Help with Nortons Equivalent. 2 sources... confused lots.

[Ledwardz]

Hi, ive attached the circuit i have. My first problem was;


do i work out the norton resistances from A-B in which case the Norton resistance is in parallel and equal to (15/8)k

or are the resistances from the 10v which gives a total resistance of 8K

meh..... its pointless me trying to explain hopefully the attachment will make it a little more clear im confused on a few things. any help appreciated.

(also, sorry for the 'crapness' of the photo, my scanner is broken)
its a 10v supply to the left and 1ma supply to the right with a 3k resistor to the top left and 5k between the 2 supplies.

Thanks for looking.

 

[(*steve*)]

I think you may have to try attaching that diagram a second time...

 

[Ledwardz]

erm it should b there now dnt know y tht didn't work the first time.

 

[Ledwardz]

actually this is probably more clear..... should have posted this first.

 

[barathbushan]

you have to convert the voltage source in series with the 3k resistors , then the circuit turns out to be two current sources and two resistors in parallel , you add up the current sources and compute the parallel resistor combination as shown in the diagram

I think this thread must have been started in the homework section

 

[Ledwardz]

you have to convert the voltage source in series with the 3k resistors , then the circuit turns out to be two current sources and two resistors in parallel , you add up the current sources and compute the parallel resistor combination as shown below

I think this thread must have been started in the homework section

Thats brilliant. Thanks. could you tell me why you didn't include the 5k resistor when working out the current for the 10v supply to get the current for it?

 

[barathbushan]

That's because the 5k wasn't in series with the voltage source , which means it doesn't come into the picture when the 10v voltage source was converted to 3.3ma current source

here's a golden rule in solving network problems

Voltage with a series resistor , and a current source with a shunt resistor are interconvertible

 

[Ledwardz]

That's because the 5k wasn't in series with the voltage source , which means it doesn't come into the picture when the 10v voltage source was converted to 3.3ma current source

here's a golden rule in solving network problems

Voltage with a series resistor , and a current source with a shunt resistor are interconvertible

cheers!