# branch current issue

Discussion in 'Homework Help' started by acelectr, Sep 1, 2010.

1. ### acelectr Thread Starter Member

Aug 28, 2010
73
0
Hi guys. I worked a lot about this branch current problem on this link http://www.allaboutcircuits.com/worksheets/dcbranch.html I couldn't do the second question with the 5mA current source. The result that shown is 1.9mA but I'm keep getting 2.72 or somethng. I'm skipping something and I can't find it.

Appreciate any help

2. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115
But 1.9mA is correct answer.
So maybe you show as your solution.

3. ### acelectr Thread Starter Member

Aug 28, 2010
73
0
No 1.9 is not what I've found. that is the problem. in short how can one could solve this problem pls I worked on it a while and still couldn't get in to a conclusion. waiting for backup...

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115
But if you show your approach to the problem then we can tell you in which part you made error.
And I to check the answer, written nodal equation.
5mA = (6V - Va)/1K + (7.2V - Va)/1K

5. ### acelectr Thread Starter Member

Aug 28, 2010
73
0
firstly what I did was to choose the top middle node and determine 3 current directions arbitrarily and tried to apply simultaneous equation method. Well as it is written simultaneous equation is not really needed in order to solve the problem, because the lack of unknown variables. what ever later I applied the kirc's voltage law and see one of the resistors should has a 6 voltage drop across it as the other one has 7.2. I think this is the point where things don't make sense. Because applying ohm's law according to those values didn't really help me. But I also traced each loop and this is what I get. Where do I make the mistake? Though the nodal equation that you wrote, kinda make sense but if I would get an explanation about the contradiction that I'm facing with I would really have a relief.

6. ### acelectr Thread Starter Member

Aug 28, 2010
73
0
firstly what I did was to choose the top middle node and determine 3 current directions arbitrarily and tried to apply simultaneous equation method. Well as it is written simultaneous equation is not really needed in order to solve the problem, because the lack of unknown variables. what ever later I applied the kirc's voltage law and see one of the resistors should has a 6 voltage drop across it as the other one has 7.2. I think this is the point where things don't make sense. Because applying ohm's law according to those values didn't really help me. But I also traced each loop and this is what I get. Where do I make the mistake? Though the nodal equation that you wrote, kinda make sense but if I would get an explanation about the contradiction that I'm facing with I would really have a relief.

7. ### debjit625 Well-Known Member

Apr 17, 2010
790
186
Ok as per the picture I assume some notation
R1 = The 1000 ohms on 6V side
E1 = R1's voltage drop
I1 = Current through R1 which we will find
V1 = 6V
R2 = 1000 ohms .....
E2 = Voltage drop of R2
I2 = Current through R2 i.e.. (5-I1 given)
V2 = 7.2V
I3 = 5mA
Let E be the voltage drop across the stuff through which I3 is flowing
Now
KVL for left side
E1 + E - 6 = 0 ------- eq1
or I1*1000 + E - 6 = 0
KVL for right side
E2 + E - 7.2 = 0 ------- eq2
or 5 - I1 * 1000 + E - 7.2 = 0

Now we got two equation
I1*1000 + E - 6 + 0 = 0 --- eq1
-I1*1000 + E -7.2 + 5 = 0 --- eq2
Solving for E we get
E = 4.1
Putting the value of E in equation 1 we get
E1 + E - 6 = 0
E1 = 1.9
By ohms law
I = V/R
I1 = 1.9/1000
I1 = 0.0019 A i.e.. 1.9 mA
and
I2 = 5 - I1
I2 = 3.1 mA

Good luck

acelectr likes this.
8. ### acelectr Thread Starter Member

Aug 28, 2010
73
0
Thanks a lot man, I was waiting a while for this. Appreciate it. Though I thought that there was only an extra 5mA current and no resistor or no existence of any voltage there. Also I am kind of confused that here adding this 5mA current source reduced the currents at all points in the circuit. Kind of confused about these parts but thanks again