Norton’s Theorem Equivalent Circuit

What is Norton’s Theorem

Norton’s theorem states that:

A linear two-terminal electrical circuit can be simplified into a circuit consisting of a current source IN connected parallel with an equivalent resistance, RN with the observed terminal.

The purpose of this Norton analysis is to make an equivalent circuit with a current source connected parallel with its equivalent resistance.

norton's theorem

Equation below can help us on how to find Norton current:

    \begin{align*}i=-\frac{V}{R_{N}}+i_{sc}\end{align*}

How to Find Norton Equivalent Circuit

Assume that we have a linear as shown below:

norton's theorem

We can redraw the circuit above into Norton equivalent circuit example:

norton's theorem

Our next main focus are finding the value of:

  • Norton equivalent resistance (RN)
  • Norton equivalent current (IN)

If you have learned about the source transformation theorem, we will know that Thevenin equivalent resistance, RTh and Norton equivalent resistance, RN are equivalent because they don’t affect the voltage-current relationship since they are used on linear circuits.

    \begin{align*}R_{N}=R_{Th}\end{align*}

In order to calculate the Norton equivalent current IN, we calculate the current flowing through short-circuited terminal a-b.

norton's theorem

The short-circuit current flowing from terminal a to b is short circuit current (isc) is the same with Norton equivalent current (IN). Hence,

    \begin{align*}I_{N}=i_{sc}\end{align*}

Keep in mind that we treat the independent and dependent source the same with we do in Thevenin’s theorem. Since Thevenin and Norton are equivalent, hence

    \begin{align*}I_{N}=\frac{V_{Th}}{R_{Th}}\end{align*}

This is basically the source transformation theorem. Because of this, source transformation is also known as Thevenin-Norton transformation. Since VTh, IN, RTh are related each other, we conclude that we need:

  • Open circuit voltage across terminal a-b, voc
  • Short circuit current at terminal a-b, isc
  • Equivalent resistance at terminal a-b when all independent sources are turned off, RN

Using the basic Ohm’s Law we can use the equations below:

    \begin{align*}V_{Th}&=v_{oc}\\I_{N}&=i_{sc}\\R_{Th}&=\frac{v_{oc}}{i_{sc}}=R_{N}\end{align*}

Procedure of Norton’s Theorem

Below is the step of Norton circuit analysis.

  1. Find and determine terminal a-b where a parameter is observed.
  2. Remove the component on that terminal, make it short circuit to the terminal a-b, and calculate the current at that point a-b (Iab=Isc=IN). This is known as I Norton or Norton equivalent current.
  3. If all the sources are independent sources, then find the equivalent resistance when all the sources are turned off and replaced by their inner resistances (Rab=RN=RTh):
    • Independent voltage source is replaced by a short circuit.
    • Independent current source is replaced by an open circuit.
  1. If there is a dependent source, to find the Norton equivalent resistance we can use:

    \begin{align*}R_{N}=\frac{V_{oc}}{I_{N}}\end{align*}

  1. In order to find the Voc at terminal a-b, make that terminal open circuit and find the voltage across that terminal (Vab=Voc).
  2. Redraw the Norton equivalent circuit consisting of the Norton equivalent current source, Norton equivalent resistance, and the component we remove in Step.(2).

Norton’s Theorem with Independent Sources

1. Find the value of i with Norton’s theorem!

norton's theorem

Solution:

Determine the point a-b on R where i is observed. Calculate the isc=IN when R=4Ω is removed:

norton's theorem

With mesh analysis:

From loop I1:

(1)   \begin{align*}I_{1}=6A\end{align*}

From loop I2:

(2)   \begin{align*}8(I_{2}-I_{3})+4I_{2}&=0\\3I_{2}-2I_{3}&=0\\I_{3}&=\frac{3I_{2}}{2}\end{align*}

From loop I3:

(3)   \begin{align*}-5+8(I_{3}-I_{2})=0\\8(I_{3}-I_{2})=5\end{align*}

Substitute Equation.(2) into Equation.(3):

    \begin{align*}8(\frac{3I_{2}}{2}&-I_{2})=5\\4I_{2}=5\quad&\rightarrow\quad I_{2}=\frac{5}{4}A\end{align*}

Thus,

    \begin{align*}i_{sc}=i_{N}&=I_{1}-I_{2}\\&=6-\frac{5}{4}=\frac{19}{4}A\end{align*}

Find RN when all of the independent sources are turned off (replaced with their inner resistances), from the perspective of point a-b:

norton's theorem

The Norton equivalent resistance is

    \begin{align*}R_{N}=4\Omega\end{align*}

Norton equivalent circuit:

norton's theorem

The value of i is

    \begin{align*}i&=\frac{4}{4+4}i_{N}\\&=\frac{4}{8}\times\frac{19}{4}=\frac{19}{8}A\end{align*}

2. Find the value of V with Norton’s theorem!

norton's theorem

Solution:

We remove the 40Ω resistor and make it short-circuit

norton's theorem

Find the equivalent resistance of the parallel resistors:

    \begin{align*}20\Omega//12\Omega\rightarrow R_{p}=\frac{20\times12}{20+12}=\frac{15}{2}\Omega\\\end{align*}

The voltage across terminal a-b is equal to the voltage across the Rp:

    \begin{align*}V_{1}=\frac{R_{p}}{R_{p}+5}\times18=\frac{54}{5}V\end{align*}

Find isc:

    \begin{align*}i_{sc}=i_{N}=\frac{V_{1}}{20}=\frac{27}{50}A\end{align*}

Find RN at point a-b

norton's theorem

The Norton equivalent resistance is

    \begin{align*}5\Omega//12\Omega\rightarrow R_{p}=\frac{5\times12}{5+12}=\frac{60}{17}\Omega\\R_{N}=R_{p}+20=\frac{400}{17}\Omega\end{align*}

The Norton equivalent circuit is

norton's theorem

The value of V is

    \begin{align*}R_{N}//40\Omega\rightarrow R_{p}=\frac{400}{27}\Omega\\\end{align*}

Thus

    \begin{align*}v=i_{N}\times R_{p}=\frac{27}{50}\times\frac{400}{27}=8V\end{align*}

3. Find the value of i with Norton’s theorem!

norton's theorem

Solution:

We remove the voltage source:

norton's theorem

Find isc

    \begin{align*}I_{48\Omega}=\frac{24}{48+24}\times6=2A\\I_{12\Omega}=\frac{24}{24+12}\times6=4A\end{align*}

Thus

    \begin{align*}i_{sc}=i_{N}=I_{12\Omega}-I_{48\Omega}=4-2=2A\end{align*}

Turn off all the independent sources to find the RN:

norton's theorem

The Norton equivalent resistance RN

    \begin{align*}R_{s1}=24+48=72\Omega\\R_{s2}=24+12=36\Omega\\R_{N}=\frac{R_{s1}\times R_{s2}}{R_{s1}+R_{s2}}=24\Omega\end{align*}

The Norton equivalent circuit:

norton's theorem

Thus

    \begin{align*}i_{1}=\frac{24}{24}=1A\\\mbox{then : }i=i_{N}+i_{1}=3A\end{align*}

Norton’s Theorem with Dependent Sources

1. Find the value of i with Norton’s theorem!

norton's theorem

Solution:

We remove the observed component

norton's theorem

 

The value of isc:

    \begin{align*}&v_{1}=3V\\&\Sigma v=0\\&-4v_{1}+6i_{sc}=0\\&-4.3+6i_{sc}=0\\&i_{sc}=\frac{12}{6}=2A\end{align*}

In order to find RN, we need to find Voc first

norton's theorem

 

The value of Voc is

    \begin{align*}&v_{1}=3V\\&V_{ab}=V_{oc}=\frac{12}{12+6}\times4v_{1}=8V\end{align*}

Hence the Norton equivalent resistance is

    \begin{align*}R_{N}=\frac{V_{oc}}{i_{sc}}=\frac{8}{2}=4\Omega\end{align*}

The Norton equivalent circuit is

norton's theorem

 

Thus

    \begin{align*}i=\frac{4}{4+4}\times2A=1A\end{align*}

2. Find the value of i with Norton’s theorem!

norton's theorem

Solution:

Remove the observed component

norton's theorem

Find the isc

    \begin{align*}&\Sigma v=0\\&2i_{sc}+3(i_{sc}+6)-12=0\\&5i_{sc}+6=0\rightarrow i_{sc}=\frac{-6}{5}A\end{align*}

Find the RN from Vab when point a-b is open circuit:

norton's theorem

The Vab is

    \begin{align*}V_{ab}=V_{oc}=12-(3\times6)=-6V\end{align*}

The Norton equivalent resistance, RN

    \begin{align*}R_{N}={V_{oc}}{i_{sc}}=\frac{-6}{-6/5}=5\Omega\end{align*}

The Norton equivalent circuit:

norton's theorem

Thus,

    \begin{align*}i=\frac{5}{5+1}\times\frac{-6}{5}=-1A\end{align*}

3. Find the voltage value V with Norton’s theorem!

norton's theorem

Solution:

Remove the observe component

norton's theorem

Find isc

    \begin{align*}&\Sigma v=0\\&-6+2i_{1}+\frac{i_{1}}{2}=0\\&\frac{5i_{1}}{2}=6\rightarrow i_{1}=\frac{12}{5}A\\&\mbox{then : }i_{sc}=\frac{i_{1}/2}{6}=\frac{1}{5}A\end{align*}

Find the Vab

norton's theorem

Thus

    \begin{align*}&V_{ab}=V_{oc}=\frac{i_{2}}{2}\\&\Sigma v=0\\&-6+2i_{2}+\frac{i_{2}}{2}=0\\&\frac{5i_{2}}{2}=6\quad\rightarrow\quad i_{2}=\frac{12}{5}A\\&\mbox{then : }V_{oc}=\frac{i_{2}}{2}=\frac{6}{5}A\end{align*}

Hence the Norton equivalent resistance

    \begin{align*}R_{N}=\frac{V_{oc}}{i_{sc}}=\frac{6/5}{1/5}=6\Omega\end{align*}

The Norton equivalent circuit

norton's theorem

Thus

    \begin{align*}V&=R_{p}\times\frac{1}{5}A\\&=\frac{3}{2}\times\frac{1}{5}=\frac{3}{10}V\end{align*}

Norton’s Theorem Examples

1. Redraw the circuit below into its Norton equivalent circuit at terminals a-b.

norton's theorem

Solution:

Replace all the independent sources with their inner resistances.

norton's theorem

 

From this circuit we will get Norton resistance

    \begin{align*}R_{N}&=5||(8+4+8)\\&=5||20=\frac{20\times5}{25}=4\Omega\end{align*}

In order to find the Norton current IN, we make terminal a-b short circuit as shown in circuit below,

norton's theorem

We ignore the 5Ω resistor because it is parallel with a short circuit. Using mesh analysis, we get

    \begin{align*}i_{1}=2A,\qquad20i_{2}-4i_{1}-12=0\end{align*}

From equations we obtained above, we get

    \begin{align*}i_{2}=i_{sc}=I_{N}=1A\end{align*}

Because Norton is equivalent to Thevenin, we can get Norton current IN from VTh/RTh. We can get VTh if we open circuit the terminal a-b as shown below.

norton's theorem

Using mesh analysis, we get

    \begin{align*}i_{3}&=2A\\25i_{4}-4i_{3}-12=0\qquad&\rightarrow\qquad i_{4}=0.8A\end{align*}

And

    \begin{align*}v_{oc}=V_{Th}=5i_{4}=4V\end{align*}

Thus

    \begin{align*}I_{N}=\frac{V_{Th}}{R_{Th}}=\frac{4}{4}=1A\end{align*}

The Norton equivalent circuit is

norton's theorem

2. Find Norton resistance RN and Norton current IN using Norton’s theorem from the circuit below at terminal a-b.

norton's theorem

Solution:

We replace the independent voltage source with short circuit and connect terminal a-b with voltage source vo = 1 V as shown below,

norton's theorem

Ignore the 4Ω resistor because it is in parallel with a short circuit. Hence voltage source vo, dependent current source, and 5Ω resistor are in parallel. Thus, ix = 0.

From node a,

    \begin{align*}i_{o}=\frac{1V}{5\Omega}=0.2A\end{align*}

And

    \begin{align*}R_{N}=\frac{v_{o}}{i_{o}}=\frac{1}{0.2}=5\Omega\end{align*}

In order to get Norton equivalent current IN, we make terminal a-b short circuit to find the current isc as shown below

norton's theorem

From the circuit above all the components are in parallel. Thus,

    \begin{align*}i_{x}=\frac{10}{4}=2.5A\end{align*}

At node a, use KCL

    \begin{align*}i_{sc}=\frac{10}{5}+2i_{x}=2+2(2.5)=7A\end{align*}

Hence,

    \begin{align*}I_{N}=7A\end{align*}

Frequently Asked Questions

What does Norton theorem states?

Norton’s theorem states that a linear two-terminal circuit can be replaced by an equivalent circuit consisting of a current source in parallel with a resistor

What is Norton’s theorem formula?

Norton’s theorem formula consists of a equivalent current source connected parallel with the equivalent resistance and desired component. We can use the Ohm’s Law.

What is Norton theorem with example?

A linear two-terminal electrical circuit can be simplified into a circuit consisting of a current source IN connected parallel with an equivalent resistance, RN with the observed terminal.

Is Norton theorem same as Thevenin?

If you have learned about the source transformation theorem, we will know that Thevenin equivalent resistance, RTh and Norton equivalent resistance, RN are equivalent because they don’t affect the voltage-current relationship since they are used on linear circuits.

How do I get Norton equivalent?

Remove the component on that terminal, make it short circuit to the terminal a-b, and calculate the current at that point a-b (Iab=Isc=IN). This is known as I Norton or Norton equivalent current.

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