parallel resistance Rs of a current source

[PG1995]

Hi

I have always thought of a current source as a very, very high voltage supply with a very, very large resistor, Rs, in series (I mean this large resistor would make series connection with the load resistor, RL). The larger the series resistor, Rs, the less effect the change in load resistor has on the load current. The Rs should be a variable resistor which is adjusted to get the wanted current through the RL. In other words, a current source is also a special type of voltage source.

In a regular voltage source the Rs is very small which can be adjusted to vary the voltage across the RL. If you want more volts to appear across the RL then you need to decrease the resistance of Rs.

Please correct the stuff above if you deem something wrong. Thanks

Now I'm coming to the main question(s).

Please have a look on the linked diagram (or attached one):
https://img843.imageshack.us/img843/3518/imgan.jpg

A current source is represented as shown in Fig. 3 in the linked and attached diagram with the Rs in parallel configuration. I don't get it. Why? As I say above, in my view, a current source is a special kind of voltage source. They say, Rs, is an internal resistance and makes a parallel connection (as shown in Fig. 3). I don't get it. Do you get where I'm having problem? Please help me. Thanks

Regards
PG

 

[FvM]

The (ideal) current source with parallel Rs is an exact equivalent to a voltage source with series Rs. I you observe the source behavior from terminal a/b, you can't decide which of both circuits actually is inside. Although an ideal current source is an abstraction, it's not less real than an ideal voltage source.

 

[srikanth1555]

I am just tryin to get the things easy for you to understand...


consider the ideal current source which doesnt consist of any resistance either in seriers or in parallel with it....
the ideal current source should now provide the constant current to the load...even if you attach a resistance in series with this source the same current will flow through the load.....now if you consider the practical current source,it must be providing less current to the load than required because of some losses...the current gets divided only if you connect a resistance in parallel with the source...and current remains same even you connect the resistance in series with the current source.....so the practical current source can be obtained by connecting a resistance Rs in shunt with the ideal current source....
I hope that his may be helpful for you

 

[PG1995]

Thank you, FvM, Srikanth.

Best regards
PG