# Various Simple Questions

Discussion in 'Homework Help' started by Hitman6267, May 27, 2010.

1. ### Hitman6267 Thread Starter Member

Apr 6, 2010
82
0
I'm solving a test so I'm going to have a lot of questions. If they're simple enough I'm going to post them here instead of making new threads.

Question 1:

The solution says the current through R is 6A. Is it like that because the current sources are attached at one end so the current at that end is their sum ?

2. ### Markd77 Senior Member

Sep 7, 2009
2,796
595
4 and 2 Amps have to flow through the 2ohm resistors and there is only one place it can go.

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3. ### Hitman6267 Thread Starter Member

Apr 6, 2010
82
0

If I want to calculate the current passing the 5 ohm resistor I have to apply current divider.

The correct way of doing it is to consider the resistors I circled in red in the current divider formula and not only the 4 ohm and 5 ohm alone.

Why is that ?
In what case would I need to only consider the 4 and 5 ohm only ?

4. ### Markd77 Senior Member

Sep 7, 2009
2,796
595
The way to do these is to find combinations of resistors that could be replaced by a singe one. In this case the first thing is to parallel the 6 and the 2, then series that with the 0.5. Then parallel that with the 4. It might help to redraw the circuit at each stage.

5. ### Hitman6267 Thread Starter Member

Apr 6, 2010
82
0
Yes i know that, I have the solution.

My question is why we can't calculate the current through the 5ohm like follows:

i= 0.5+4 /0.5 XIs

When do we only include the resistors we're interested in ? And when do we have to take Req

6. ### Markd77 Senior Member

Sep 7, 2009
2,796
595
Presumably you have to find Io given Is. That means you have to find the voltage at Is which means that you need to find the equivalent resistance of the whole circuit. Each branch of the circuit has a different current so I don't think there is an easier way.

7. ### Hitman6267 Thread Starter Member

Apr 6, 2010
82
0
I'm only interested in current divider, I don't have a goal to reach (i.e. find Io). There is something wrong with the way I'm applying current divider.

I just remembered that I've already asked that question in another thread. I didn't totally get the answer I got there so I'm going to re-ask. Here if you're interested

8. ### Hitman6267 Thread Starter Member

Apr 6, 2010
82
0

If we use source transformation to transform Vs into a current source parallel with a resistor.

Why are we allowed to substitute the current source and the 4 mA current source with one equal to their sum ? Don't they need to be in series so we can do that ?

9. ### salmanshaheen_88 Active Member

Mar 5, 2009
88
1
Use mesh and nodal analysis

10. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
If two ideal current sources of different value are connected in series the resultant current is indeterminate - both sources are trying to drive different currents, which is a nonsense. The same is true of connecting two ideal voltage sources in parallel - the resultant is again indeterminate.

Last edited: May 28, 2010
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11. ### Hitman6267 Thread Starter Member

Apr 6, 2010
82
0
oh yea, you're right. I mixed things up. We can add them up because they're in parallel.

We add voltages sources in series and current sources in parallel.