Norton's Theorem Poker

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1. Norton's Theorem Poker Game
2. Norton's Theorem Poker Rules
3. Norton S Theorem Procedure
4. Norton S Theorem Problems
5. Norton's Theorem Poker Games
6. Norton's Theorem Poker Practice

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)R₄ = 0[6 – (I₁ – I₂)R₄] = 0

• As we know the value of (I₂) which is (-4A) by putting this value in the above equation we have.

[6- (I₁ – (-4))] =0

• Solving this equation we have.

I₂ = -2.5 amperes

• Now we apply KVL at loop three, its crossponding equation is.

(I₃)(R₁) – (I₃ –I₂)(R₃) = 0(-4)(I₃)- (6)(I₃ +4) = 0(-10)(I₃) =24I₃ = -2.4 amperesSo, (I_n) = (I₁ –I₃)= (-2.5 + 2.4)= 0.1 amperes

• It is the current which is moving from A to B.
• Now we calculate the equivalent resistor which (R_N). For this resistor, you should swap all supplies in the circuitry with their interior resistors.
• The value of the net resistance at the points a and b is

R_N = (10 x 4)/ (10+4)= 2.85Ω

• Now we add the current source (I_N) with the resistor (R_n) to make Norton circuitry which is shown in the diagram.
• To measure the output variable, we now add the load resistor at the load points.
• The load current I_L will be.

I_L = (I_N) × [R_N / (R_L + R_N)]= (0.1) x [ (2.85)/(2+2.85)]= 0.05 amperes

• For the variable value of the resistor, the current is given here.

When (R_L) = (8 ohm)I_L = (0.1) × [(2.85) / (8 + 2.85)]0.02 Amperes

Working of Norton Theorem

• As we have discussed different steps to apply the Norton theorem. Now we apply these steps practically on a given circuit.
• In the specified figure, we construct a circuit which has three resistances and two voltage sources.
• To apply the Norton theorem, first of all, remove the centered forty-ohm resistor and join the terminals A and B then we get circuit shown in the figure, represented by A.
• When we join two points A and B, then 2 resistances become parallel to each other then we can find the current passing through this 2 resistance.

I1=10v/10ohm = 1A,I2 =20v/20ohm =1A

• The value of the short current is the sum of these two currents.

I (Short-Circuit) =I1 + I2= 2A

• Now if we remove the voltage sources from the circuit and connect open points with each other and also open the connection among the point A and point B.
• Now the 2 resistances are efficiently linked with each other in parallel.
• The values of interior resistance (Rs) can be found by the sum of resistance at point A and B, the circuitry for this procedure is shown in the figure and represented by B.

R_T =(R1 x R2)/(R1+R2) =6.67 ohm

• After finding the value of the equivalent resistor (Rs) and short circuit current (Is), we make a Norton corresponding circuit. Which is represented in the picture by C.
• Now we have made Norton equivalent circuit, but we have to resolve circuit for the forty-ohm resistance which is connected across the point A and point B. this circuitry is signified in the figure by D.
• We can see in diagram (D) that the two resistors (Rs) and (RL) are now connected in parallel so we find the values of total resistance Rt.

Rt= (R1 x R2)/(R1+R2) = 5.72 ohm

• The value of the voltage at point A and B with load resistance is found by the formula.

V_A-B = I x R =2 x 5.72=11.44V

• The value of current across load resistance is;

I= V/R=11.44/ 40 =0.29A

Limitations of Norton’s Theorem

• This formula is appropriate for the linear modules like resistors.
• It’s not for such modules which are not linear like diodes, the transistor.
• It also not operate for such circuitries which has magnetic locking.
• It also not work for such circuitries which has loaded in parallel with dependent supplies.

It is the detailed article on the Norton Theorem if you have any question about it ask in comments. Thanks for reading. Take care until the next tutorial.

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Norton's Theorem Poker Game

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Syed Zain Nasir

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I am Syed Zain Nasir, the founder of The Engineering Projects (TEP).I am a programmer since 2009 before that I just search things, make small projects and now I am sharing my knowledge through this platform.I also work as a freelancer and did many projects related to programming and electrical circuitry. My Google Profile+

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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

Norton's Theorem Poker Rules

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.

Norton S Theorem Procedure

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.

Norton S Theorem Problems

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.

Norton's Theorem Poker Games

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.

Norton's Theorem Poker Practice

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

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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