Hello Could somebody PLEASE help me solving this circuit.
I tried it x times and i just dint get the write answer. We are supposed to solve this circuit using KCL at the top centre node. cheers

The answer should be 3A for ix

 

Can you scan you handiwork and post it here so that we can see where you may be running into your problem?

Being homework, it is always good to present your efforts at solving the problem so that we can make suggestions on ways to help you over your snag.

hgmjr

 

sure, although i can't scan it i can type what i did: it is in the document.
thanx for ur help

 

You need to revisit your equations based on the direction of the branch currents you have assumed.

The signs in your KCL equation appear to indicate that you are assuming the branch current i to be flowing from the positive terminal of the 10V source to the node you have chosen to label va. The assumed direction for the branch current you have labeled ic is from the positive terminal of the dependent voltage source toward the node labeled va.

For example, since the signs in your KCL equation have set the direction of the branch current i flowing from the 10V source to the node labeled va then the signs in your equation for i need to be written with that in mind.

hgmjr

 

so you reckon i am on the right track.

 

so you reckon i am on the right track.

Yes indeed. You just need to rethink the signs in your equations.

HINT: The voltage Va at the junction of the three resistors is going to be an integer value.

hgmjr

 

Ok. Here is a stronger HINT.

Taking your first equation as an example:

\(i = \frac{v_{a}-10}{5}\)

Notice that the numerator is written as \(v_{a}-10\). This ordering of the two values is consistent with the direction of current \(i\) flowing from the node labeled \(v_{a}\) toward the positive terminal of the 10 volt dc source. Your KCL equation is written such that the current \(i\) is flowing from the positive terminal of the 10 volt dc source toward the node \(v_{a}\).

To be consistent with the direction of current flow indicated by the way you have formed the KCL equation, your equation for current \(i\) should be written:

\(i = \frac{10-v_{a}}{5}\)

This is because the potential that is assumed more negative \(v_{a}\) must be subtracted from the potential that is assumed more positive 10 to yield a positive value for current \(i\) as indicated in your KCL equation.

Try applying this same logic to the remaining two branch current equations and see how that affects the calculated value of \(v_{a}\) .

Be careful when forming the equation for \(i_{x}\). since the voltage source in that branch is negative.

hgmjr

 

thanx man!!

 

He says ix is supposed to be 3A. I don't get that answer either.