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# 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 I*

_{N }and a Parallel Resistance R_{N}.**Simple Steps to Analyze Electric Circuit through Norton’s Theorem**

- Short the load resistor
- Calculate / measure the Short Circuit Current. This is the Norton Current (I
_{N}) - Open Current Sources, Short Voltage Sources and Open Load Resistor.
- Calculate /measure the Open Circuit Resistance. This is the Norton Resistance (R
_{N}) - Now, Redraw the circuit with measured short circuit Current (I
_{N}) in Step (2) as current Source and measured open circuit resistance (R_{N}) 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. - Now find the Load current flowing through and Load Voltage across Load Resistor by using the Current divider rule. I
_{L}= I_{N}/ (R_{N }/ (R_{N}+ R_{L}))

**Example:**

Find R_{N}, I_{N}, 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 (I

_{N}).We have shorted the AB terminals to determine the Norton current, I

_{N.}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).

R

_{T }= 2Ω + [(3Ω x 6Ω) / (3Ω + 6Ω)] → I_{T}= 2Ω + 2Ω = 4Ω.R

_{T }= 4ΩI

_{T}= V / R_{T}I

_{T}= 12V / 4ΩI

_{T}= 3A..Now we have to find I

_{SC}= I_{N}… Apply CDR… (Current Divider Rule)…I

_{SC}= I_{N}= 3A x [(6Ω / (3Ω + 6Ω)] = 2A.**I _{SC}= I_{N }= 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 (R

_{N})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)

R

_{N}= 3Ω + [(6Ω x 2Ω) / (6Ω + 2Ω)]R

_{N}= 3Ω + 1.5Ω**R**

_{N}= 4.5Ω**Step 5.**

Connect the R

_{N}in Parallel with Current Source I_{N}and 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.

Load Current through Load Resistor…

I

_{L}= I_{N}x [R_{N}/ (R_{N}+ R_{L})]= 2A x (4.5Ω /4.5Ω +1.5Ω) → = 1.5A

**I**

_{L}= 1. 5AAnd

Load Voltage across Load Resistor…

V

_{L}= I_{L}x R_{L}V

_{L}= 1.5A x 1.5Ω**V _{L}= 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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WH Hayt and JE Kemmerly

Engineering circuit analysis

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

Great article! Norton's Theorem can certainly be confusing for a beginner.

great and thanku<br />

Thanks a lot

Nortons theorem

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

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

May you please help with the best book I can use for electrical engineering technology

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

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

Thanks sir a lot of thanks

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

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

Thank you so much

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

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

Thanks for your clear information 👍😊

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