# Norton’s Theorem

#### What is Norton’s Theorem ?

Fig.1 shows a complex network enclosed in a box with two terminal A and B brought out. The network in the box may consist of any number of resistors and energy sources connected in any manner.

Fig.1

According to Norton, the entire circuit enclosed in the box can be replaced by a single current source I_{N }in parallel with a resistance R_{N }as shown in Fig.2.

Fig.2

The resistance is same as Thevenin resistance R_{TH }.The value of I_{N }is determined by using Norton’s Theorem.

Once Norton’s equivalent circuit is obtained, then current thorough the load resistance R_{L} connected across AB can be found out.

Norton’s theorem as applied to d.c. circuit may be stated as below :

**Any linear, bilateral network having two terminals can be replaced by an equivalent circuit consisting of a current source of current output I _{N }in parallel with a resistance R_{N}.**

- The output I
_{N }of the current source is equal to the current through AB when A and B are short circuited. - The resistance R
_{N }is the resistance of the network measured between terminal A and B with load removed and voltage sources replaced by their internal resistances. Ideal voltage sources are replaced with short circuits and ideal current sources are replaced with open circuits.

**Norton’s theorem is converse of Thevenin’s theorem in the following ways:**

- Norton equivalent circuit uses a current source instead of voltage source as in case of Thevenin equivalent circuit.
- The resistance R
_{N }(same as R_{TH}) in parallel with the source instead of being in series with it as in case of Thevenin equivalent circuit.

Let’s analyse the application of Norton’s theorem by taking an example as shown in Fig.3.

Fig.3

##### Step 1: Find the value of I_{N}

Here, the circuit behind the terminal AB can be replaced by a current source I_{N }in parallel with a resistance R_{N }by applying Norton’s theorem.

The output I_{N }of the current generator is equal to the current that would flow when terminal A and B are short circuited as shown in Fig.4.

Fig.4

Here, we can observe that terminal AB are shorted and resistance R_{2 }and R_{3 }are in parallel and this parallel combination is in series with the resistance R_{1} .

So, the total resistance of the circuit is given by :

R’ = R_{1 }+[( R_{2 }× R_{3}) /( R_{2 }+ R_{3})]

= (R_{1}R_{2}+R_{1}R_{3}+R_{2}R_{3})/( R_{2 }+ R_{3})

Hence, Source Current I’ = V / R’

= [V× ( R_{2 }+ R_{3})]/ (R_{1}R_{2}+R_{1}R_{3}+R_{2}R_{3})

Short- Circuit Current I_{N }= Current flowing through R_{2 }(as shown in Fig.4)

= I’ × [ R_{3}/( R_{2 }+ R_{3})]

= (V R_{3}) / (R_{1}R_{2}+R_{1}R_{3}+R_{2}R_{3})

##### Step 2 : Find the value of R_{N}

To find R_{N}, remove the load R_{L }and replace the battery by a short because its internal resistance is assumed to be zero.

This is shown in Fig. 5

Fig.5

R_{N }= Resistance at terminal AB in Fig.5

= R_{2 }+ [(R_{1}R_{3})/(R_{1 }+ R_{3})]

Now as the values of I_{N }and R_{N }are determined, we can draw the Norton equivalent circuit which is shown in Fig.6.

Fig.6