In some ways **Norton’s Theorem** can be thought of as the opposite to “Thevenins Theorem”, in that Thevenin reduces his circuit down to a single resistance in series with a single voltage.

Norton on the other hand reduces his circuit down to a single resistance in parallel with a constant current source.

**Nortons Theorem** states that “*Any linear circuit containing several energy sources and resistances can be replaced by a single Constant Current generator in parallel with a Single Resistor*“.

As far as the load resistance, R_{L} is concerned this single resistance, R_{S} is the value of the resistance looking back into the network with all the current sources open circuited and I_{S} is the short circuit current at the output terminals as shown below.

The value of this “constant current” is one which would flow if the two output terminals where shorted together while the source resistance would be measured looking back into the terminals, (the same as Thevenin).

For example, consider our now familiar circuit from the previous section.

To find the Nortons equivalent of the above circuit we firstly have to remove the centre 40Ω load resistor and short out the terminals A and B to give us the following circuit.

When the terminals A and B are shorted together the two resistors are connected in parallel across their two respective voltage sources and the currents flowing through each resistor as well as the total short circuit current can now be calculated as:

If we short-out the two voltage sources and open circuit terminals A and B, the two resistors are now effectively connected together in parallel. The value of the internal resistor Rs is found by calculating the total resistance at the terminals A and B giving us the following circuit.

Having found both the short circuit current, Is and equivalent internal resistance, Rs this then gives us the following Nortons equivalent circuit.

Ok, so far so good, but we now have to solve with the original 40Ω load resistor connected across terminals A and B as shown below.

Again, the two resistors are connected in parallel across the terminals A and B which gives us a total resistance of:

The voltage across the terminals A and B with the load resistor connected is given as:

Then the current flowing in the 40Ω load resistor can be found as:

which again, is the same value of 0.286 amps, we found using KirchoffÂ´s circuit law in the previous tutorials.

The basic procedure for solving a circuit using **Nortons Theorem** is as follows:

**1.**Remove the load resistor R_{L}or component concerned.**2.**Find R_{S}by shorting all voltage sources or by open circuiting all the current sources.**3.**Find I_{S}by placing a shorting link on the output terminals A and B.**4.**Find the current flowing through the load resistor R_{L}.

In a circuit, power supplied to the load is at its maximum when the load resistance is equal to the source resistance. In the next tutorial we will look at Maximum Power Transfer. The application of the maximum power transfer theorem can be applied to either simple and complicated linear circuits having a variable load and is used to find the load resistance that leads to transfer of maximum power to the load.

Error! Please fill all fields.

If we short-out the two voltage sources and open circuit terminals A and B, I thought the two resistors are now effectively connected together in series, how are they in parallel?

They are in parallel, please redraw from the point of A and B.

It is very useful to learn electrical

I want to learn about electronics.

Piece of Pie !

thank you for the tutorial.

01839720380

welcome

good definitions , very useful for students like me.

Thanks a lot it really helped

helpful 4 beginners lyk me……..

gud explanation

Hi

Would the calculations above still work if either the 10v or 20v battery polarity was reversed so that “I – short-circuit” was either “I1 +(- I2)” or “I2 + (- I1)”? I’m not sure if that condition would occur in real-life but I have seen it on numerous example circuit on the net. (I use them to test what I’ve learnt).

Hello Rod, interesting question. My thoughts are that while mathematically reversing the 10v or 20 v supply would mean the use of a negative voltage (the voltage times by -1), I do not think the circuit will produce the same load voltage (11.44v) and current (2.86A) result as calculated above. I think that this is because the two supplies would effectively be in series around the outer loop producing a current of +1A in one direction and -1A in the other (depending upon which voltage was reversed) cancelling each other out as you have stated as 1 + (-1) = 0. Obviously reversing both voltages would give the same but negative values, that is -11.44v and -2.85A.