Norton’s Theorem Poker
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*Norton S Theorem Procedure
*Norton’s Theorem Poker Games
*Norton’s Theorem Poker Rules
*Norton S Theorem Problems
*The Norton theorem reduces the networks equivalent to the circuit having one current source, parallel resistance, and load. Norton’s theorem is the converse of Thevenin’s Theorem. Its formation of the equivalent current source instead of an equivalent voltage source as in Thevenin’s theorem.
*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.
Game Theory and Poker. John Nash developed game theory as a branch of mathematics at Princeton University around 1950. As poker has become more popular over the last 15 years or so, players have improved dramatically, to the point where it’s very difficult to consistently to beat the game without game theory knowledge in your corner.1 – I2)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)(RNorton S Theorem Procedure3) = 0(-4)(I3)- (6)(I3 +4) = 0(-10)(I3) =24I3 = -2.4 amperesSo, (InNorton’s Theorem Poker Games) = (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 (RN). 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 isRN = (10 x 4)/ (10+4)= 2.85Ω
*Now we add the current source (IN) with the resistor (Rn) 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 IL will be.IL = (IN) × [RN / (RL + RN)]= (0.1) x [ (2.85)/(2+2.85)]= 0.05 amperes
*For the variable value of the resistor, the current is given here.When (RL) = (8 ohm)IL = (0.1) × [(2.85) / (8 + 2.85)]0.02 AmperesWorking 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.RT =(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.Norton’s Theorem Poker RulesVA-B = I x R =2 x 5.72=11.44V
*The value of current across load resistance is;I= V/R=11.44/ 40 =0.29ALimitations 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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Huge jackpots high roller slots. 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+Follow
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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. Monopoly party train slot machine app.
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.
Casino montreal poker tournament schedule. 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.
Register here: http://gg.gg/vig1j
https://diarynote-jp.indered.space
V- (I
*Norton S Theorem Procedure
*Norton’s Theorem Poker Games
*Norton’s Theorem Poker Rules
*Norton S Theorem Problems
*The Norton theorem reduces the networks equivalent to the circuit having one current source, parallel resistance, and load. Norton’s theorem is the converse of Thevenin’s Theorem. Its formation of the equivalent current source instead of an equivalent voltage source as in Thevenin’s theorem.
*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.
Game Theory and Poker. John Nash developed game theory as a branch of mathematics at Princeton University around 1950. As poker has become more popular over the last 15 years or so, players have improved dramatically, to the point where it’s very difficult to consistently to beat the game without game theory knowledge in your corner.1 – I2)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)(RNorton S Theorem Procedure3) = 0(-4)(I3)- (6)(I3 +4) = 0(-10)(I3) =24I3 = -2.4 amperesSo, (InNorton’s Theorem Poker Games) = (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 (RN). 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 isRN = (10 x 4)/ (10+4)= 2.85Ω
*Now we add the current source (IN) with the resistor (Rn) 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 IL will be.IL = (IN) × [RN / (RL + RN)]= (0.1) x [ (2.85)/(2+2.85)]= 0.05 amperes
*For the variable value of the resistor, the current is given here.When (RL) = (8 ohm)IL = (0.1) × [(2.85) / (8 + 2.85)]0.02 AmperesWorking 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.RT =(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.Norton’s Theorem Poker RulesVA-B = I x R =2 x 5.72=11.44V
*The value of current across load resistance is;I= V/R=11.44/ 40 =0.29ALimitations 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.
JLCPCB – Prototype 10 PCBs for $2 (For Any Color)
China’s Largest PCB Prototype Enterprise, 600,000+ Customers & 10,000+ Online Orders Daily
How to Get PCB Cash Coupon from JLCPCB: https://bit.ly/2GMCH9wWhat is the Maximum Power Transfer Theorem Next-Website AuthorSyed Zain Nasir@syedzainnasir
Huge jackpots high roller slots. 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+Follow
Get ConnectedNorton S Theorem ProblemsLeave a ReplyLeave a Reply You must be logged in to post a comment.
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. Monopoly party train slot machine app.
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.
Casino montreal poker tournament schedule. 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.
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