# Thevenized or Nortonized circuit is a reduced form of a original circuit

#### PG1995

Joined Apr 15, 2011
832
Hi

A Thevenized or Nortonized circuit is a reduced form of a original circuit; in other words a simple form of a complex one. For instance, a circuit with 20 resistors could be reduced to an equivalent circuit which uses no more than one resistor. Well, why don't we make a simple circuit instead of complex one? What's the need of wasting money on 20 resistors when only one can do? Would you please tell me? Thanks.

Regards
PG

#### kubeek

Joined Sep 20, 2005
5,790
First, this is true only for linear sources, that is resistors and voltage and current sources. Most circuits don´t have just that, they have capacitors, inductors and semiconductors so they can´t be simplified like that.

You may need 20 resistors for example when the heat dissipated per resistor is too high for the part. Other example are high voltage dividers, where the maximum permissible voltage across the resistor makes you use more in series.

Joined Dec 26, 2010
2,148
Models of this kind are useful because they allow us to predict how things will behave. If the behaviour of a circuit viewed from a couple of terminals can be represented by a simpler network, so much the better as far as ease of calculation is concerned. That said, such a model is only a mathematical equivalent of the behaviour seen at a given point. There is no reason to suppose that generally you could physically replace a circuit by (say) a Thévenin equivalent.

Apart from anything else, the equivalent circuit would only replace the behaviour seen from the chosen point, without replicating the operation of other parts of the circuit, which may be necessary in practice. The equivalent circuit may also be impossible to build. For instance, suppose a 12V battery is built of eight cells in series, each of 1.5V emf and 0.25Ω internal resistance. Its equivalent circuit could have a single voltage source of 12V in series with 2Ω resistance, but you can't make that in practice.

Another example might be an equivalent for the output of a mains power supply circuit, or perhaps an amplifier. Such a circuit may need many parts to do its job properly. Again, you might represent the output by a simple voltage source and a resistance (or possibly a complex impedance), but you can't just magically call the equivalent circuit into being. The power supply regulation or the gain of the amplifier may be the result of many circuit components working together, and the equivalent circuit is just a way of imitating that.

A student may of course be given complicated networks of sources and impedances to be worked through as learning exercises. Some of these may have real-world relevance, e.g. bridge circuits, while others may be more contrived.

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#### debjit625

Joined Apr 17, 2010
790

#### PG1995

Joined Apr 15, 2011
832

Would it be possible for any of you to show me a simple circuit which makes it clear that Thevenin equivalent circuit cannot replace the original one? It would be very kind of you.

Regards
PG

#### kubeek

Joined Sep 20, 2005
5,790
For example this. Even if the diodes are ideal, so that the voltage drop on them or Vf equals zero volts, the circuit cannot be described by equivalent thevenin circuit.

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#### PG1995

Joined Apr 15, 2011
832
For example this. Even if the diodes are ideal, so that the voltage drop on them or Vf equals zero volts, the circuit cannot be described by equivalent thevenin circuit.
Thanks, kubeek.

The circuit cannot be Thevenized or Nortonized because it contains diodes. Correct? Please let me know. Thanks.

Joined Dec 26, 2010
2,148
More generally, it is because the diode circuit is non-linear. A Thevenin equivalent circuit might be made up to cover a very limited range of operation for such a circuit, but it could never provide a complete description. It could therefore never act as a physical equivalent over the full range of possible operating conditions

http://en.wikipedia.org/wiki/Thévenin's_theorem#Practical_limitations

#### debjit625

Joined Apr 17, 2010
790
PG1995 said:
Would it be possible for any of you to show me a simple circuit which makes it clear that Thevenin equivalent circuit cannot replace the original one? It would be very kind of you.
I think you still didn’t understand, otherwise you would not ask this question...
Ok let’s correct your first question...that might make any sense to you.....

PG1995 said:
For instance, a circuit with 20 resistors could be reduced to an equivalent circuit which uses no more than one resistor.
Its not just about reducing resistors, like reducing from parallel or series network of resistors, don’t forget you are reducing for a voltage source and a series resistor with it. You are doing so because in a big network of resistors and different voltage source you load(a simple resistor in the network) is connected now if you change the load the whole system will be affected as the new load may use more or less current and it's voltage drop will be proportional to its current and value of resistance (V = I x R ,ohms law).So to know about its voltage drop or current you have to recalculate the whole network ,may be 100's of resistor and couple of voltage source.
So in Thevenin equivalent circuit we reduce the whole network with a single voltage source and a series resistor so that if you change the load ,we will only have to calculate a circuit with only two resistors one of them is the load and one voltage source, it will be easier to do it rather solving a big network.
In my earlier post I have provide you some link read it properly you will understand.

Remember when you will understand this stuff properly these statements will never arise "Thevenin equivalent circuit cannot replace the original one" or "Thevenin equivalent circuit can replace the original one”. They really doesn’t means anything in context of network theorems.

There are many passive circuits (networks) which can’t be Thevenized or Nortonized.But that’s different it just can’t be solved using those two theorem.

Good Luck

#### t_n_k

Joined Mar 6, 2009
5,455

Would it be possible for any of you to show me a simple circuit which makes it clear that Thevenin equivalent circuit cannot replace the original one? It would be very kind of you.

Regards
PG
An obvious example would be BJT common emitter amplifier voltage divider bias. One can replace the Base side voltage divider bias network with an equivalent Thevenin voltage source and resistance. So you saved on having one resistor. Trouble is you then need another DC source (while still needing the main amplifier Vcc supply) for the Thevenin source. Nothing is gained.

#### steveb

Joined Jul 3, 2008
2,436
PG1995 said:
Would it be possible for any of you to show me a simple circuit which makes it clear that Thevenin equivalent circuit cannot replace the original one? It would be very kind of you.
... Trouble is you then need another DC source ...
And let's not forget the fact that this new and additional DC source we would have just specified is an "ideal" source, which must be purchased at "Ideal Sources Are Us, Inc.", and they always seem to be out of stock.

#### PG1995

Joined Apr 15, 2011
832
I think you still didnt understand, otherwise you would not ask this question...
Ok lets correct your first question...that might make any sense to you.....

Its not just about reducing resistors, like reducing from parallel or series network of resistors, dont forget you are reducing for a voltage source and a series resistor with it. You are doing so because in a big network of resistors and different voltage source you load(a simple resistor in the network) is connected now if you change the load the whole system will be affected as the new load may use more or less current and it's voltage drop will be proportional to its current and value of resistance (V = I x R ,ohms law).So to know about its voltage drop or current you have to recalculate the whole network ,may be 100's of resistor and couple of voltage source.
So in Thevenin equivalent circuit we reduce the whole network with a single voltage source and a series resistor so that if you change the load ,we will only have to calculate a circuit with only two resistors one of them is the load and one voltage source, it will be easier to do it rather solving a big network.
In my earlier post I have provide you some link read it properly you will understand.

Remember when you will understand this stuff properly these statements will never arise "Thevenin equivalent circuit cannot replace the original one" or "Thevenin equivalent circuit can replace the original one. They really doesnt means anything in context of network theorems.

There are many passive circuits (networks) which cant be Thevenized or Nortonized.But thats different it just cant be solved using those two theorem.

Good Luck
Hi Debjit

Thank you for the help. I have checked out those links but I'm sorry to say I'm still confused! The screenshot attached is from one of those links. See, in both circuits 2 ohm load resistor is used but the thevenized circuit is simple and clutter-less. I have vague picture that why thevenized circuit cannot replace real circuit but I need some light to see thru that vagueness.

An obvious example would be BJT common emitter amplifier voltage divider bias. One can replace the Base side voltage divider bias network with an equivalent Thevenin voltage source and resistance. So you saved on having one resistor. Trouble is you then need another DC source (while still needing the main amplifier Vcc supply) for the Thevenin source. Nothing is gained.
Hi t_n_k

Thank you for your reply. Could you please attach a diagram to graphically show what you are trying to tell me? It would be little easy for me to grasp your point. If it's not possible then it's absolutely okay. I still appreciate your input and try to understand what you said. Thanks.

Best wishes
PG

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#### steveb

Joined Jul 3, 2008
2,436
I have vague picture that why thevenized circuit cannot replace real circuit but I need some light to see thru that vagueness.
Can you clarify your question? Are you asking why a Thevenin circuit can not replace an actual circuit in the real world?

#### PG1995

Joined Apr 15, 2011
832
Can you clarify your question? Are you asking why a Thevenin circuit can not replace an actual circuit in the real world?
Yes, you guessed it right!

#### steveb

Joined Jul 3, 2008
2,436
Yes, you guessed it right!
OK, so think about what you are asking. A Thevenin equivalent voltage source is made of two components; an ideal voltage source and an ideal resistor. Let's say you want to make a real voltage source. Perhaps you use a battery, attach wires and connectors and put it in a box. Or, perhaps you make an AC generator, rectify and filter the output and use a voltage regulator and capacitors, and again put it all in a box. How could you possibly replace these or any other real device by an Thevenin equivalent?

Let's go through it logically. The ideal voltage source, which is one of the two components in a Thevenin equivalent circuit, is an ideal voltage source, which does not exist in the real world. Further, if it did exist, why would you add a resistor in series with it and make the total system a non-ideal voltage source? We have to make voltage sources with real devices, and the end result is always an imperfect voltage source which includes parasitic resistance, inductance and capacitance. We do our best to minimize these parasitics and other nonideal behavior (ripple, nonlinearity, etc.), but there is always some practical limit.

I think t_n_k was cutting you some slack and allowing you to define a nearly ideal voltage source that approximates an ideal voltage source well enough for your purposes. Think of a battery with very very low source resistance. There is no way to remove the source resistance, but it might be negligible in your application. Here he was making the point that you don't need to use two of these nearly ideal batteries, but you can use just one. But, when you analyze it, you are free to make a Thevenin equivalent representation.

To some extent, your question is a little bit like asking why winged stallions and unicorns are not ever entered in horse races. My response above was a serious response to your question. You can't buy an ideal voltage source, because they don't exist, hence you can't use them to replace anything in the real world. What you can do is use them abstractly as a tool to simplify analysis and help in understanding complex systems by reducing them to simpler systems.

#### steveb

Joined Jul 3, 2008
2,436
The screenshot attached is from one of those links. See, in both circuits 2 ohm load resistor is used but the thevenized circuit is simple and clutter-less. I have vague picture that why thevenized circuit cannot replace real circuit but I need some light to see thru that vagueness.
So hopefully my above discussion helps to some extent. But, let's go further from another angle. In your attached diagram you are asking, "Why use a cluttered ciruit instead of a simple one?". What if those resistances represent light bulbs in the real world? These light bulbs may be physically placed where you need them (reading lights, warning indicators etc.). Perhaps the ideal voltage sources in the diagram are representing real sources that are not really ideal, but maybe they have source resistances much much smaller than the light bulb resistances.

So, replacing all this with one very good voltage source and one resistor does nothing useful. The resistor does not emit light and you have lost all the functionality of your multi light bulb system.

Still, the Thevenin circuit might prove useful to you. Maybe you would like to know how much power is dissipated inside your room or box. The Thevenin equivalent circuit can give you that answer. Maybe instead of light bulbs, those resistances are heaters. Then you would really care how much power is dissipated.

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#### steveb

Joined Jul 3, 2008
2,436
Hopefully, the above two posts clarify, but I have even one more answer which might get to the real heart of why you are asking this question.

Sometimes teachers and textbooks give pointless questions, at least from a practical point of view. They might be interested in giving a problem that allows you to work out the mathematics, but there might be absolutely no point or use in the exercise. This can sometimes be a cause of confusion. So, you get a problem with a random bunch of restistors and sources and you make a Thevenin equivalent, and you make a Norton equivalent, and then everyone stands up and cheers without ever asking, what the point of the process is. Both the complex circuit and the simple circuit never had a use and never represented anything physical in the real world.