# Resolving those simple circuits

#### activee

Joined Jan 16, 2014
39
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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#### pwdixon

Joined Oct 11, 2012
488
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.

#### pwdixon

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

#### t_n_k

Joined Mar 6, 2009
5,455
Pretty sure the answer should be -0.5mA for I2. Are you sure you drew the circuit correctly?
That's incorrect. I2 is -250uA as stated.

#### pwdixon

Joined Oct 11, 2012
488
That's incorrect. I2 is -250uA as stated.
I'll do the maths again.

#### crutschow

Joined Mar 14, 2008
31,114
It is simpler to do the Thevenin equivalent of the two voltage source branches in parallel. Then use that to calculate I2.

#### syed_husain

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

#### activee

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

#### Fibonacci

Joined May 23, 2014
25
thanks I found where I was doing wrong.
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.