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>> In the next few videos, we're[br]going to extend the concepts of

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equivalent circuits into the phasor domain,

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in terms of impedances and[br]phasor voltages and phasor currents.

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So to do that,

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we're going to start by looking[br]at the source transformations,

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transforming a voltage source with

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a series impedance into a current source[br]with a parallel impedance and back.

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Then we'll also extend the concept of

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Thevenin equivalent circuits to include[br]phasors and complex impedances.

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So by review, what we mean when we[br]say two circuits are equivalent,

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we mean that they have in this case[br]the same terminal characteristics.

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By that we mean that[br]an external circuit connected to

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a voltage source with a series[br]impedance will experience

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the same voltage and current[br]as that same external circuit

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would experience if it were connected[br]to a parallel current source,

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connected in parallel with an impedance.

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Both instances, we're going to have[br]the source impedance be the same value,

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and what we wanted to do is[br]determine the relationship between

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V sub s and I sub s.

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So that loads or

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external circuits connected to either

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of these would not be able[br]to tell the difference.

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So they have the same[br]terminal characteristics.

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That means same voltage and same current.

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To accomplish that, this load here[br]is going to experience the same V,

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what reference V12, the voltage from

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node one to node two in both the circuits.

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In other words, this V12 and[br]this V12 will be the same.

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So to do that,

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we need to have the open-circuit voltage.

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The voltage that you would[br]experience if there was

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no load connected here to be the same.

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In this case, since it's[br]open circuit there'll be

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no current flowing through here,

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and the voltage V12 will simply equal V,

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the open circuit will equal

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V sub s. Down here when[br]the terminals one and two are open,

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no current is coming this way.

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So in this case, all of[br]the current from the source is

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going through this parallel impedance,

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and the voltage that you[br]would then measure here,

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this V open circuit,

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would equal just I sub s times[br]Z sub s. From this then,

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we can write directly what[br]the relationship needs to be.

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In order for these two open-circuit[br]voltages to be the same,

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this V_OC which is V sub s,

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must equal this open- circuit voltage here,

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I sub s times Z sub s. So in transforming

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a current source with

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a parallel impedance into[br]a voltage source with a series impedance,

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the voltage source here[br]would be equal to I sub

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s times Z sub s. Simply rearranging it,

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we can come up with[br]an expression for I sub s in

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terms of V sub s. That would[br]be I sub s equals V sub s

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over Z sub s. So if we had[br]a series voltage source and impedance,

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we could replace those with a parallel[br]current source and impedance.

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If I sub s here was equal to[br]the quantity V sub s divided by Z sub s,

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we get a little bit better[br]feel for that by looking

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at what is referred to as[br]the short circuit current.

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If you short this out here[br]and call it I short circuit,

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we should expect to experience[br]the same I short circuit,

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the same current through this short here.

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Well in this circuit here,

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I short circuit, in other words[br]zero resistance there,

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the current there is just[br]going to be V sub s divided by

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Z sub s. On the other hand down[br]here with this being shorted,

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it shorts out the impedance and so[br]none of the current goes through here.

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The short-circuit current then would[br]be simply I sub s or I short circuit

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equals I sub s. Here we then see that in[br]order for these two to be equivalent,

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I sub s equals V sub s over[br]Z sub s as we saw there.