# 2 current sources. circuit analysis

#### jericoperlas

Joined Oct 11, 2017
7
I need help understanding this circuit.
whats the V, I, and P at the resistor? and why

#### WBahn

Joined Mar 31, 2012
26,398
What is your best attempt at answering the question (including the why part)?

#### RBR1317

Joined Nov 13, 2010
691
Would it be any easier to understand if the circuit were redrawn from a vertical to a horizontal layout?

#### jericoperlas

Joined Oct 11, 2017
7
I = 5A
V = 100V
P = 500W

#### jericoperlas

Joined Oct 11, 2017
7
What is your best attempt at answering the question (including the why part)?

I = 5A
V = 100V
P = 500W
this is what i originally thought. and then i put it into a circuit simulator and its saying the 4 A current source wouldn't output. that's whats throwing me off. hence the color grey in the photo.

#### WBahn

Joined Mar 31, 2012
26,398
I = 5A
V = 100V
P = 500W
this is what i originally thought. and then i put it into a circuit simulator and its saying the 4 A current source wouldn't output. that's whats throwing me off. hence the color grey in the photo.
What simulator are you using?

It appears to be one of these online easy-to-use simulators. These often have problems.

I would recommend downloading and learning how to use LTSpice. It's free and it's a real simulator intended for real work. While you can get up and going very quickly with it, there IS a learning curve that has to be dealt with as you go. The good news is that there's a lot of folks here that are very proficient with it (not to mention user forums dedicated to it).

#### jericoperlas

Joined Oct 11, 2017
7
What simulator are you using?

It appears to be one of these online easy-to-use simulators. These often have problems.

I would recommend downloading and learning how to use LTSpice. It's free and it's a real simulator intended for real work. While you can get up and going very quickly with it, there IS a learning curve that has to be dealt with as you go. The good news is that there's a lot of folks here that are very proficient with it (not to mention user forums dedicated to it).
So am I right?

#### WBahn

Joined Mar 31, 2012
26,398
So am I right?
Yes, you are correct.

So, if nothing else, I would recommend never using that simulator again -- you can't have any faith that it will yield correct results.

#### Vahram Hambardzumyan

Joined Jul 29, 2017
7
Generally two current sources connected in parallel can provide current flow into load equal to the sum of these two current sources.
In series current sources are not defined, and there is no guarantee what you will have.

In your case following calculation looks right:

I = 5A
V = 100V
P = 500W

#### MrAl

Joined Jun 17, 2014
8,984
Yeah that's weird about the simulator because everybody who analyzes circuits knows that currents add like that.
It must have gotten confused or the designers took too many short cuts to a solution.
Now voltage sources are a different story, two of them in parallel are not allowed unless maybe they are the exact same voltage.

#### manson1972

Joined Sep 14, 2015
3
You connected the resistor terminals to the 1A source directly. That's why the "connections" for the 4A source are red circles. You need to connect the resistor terminals to the 4A source first, and then to the 1A source.

#### WBahn

Joined Mar 31, 2012
26,398
Yeah that's weird about the simulator because everybody who analyzes circuits knows that currents add like that.
It must have gotten confused or the designers took too many short cuts to a solution.
Now voltage sources are a different story, two of them in parallel are not allowed unless maybe they are the exact same voltage.
Even then it would get many simulators fits because the solution is indeterminate -- pick any current you want in one source and you can find a current in the other source that is a solution.

And the same indeterminancy issue exists with two identical current sources in series -- the voltage across one can be anything as long as the voltage across the other compensates.

If the simulator assumes at least some internal resistance, then the solution can converse. but the same is true for current sources in series.

#### crutschow

Joined Mar 14, 2008
29,786
Below is the LTspice simulation:
Results as expected.