Norton’s Theorem Simplified

Norton’s Theorem Simplified

This is another useful theorem used in simplifying complex/linear  circuits just  like Thevenin’s Theorem. 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 as:

It is possible to simplify any complex circuit/linear circuit with Current and Voltage source with an equivalent circuit containing a single Current Source IN and RN  connected in parallel to a load.

 

Simple Steps To Solve Electric Circuit By Norton’s Theorem

1. Short the load resistor

2. Calculate the Short Circuit Current. This is the Norton Current (I N )

3. Short Voltage Sources, open load resistor and calculate the Open Circuit Resistance. This is the  Norton Resistance (R N )

4. Now, Redraw the circuit with the calculated short circuit Current (I N ) in Step (2) as current Source and the calculated open circuit resistance (R N ) in step (3) connected in  parallel to the load resistor that was short  in Step (1). This is the Equivalent Norton Circuit of that Linear Electric Network or Complex circuit which had to be simplify and solved. 

5. Finally find the Load current  and Load Voltage across Load Resistor by using the Current divider rule or this formula:I_{{L}}=\frac{I_{N}R_{N}}{R_{L}+R_{N}}

Example:

Find the current flowing through and Voltage across the load resistor in the circuit below by using Norton’s Theorem.

Norton's Theorem Circuit

Step 1

Short the load circuit

Norton's Theorem

Step 2

Calculate the Short Circuit Current.

Using Current Divider Rule

I_{{6}}=\frac{I_{T}R_{T}}{R_{6}} ;

I_{{T}}=\frac{12}{4}=3A ;

2+(6//3)=4\Omega \Rightarrow R_{{T}} ( Resistor 6 and 3 are in parallel and in series to resistor 2)

I_{{6}}=\frac{3\times4}{6}=2A\Rightarrow I_{{N}}

Step 3

Short Voltage Sources, open load resistor and calculate the Open Circuit Resistance. This is the  Norton Resistance (R N )

Norton's Theorem

 

 

 

 3+(2//6)=4.5\Rightarrow R_{{N}} ….calculation of this, is started from the direction of the arrow.

Step 4

Redraw the circuit with the calculated short circuit current.

Norton's theorem

 

Step 5

Finally using this formula I_{{L}}=\frac{I_{N}R_{N}}{R_{L}+R_{N}}

I_{{L}}=\frac{2\times4.5}{2+4.5}=1.4A……Load current

From Ohm’s Law:

V_{{L}}=I_{{L}}R_{{L}}

 

V_{{L}}=1.4\times2\Rightarrow2.8V…….Load Voltage

You will get the same answer as above if you solve the circuit by using Thevenin’s Theorem

 

 

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