# Resolving those simple circuits

Discussion in 'Homework Help' started by activee, Jun 11, 2014.

1. ### activee Thread Starter Member

Jan 16, 2014
39
0
Hello,
I did this exercice now 3 times and for I2 i got I2 = -270µA but my book says it's 250µA. Is there an error ? what I wrote on the page I put in attached file is not correct. there is a sign error somewhere.

It would be nice if someone could point out my mistake(s). Thanks in advance.

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Last edited: Jun 11, 2014
2. ### pwdixon Member

Oct 11, 2012
488
56
Without doing any calculations at all it is easy to see that the current I2 will be negative as defined in your schematic as the 10V will dominate and push current upwards through R2. After that it's maths.

3. ### pwdixon Member

Oct 11, 2012
488
56
Pretty sure the answer should be -0.5mA for I2. Are you sure you drew the circuit correctly?

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
That's incorrect. I2 is -250uA as stated.

5. ### pwdixon Member

Oct 11, 2012
488
56
I'll do the maths again.

6. ### crutschow Expert

Mar 14, 2008
12,538
3,072
It is simpler to do the Thevenin equivalent of the two voltage source branches in parallel. Then use that to calculate I2.

7. ### syed_husain Active Member

Aug 24, 2009
61
5
I got I2 = -250 uA. Try calculate the circuit using Superposition method.

8. ### activee Thread Starter Member

Jan 16, 2014
39
0
thanks I found where I was doing wrong.

9. ### Fibonacci New Member

May 23, 2014
25
5
The answer is I2= -250μA and Uab=-1.25 V, as has been established. The simplest way is by using mesh method and the solution of simultaneous equations resulting. In this simple case is an array of 2 x 2, which casts I1=625μA and I3=875 μA, from the circuit I2=I1-I3.