Thevenins and nortons circuit help

[Arouse1973]

It's a bit small I can't make it out. First have ago at the Thevenins question first. I'll work on the Norton's for you. But as this is homework I can't give you the answer.
Adam

 

[Arouse1973]

So for the Norton equivalent you need to short out the load resistor. Calculate the short circuit current. Remove the short and the load and short out the 12 Volts. Calculate the total resistance. Now Norton circuits don't have a voltage they have something else, see if you can find out what that is. Also the resistor is connected differently to the Thevenins resistor, also see if you can work it out.
Adam

 

[Laplace]

...if I could work out the resistance then the rest will follow?

Ohm's Law relates 3 items: voltage, current, resistance. Given any two the third can be calculated.

Equivalent linear circuits have 3 items: Thevenin voltage, Norton current, Equivalent resistance. Ohm's Law applies here.

Does R5 have a value?

 

[lp2014needhelp]

Laplace R5 is 143

 

[lp2014needhelp]

Adam its not homework its extra work i'm doing to try and improve my understanding ready for when I start college next march. I asked the tutor for some extra circuits but not this hard.

 

[Laplace]

Took your circuit for finding the equivalent resistance and redrew it. Is that what you used for finding Rth?

 

[Arouse1973]

Nice one Laplace!

 

[Arouse1973]

Adam its not homework its extra work i'm doing to try and improve my understanding ready for when I start college next march. I asked the tutor for some extra circuits but not this hard.

Oh Ok we can work through this together.
Adam

 

[Arouse1973]

Ok for the Norton I think it is this. Laplace will correct me here if I am wrong.
Adam

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Fig 1. Original circuit with load resistor

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Fig 2. Short out load and calculate short circuit current. Note this is not the same as the total current from the source. It's what is going through the short circuit.

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Fig 3. Remove loads from output and short voltage source or make it Zero Volts it's the same thing.

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Fig 4. Draw circuit diagram to represent Norton circuit consisting of a current source and parallel resistor. Set the current source to the short circuit current you just worked out and the resistors value which equals the equivalent resistance of the circuit you just worked out. Now connect your load across the output as the original circuit shows and that's it.

 

[Laplace]

I was looking at finding the equivalent resistance of the original circuit and noticed the configuration is an unbalanced bridge. There is no easy way to calculate an unbalanced bridge by combining resistance in series and parallel. One needs to write the node equations, find the two bridge voltages and work from there.

 

[Laplace]

Looking for the total current flowing through a bridge. Attached image shows the node equations for the two bridge voltages. Total current into the bridge is Ii=(1-Vab)/Ra+(1-Vcd)/Rc where Vi has been arbitrarily set to 1 for convenience. Equivalent resistance is Req=Vi/Ii.

 

[Arouse1973]

Laplace Your a tech nutter. Thats a compliment by the way I thought there was something strange about that circuit.
Thanks Adam

 

[Ratch]

Laplace,

I think you realize that the topology of the schemat in post #26 is not correct. R6 is missing.

I also think the best way to solve this problem is to convert the delta resistor configuration of R2,R3, and R6 into a Y configuration. That is easy to do. Then the open circuit voltage and short current of R4 are easily found without all the heavy algebra of three node equations or three loop equations.

Ratch

 

[Laplace]

I think you realize that the topology of the schemat in post #26 is not correct. R6 is missing. I also think the best way to solve this problem is to convert the delta resistor configuration of R2,R3, and R6 into a Y configuration.

Post #26 dealt with the image file from post #20, which was a different problem.

I must be just a one-trick pony. Node equations are something I can write generally without making a mistake. The algebra is not a problem when using a symbolic algebra engine. However, simplifications using clever transformations usually leave me stranded somewhere up that infamous 'river of excrement'.

 

[The Electrician]

lp2014needhelp, consider the schematic shown in post #6.

Calculate the equivalent resistance of R2, R3 and R6; replace those three with a resistor labeled Rx. Now the current out of the battery is Ib = 12/(R1+Rx+R5)

The voltage across Rx is Vx = Ib*Rx. Now R3 and R6 form a voltage divider and the voltage across R6 can be calculated as V6 = Vx*(R6/(R6+R3)

The voltage across R5 is V5 = Ib*R5

Vth is the voltage from ground to the junction of R3 and R6. That voltage is the sum of V5 and V6. What do you get if you carry out all these calculations? To help us help you, show each step of all the calculations; carry at least 5 digits in all your calculations.

 

[The Electrician]

Looking for the total current flowing through a bridge. Attached image shows the node equations for the two bridge voltages. Total current into the bridge is Ii=(1-Vab)/Ra+(1-Vcd)/Rc where Vi has been arbitrarily set to 1 for convenience. Equivalent resistance is Req=Vi/Ii.

View attachment 17718

You might find it interesting to have a look at this:

http://www.clever4hire.com/throwawa...ngineering-notes/Weston2-5.pdf?attredirects=0

Look on page 6 under the heading "Wheatstone bridge--List of circuit equations".

 

[Laplace]

You might find it interesting to have a look at this:

Interesting, indeed. And done in 1947 before the advent of modern mathematical computational tools that can give the same results in just minutes of interaction with the PC. I wonder how many reams of paper the author used to derive those simplifications? The year 1947 seems like just a short time before I first learned algebra.