Norton's theorem
now I look at another theorem which can be considered the dual of the Thevenin theorem it's known as Norton's theorem just like Seventeen's theorem gave us a
representation of a complicated circuit add two terminals with a single voltage source and a single series resistance Norton's theorem will also
give us an equivalent circuit add two terminals which consists of a single current source and a single resistance in parallel okay and the derivation of this equivalent is similar to the derivation of the Thevenin equivalent okay so let's say we have a network end consisting of independent sources
I will show them by a single source like this resistors I had linear controlled sources and it's connected to some other circuit okay let me call this my circuit now what we wanted was a representation of this circuit n at the terminals 1 & 1 prime so it
can be a stand-in for the circuit regardless of what you connect to it now if you recall while deriving Thurmond is equivalent we replace this with a current source based on substitution theorem what we'll do here instead is let me again say this voltage happens to be VL
and this current is il I will replace it with a voltage source instead okay so I'll substitute this whole thing by the voltage source whose value is the voltage that actually occurs in the circuit well as before we see that the derivation will be independent of whatever we
LS so it applies to any circuit okay anything that you connect again this is the circuit for which we have come building the equivalent and this circuit needs to have only independent sources linear resistors and linear control sources this can be anything it doesn't matter what it is
okay because whatever it is between these two terminals it can be substituted by a voltage or current source of that specific value whatever voltage or current it's drawing and finally whatever we are going to derive for this the equivalent for this will be independent of the value of
wheel okay so let's take this and proceed as before now let's say I were given this circuit and asked to compute IL now there are independent sources inside and there is one independent source outside which is real ok how would I go about solving for il I could
use superposition I can think of this as superposition of two cases I will indicate a plus here to indicate superposition and what are the two cases one in which the internal sources are active and really set to 0 that is VL is replaced by a short circuit I
will compute il 1 okay and the other case is where the internal independent sources are all nulled or deactivated I will have only this VL to be active and I compute il to my il in the original case would be the sum of these two cases il 1
plus il 2 basically I have these independent sources VL and all of the internal independent sources in one case I activate all the internal independent sources but null VL and in the other case I deactivate all of the independent sources and activate only VL okay now let's go
step by step what is this in this case what I've done is to have all the internal independent sources to be active okay so these are all active and I terminate the output or this terminals 1 & 1 prime where I want to find the equivalent with a
short-circuit okay because when I said we L equal to 0 that's what I get I have a short circuit and I measure the current through the short circuit okay here I called it il one but this is usually called I n the Norton current okay it's the northern
current or the short-circuit current okay so I terminate one one prime with a short circuit and measures the current in the short circuit in the right direction okay going from 1 to 1 prime that gives me the short-circuit current or the norton current okay and let me look
at the second case where all the internal sources Arnold but all the internal sources are not independent sources are null we have only linear resistors and linear control sources and we know that such a network consisting of linear resistors and linear control sources looks like a resistor from
these two terminals okay and I'm considering the case with this world and looking in here it always looks like some resistor what it means is that if I applied some we test and measure I test retest by I test would be some constant which is given by our
equivalent in this particular case I will call it RN the northern resistance okay so this is equivalent to having an order of resistance RM okay and it's driven by this VL and I'm measuring il-2 in that direction okay so clearly il-2 would be minus VL divided by r
and okay and I have to superpose this with the short-circuit current to get the total current I'll do that later if you recall what we calculated for the Thevenin equivalent in terminal equivalent also there was a resistance rth it was exactly the same thing we deactivated the internal
sources and found the resistance looking in the only difference is there we were applying a current and here we happen to be applying a voltage but it doesn't matter whether you apply a voltage or current the ratio of ITA's by I just remains exactly the same so although
I call it RN just to be consistent with the notation for Norton's equivalent it is exactly equal to R th the terminal resistance the Thevenin resistance and northern resistance r1 and the same thing okay so now let's go about our superposition with the full circuit that is with
the independent sources active and we have all the components in place if we apply VL we have a certain current il and from the two cases that we took for superposition we know that this is equals I am the short-circuit current or the Norton current minus VL divided
by RN ok now this entire thing can be represented using a simple equivalent ok the current n which corresponds to this particular term ok and then I have the voltage VL over there and to get this term I simply connect a resistance R M ok so clearly you
see that the current flowing here is we L divided by RN the current flowing that way is IM the short circuit current so the current flowing here is nothing but this current – that current which is i n minus VL divided by RN ok this is fine if
you recall with the Thevenin equivalent we had got well to be v th – IL times R th and we have a similar equation here there we could emulate the output voltage using a voltage v th in series with the resistance rth here we emulate the output current
il using a current source I am in parallel with a resistance RN which also is equal to R th okay and this particular circuit is the equivalent of this ad the terminals 1 1 prime okay this is 1 and 1 Prime and this is known as Norton's equivalent
circuit of the network and ok so as I said it's the dual of the Thevenin equivalent where we have a voltage source here we have a current source okay we can state Norton's theorem in a way similar to heaven and theorem any circuit with independent sources and linear
components that is linear resistors and controlled sources can be represented at two terminals by a current source and parallel with a resistor okay and what is this current source this is I n which is the current flowing in the short circuit between two terminals and this resistor R
n is nothing but a resistance looking into the two terminals with the independent sources deactivated or null that is they are said to zero okay so exactly the dual of theveninís theorem now the same circuit and can be represented by its Thevenin equivalent we th in series with
xxx or by the Norton equivalent which is I n in parallel with r n RN of course is equal to R th again the equivalence holds for the circuit between these two again the equivalent holds at these two terminals 1 and 1 Prime and what it means is
that if any circuit is connected to these two terminals 1 1 prime you connected to either the full circuit or the Thevenin equivalent or the Norton equivalent the currents and voltages at these terminals will be exactly the same and consequently those in the circuit that you connect will
also be exactly the same okay so when you are interested in modeling this circuit you can really you can either use this or that one now one thing I want to point out is the relationship between the value of ETA s and I n these two are also
exactly equivalent to each other that's what it means right because this circuit is equivalent to this which is equal into that so these two also equivalent to each other or the relationship between iron and we th can be easily found out by considering the open circuit voltage between
these two okay because this circuit should also have exactly the same open circuit voltage as this one if they are to be equivalent the open circuit voltage in this case in the Havanese case obviously equals V th no current flows through rth Vth is the open circuit voltage
in fact that's the definition of ETS and in this case if you don't have anything connected to one and one prime all of this I am flows into RM okay so the voltage across one on one prime is simply I n times RN or I n times R
th okay so this is the relationship between the feminine voltage and the norton current of a given circuit okay
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