## Norton’s Theorem

Easy Step by Step Procedure with Example (Pictorial Views)

**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.**

**Step 2.**

_{N}).

_{N.}The 6Ω and 3Ω are then in parallel and this parallel combination of 6Ω and 3Ω are then in series with 2Ω.

_{T }= 2Ω + [(3Ω x 6Ω) / (3Ω + 6Ω)] → I

_{T}= 2Ω + 2Ω = 4Ω.

_{T }= 4Ω

_{T}= V / R

_{T}

_{T}= 12V / 4Ω

_{T}= 3A..

_{SC}= I

_{N}… Apply CDR… (Current Divider Rule)…

_{SC}= I

_{N}= 3A x [(6Ω / (3Ω + 6Ω)] = 2A.

**I _{SC}= I_{N }= 2A.**

**Step 3.**

**Step 4.**

_{N})

_{N}= 3Ω + [(6Ω x 2Ω) / (6Ω + 2Ω)]

_{N}= 3Ω + 1.5Ω

**R**

_{N}= 4.5Ω**Step 5.**

_{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.**

_{L}= I

_{N}x [R

_{N}/ (R

_{N}+ R

_{L})]

**I**

_{L}= 1. 5A_{L}= I

_{L}x R

_{L}

_{L}= 1.5A x 1.5Ω

**V _{L}= 2.25V**

*Click image to enlarge*

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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

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?