Norton's Theorem Poker

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. Join 326,364+ Members & Access 4,019,197 Poker Strategy Posts, Exclusive Freerolls & Added Money Buyins at the CardsChat™ Online Poker Forum.

Definition of Norton’s Theorem

As per Norton’s theorem, we can replace a complex linear active two-port network with a single current source in parallel with a resistance.

The next question comes to our mind what would be the current and resistance of the source. First of all, we have to short circuit the ports of the network. As a result, the current flowing through the short circuit path between the ports is the current value of the source. Then we have to consider the value of the parallel resistance of the source. Because of that, we have to replace each connected source of the active network with its internal resistance. Then we calculate the equivalent resistance of the network between the said ports. This calculated value is nothing but the value of the internal resistance of the current source. Hence we connect this resistance in parallel with that imaginary source.

Explanation of Norton’s Theorem

To explain Norton’s Theorem in detail let us consider a common active linear network having one voltage source, one current source and resistances connected in certain manner.

Norton’s Current

To find out Norton’s Current, we first remove the resistance RL.

Now, we will short circuit the terminal A and B.

By applying any circuit analysis we can easily find out the current through the short circuit path between A and B. That current would be the source current of the equivalent current source.

Norton’s Resistance

Now we will open A and B and replace all the sources of the active network by their internal resistance. If the voltage sources of the network are ideal, we will replace them with short circuits. Since the internal resistance of an ideal voltage source is zero. If the current sources of the network are ideal, we will replace them with open circuits. Because the internal resistance of an ideal current source is infinite.

After doing that, we will calculate the equivalent resistance of the network across terminal A and B.

Now we can draw a current source in which source current is short circuit current between A and B. Likewise the parallel resistance is the equivalent resistance between A and B.

Lastly, we will connect back the resistance RL across the source. Now we have to calculate the current through the resistance RL. So we have seen that we can very easily find out the current through the branch AB as the circuit or network is now in its simplest form.

This calculated current through the resistance RL is exactly equal to the current which would have been flowing through the resistance RL if the entire network was connected to RL. This is the simplest explanation of Norton’s Theorem.

Norton

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

Norton's Theorem Poker Practice

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 ' INO' and a single parallel resistor, RNO. 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 'INO' in parallel with a resistance RNO.To find its Norton equivalent circuit,

  1. Find the Norton current 'INO'. Calculate the output current, 'IAB', 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:-

Norton

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 :

Norton s theorem problems

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.

Norton

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.

Norton

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

Norton's Theorem Poker Rules

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 ,

VTh=INoxRTh

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.