Current Sources in Parallel when Simplifying Circuits

[engineer_ja]

When simplifying a circuit, can a parallel branch with a current source and a resistor in series be considered as just the current source?


Also, if two parallel current sources 'point upwards' i.e. toward the positive rail, can I add them together as one source even if there are other branches in between them?


Thanks everyone!

 

[Kholdstare]

If you can ignore the resistance - yes.
Yes add them together even if other branches are present.

 

[sophiecentaur]

Yes, of course. They will both end up with the same voltage across them, each one supplying the current it's been set to deliver. (There are practical limits to that statement, of course - but only along the same lines as connecting Voltage sources in series).

 

[engineer_ja]

Thanks!

How do I know if I can ignore the resistance?

The question I am trying to do asks for the current in a particular wire of the circuit. Looking at my answer, It seems that i cannot combine things in parallel if they are on opposite sides of this wire. is that correct?

 

[Kholdstare]

If resistance is very small (close to zero ohm) ignore it like a short circuit.

You should be specific about what kind of branch you want to combine.

 

[sophiecentaur]

An ideal current source doesn't care about series resistance but in a 'real' circuit, you would need to know the details of actual resistances involved.
Post a diagram to give us a better idea.

 

[engineer_ja]

Hi, I've attached the circuit picture. To find current in xx I convert the 2 voltages to norton equivalents. Can I then combine the 2A with the left hand current source, ignoring its 5 ohm series resistance??

I then think that I cannot combine the left hand and right hand norton equivalents into one, as I 'lose' the link xx??

Thanks.

 

[Kholdstare]

Convert the voltage sources to Norton. Then combine everything except the current source branch to one current source and one parallel resistance. Then convert it back to Thevenin. Then the analysis is easy. Dont ignore the resistance.

 

[sophiecentaur]

That 2A stays as 2A - no worries there. Just do Kirchoff on the circuit and you should get the right answer. You have currents at x adding to give zero (K1) and you have three loops that you can do K2 on. I think that's enough for a solution with four equations and four unknowns.
@KS that's another approach. I don't see why mine shouldn't work too- not as clever as yours though.

 

[engineer_ja]

Thanks Guys,

@SC, that approach worked for me when I tried, but the answers gave this as an equivalent circuit, and I can't work out how they got there...!

 

[engineer_ja]

sorry file didn't attach!

 

[Jony130]

It seems to me that the solution should look like this