Short Circuit Current. This is the Norton Current (IN).
We have shorted the AB terminals to determine the Norton current, IN. The 6Ω and 3Ω are then in parallel and this parallel combination of 6Ω and 3Ω are then in series with 2Ω.
So the Total Resistance of the circuit to the Source is:-
2Ω + (6Ω || 3Ω) ….. (|| = in parallel with).
RT = 2Ω + [(3Ω x 6Ω) / (3Ω + 6Ω)] → IT= 2Ω + 2Ω = 4Ω.
RT = 4Ω
IT = V / RT
IT = 12V / 4Ω
IT = 3A..
Now we have to find ISC = IN… Apply CDR… (Current Divider Rule)…
ISC = IN = 3A x [(6Ω / (3Ω + 6Ω)] = 2A.

ISC= IN = 2A.

Step 3.
Open Current Sources, Short Voltage Sources and Open Load Resistor. Fig (4)
Step 4.
Calculate /measure the Open Circuit Resistance. This is the Norton Resistance (RN)
We have Reduced the 12V DC source to zero is equivalent to replace it with a short in step (3), as shown in figure (4)  We can see that 3Ω resistor is in series with a parallel combination of 6Ω resistor and  2Ω resistor. i.e.:
3Ω + (6Ω || 2Ω) ….. (|| = in parallel with)
RN = 3Ω + [(6Ω x 2Ω) / (6Ω + 2Ω)]
RN = 3Ω + 1.5Ω
RN = 4.5Ω
Step 5.
Connect the RN in Parallel with Current Source INand re-connect the load resistor. This is shown in fig (6) i.e. Norton Equivalent circuit with load resistor.
Click image to enlarge

Step 6.
Now apply the last step i.e. calculate the load current through and Load voltage across load resistor by Ohm’s Law as shown in fig 7.
IL = IN x [RN / (RN+ RL)]
= 2A x (4.5Ω /4.5Ω +1.5Ω) → = 1.5A
IL = 1. 5A
And
VL = IL x RL
VL = 1.5A x 1.5Ω

VL= 2.25V

Click image to enlarge

Now compare this simple circuit with the original circuit of figure 1. Can you see how much easier it will be to measure/calculate the load current and Load Voltage for different load resistors through Norton’s Theorem? Only and only yes…

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### Electrical Technology

1. anas mamji says:

sir plz give me a name of books about circuit and analysis of network for clearing concepts.<br />

WH Hayt and JE Kemmerly
Engineering circuit analysis

2. javed usmani says:

basic electronics by b l theraja

If we have to find the current through voltage source with 10ohm resistir are connected in series then how to find the Zl i.e load impedence

2. Electrician Brisbane says:

Great article! Norton&#39;s Theorem can certainly be confusing for a beginner.

3. Anonymous says:

great and thanku<br />

4. ahmed saeed says:

Thanks a lot

5. Sai says:

Nortons theorem

6. DINABANDHU PARAI says:

I want to know the best refarance book which can give me the best concept about nortons / Thevein , super position theorem
If u can pls help me
Thank you

7. Chi m says:

Pls,why did they use (3/(3 6)) to solve for the nortons cuRrent

9. Varun Chauhan says:

Thanks dude…..i hve exam in 3 hrs……ur post’s helped me a lot!!!

1. Electrical Technology says:

I’m glad it helped you. Lots of best wishes & prayers for your success!

10. Ankur Kumar says:

Thanks sir a lot of thanks

why the CDR are not (RT / RX+RT) X IT make me confuse ?

12. kobby says:

What if u r to find the current across the 6(ohm) resister?

13. Mr. P says:

Thank you so much

14. Kamal says:

Nortom resistence is not calculated correctly mine give 4.5 ohm, you have calculated as seen from the battery

15. Sanket says:

Very informative. really good website. never thought I would get hand made notes like content on a website. Keep it up guys.

16. Marvellous says:

Thanks for your clear information 👍😊

17. Rushikesh says:

Stepwise answer is very important and you do it well .i understand the problems easily .

18. NevrOnline says:

The bestt explanation i have come across so far..thank youu mann

1. I’m glad it helped… Best regards.

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