• The Norton’s theorem is also for linear networks. The Norton’s theorem states that any number of voltage sources, current sources and resistors having two open ends can be simplified into an ideal current source and a resistor connected in parallel with the source.
• Norton’s Theorem (aka Mayer–Norton theorem) states that it is possible to simplify any linear circuit to an equivalent circuit with a single current source and equivalent parallel resistance connected to a load. The simplified circuit is known as the Norton Equivalent Circuit. Norton’s Theorem is an alternative to the Thevenin Theorem.

A)-Nortan theorem is not applicable to the circuits consists of unilateral elements or non linear elements b)-not applicable to the circuits consists of load in series or parallel with controlled or dependent sources. C)- not applicable to the cir.

Construct an electric circuit with passive components and verify Norton's theorem.

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Resistors, Battery, connection wire etc..

Norton's theorem states that a network consists of several voltage sources, current sources and resistors with two terminals, is electrically equivalent to an ideal current source ' I_NO' and a single parallel resistor, R_NO. The theorem can be applied to both A.C and D.C cases. The Norton equivalent of a circuit consists of an ideal current source in parallel with an ideal impedance (or resistor for non-reactive circuits).

The Norton equivalent circuit is a current source with current 'I_NO' in parallel with a resistance R_NO.To find its Norton equivalent circuit,

1. Find the Norton current 'I_NO'. Calculate the output current, 'I_AB', when a short circuit is the load (meaning 0 resistances between A and B). This is INo.
2. Find the Norton resistance RNo. When there are no dependent sources (i.e., all current and voltage sources are independent), there are two methods of determining the Norton impedance RNo.

• Calculate the output voltage, VAB, when in open circuit condition (i.e., no load resistor — meaning infinite load resistance). RNo equals this VAB divided by INo.
  or
• Replace independent voltage sources with short circuits and independent current sources with open circuits. The total resistance across the output port is the Norton impedance RNo.
  However, when there are dependent sources the more general method must be used. This method is not shown below in the diagrams.
• Connect a constant current source at the output terminals of the circuit with a value of 1 Ampere and calculate the voltage at its terminals. The quotient of this voltage divided by the 1 A current is the Norton impedance RNo. This method must be used if the circuit contains dependent sources, but it can be used in all cases even when there are no dependent sources.

Example 1:-

Consider this circuit-

To find the Norton’s 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:

With A-B Shorted :

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.

Find the Equivalent Resistance (Rs):

10Ω Resistor in parallel with the 20Ω Resistor

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

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:

Verification of Norton’s Theorem using the simulator,
Step1:- Create the actual circuit and measure the current across the load points.

Step 2:- Create the Norton’s equivalent circuit by first creating a current source of required equivalent current in amperes (2 A in this case), and then measure the current across the load using an ammeter.

In both the cases the current measured across the resistance should be of the same value.

More about Norton's theorem:

1.Norton's theorem and Thevenin's theorem are equivalent,and the equivalence leads to source transformation in electrical circuits.

2.For an electric-circuit,the equivalence is given by ,

V_Th=I_NoxR_Th

ie Thevenin's volage=Norton's current x Thevenin's resistance

3.The applications of Norton theorem is similar to that of Thevenin's theorem.The main application is nothing but the simplification of electrical circuit by introducing source transoformation.

In this article, we explain norton's theorem.

Norton's theorem states that any combination of power sources and resistors can be replaced with a single currentsource in parallel with a single resistor.

Norton's theorem, thus, greatly reduces and simplifies a circuit.

In the end, norton's theorem produces a single current source with a single resistance in parallel, along with the load. Like the thevenin equivalent, the norton equivalent does not take into account the resistance of the load (the output the circuit is powering). Therefore, at the end is a circuit that produces the norton current source, norton resistance, and a the load resistance.

Therefore, in the end, a current divider circuit exists between the norton equivalent resistance and the load resistance, and we can calculate how much current each gets through the current division formula.

The great thing about norton's theorem is the simplicity it creates. Basically, norton's equivalent looks at the entire circuit except the load of the circuit. It then simplifies all the power sources into a single current source and all resistors into a single equivalent resistor (in parallel with the current source). At the end, we just have the single current source, in parallel with a resistor, which is in parallel with the load (resistance). It forms a very simple current division circuit between the single equivalent resistor and the load resistance.

In thevenin's theorem, the circuit is reduced to a single voltage source in series with a single resistor, along with the load resistance, forming a voltage divider circuit.

So let's now go over an example circuit, so that you can see how norton's theorem works.

Example of Norton's Theorem

So, below, we have a circuit that we will break down.

So this circuit is a typical circuit that has a voltagesource and several resistors.

In order to produce the equivalent nortonresistance, we have to simplify all of the resistors into an equivalent resistance, excluding the load (load resistance).

To do this, we find the resistance of the 1KΩ and the 2KΩ resistor.Since they are in parallel, the equivalent resistance is 1KΩ 2KΩ = (1KΩ)(2KΩ)/(1KΩ + 2KΩ)= 667Ω.

We then add the 100Ω resistor to the 667Ω equivalent resistance. This gives us 767Ω.

So the equivalent norton resistance is 767Ω.

This is shown in the circuit below.

So now you see that we have a voltage source and 2 resistors that are in series.

This forms a voltage divider circuit.

The 12V divides up between the 767Ω and the 1KΩ resistor.

Doing the calculations, this yields 5.2V across the 767Ωresistor, since 12V(767Ω)/(767Ω + 1KΩ)= 5.2V and 6.8V across the 1KΩ resistor, since 12V(1KΩ)/(767Ω + 1KΩ)= 6.8V.

To calculate the norton current source, we take the voltage across the load (which is 6.8V) and divide it by the norton equivalent resistance (which is 767Ω). This is simply ohm's law, I= V/R= 6.8V/767Ω= 8.9mA.

This is shown in the circuit below.

So now this is norton's equivalent circuit.

You can see how the norton's equivalent circuit forms acurrent divider circuit between the norton equivalent resistance and the resistance of the load.

The norton current source is 8.9mA. So this 8.9mA divides into the 2 resistors.

Using the current division formula, we can calculate how much each branch gets.

The current division formula is, IS= RT/RX, where IS is the current from the power source, RT is the equivalentresistances of the branches in a current divider, and RX is the specific resistance of the branch you are calculating the current flow through.

So IS is 8.9mA, because that's the currentflowing from the currnet source.

RT is the equivalent resistance of all the branches of the current divider circuit. Being that there is 2 resistances in the current divider circuit, 767Ω and 1KΩ, the equivalent resistance is 767Ω 1KΩ. Doing the math, the equivalent resistance is, 767 1KΩ = (767Ω)(1KΩ)/(767Ω+1KΩ)= 434Ω.

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RX is the specific resistance of the branch you are calculating the current for.

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Therefore, the current through the 767Ω resistor is, (8.9mA)(434Ω)/(767Ω)= 5.04mA.

The current through the 1KΩ is,(8.9mA)(434Ω)/(1KΩ)= 3.86mA.

Therefore, through norton's equivalent, a current divider circuit is formed, which we can easily calculate currentsthrough various branches of the circuit.

So norton's theorem allows you to take any circuit with any number of power sources and reduce it to a single current sourcein parallel with a single resistance, along with the load. This forms a simple current divider circuit, that allows for easy analysisof the circuit.

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