Can anyone help with the attached circuit. I can't understand how to work it out!

I thought you had to turn the voltage source into a current source by dividing it by the nearest resistor? But someone else told me you don't...........

There's a beer in it for the most helpful person

 

Show your attempt.

Convert the series combination of the 7 volt source and the 7 ohm resistor to a current source in parallel with a suitable resistor (that's what you will calculate).

Then what do you think you should do next?

 

7 Volts / 7 Ohms = 1 Amp

The 1 Amp is now in parallel with the 7 Ohm resitor

Add the 1 Amp to the 6 Amps = 7 Amps

The 7 Ohm and 9 Ohm are now in parallel, so 7X9/7+9 = 63/16 = 3.9 Ohm Resistor

So the new circuit looks like this:

Now I would turn the current source back to a voltage and put the 3.9 Ohm resistor in series with the voltage source?

 

You are right.A Norton's circuit consists of a current source in parallel with a resistor .So you dont need to convert it to a voltage source again

 

Do I now need to remove the 3.9 Ohm resistor to get the final answer?

 

No you should not remove it.Your answer is right.Just use current division rule and find the current through 3 ohm ( 7 x 3.9/(3.9+3))

 

Ok thanks.

So the answer is 3.95 Amps running through the 3 Ohm resistor (as per the original question)?

 

Anyone? Please.


Tin Whisker

 

Yes, that's about right. I get 3.97297 amps through the 3 ohm resistor.

 

Thanks for your help everyone.

I will post another example if you don't mind checking my answer?

 

Answer:

6v/6 Ohms = 1 Amp

1 Amp + 5 Amps = 6 Amps

6x4/6+4= 24/10 = 2.4 Ohms

6 x 2.4/2.4 + 2 = 3.2727 Amps

 

That's correct.

 

please..reply me if my answer is wrong..... I had a thought when we replace the network having 9ohms resistor with short circuit.. in my opinion the current source must be removed since short ciruited and the answer is 1.5 amps...neglecting the current source