Simply find equivalent resistance seen by looking into A-B terminals
View attachment 75613
This is nothing more than a resistor combinations (series and parallel).

 

Also check your math. How can 1.92 * 0.54 equals 2.45??

Then do:
I1= 1.92 x 0.53
I1= 2.45 amps

I2= 1.92 x 0.47
I2= 2.39 amps

You appear to be suffering from a very common ailment -- namely you punch a bunch of keys into a calculator and just accept whatever it spits back at you without thinking. It is a very common malady -- I literally once convinced a 16 year old girl that she was only 14 because she suffered from this disease so badly.

Always, always, ALWAYS ask if your answer makes sense. You are multiplying a number that is nearly 2 by a number that is nearly 0.5 and getting a value that is nearly 2.5. Does that make sense? No. But you catch it only if you ask the question. Get in the habit of asking the question. Always. In fact, before you touch the damn calculator, estimate what you expect the answer to be. In both of these cases you should have mentally expected an answer of about 1 (even better, a bit less than 1 for I2) and if your computed answer wasn't very close to those estimates then warning bells and sirens should have gone off and you should have stopped right there until you resolved the discrepancy adequately. I'm deadly serious about this because engineering is a deadly serious business. You WILL make mistakes. You WILL enter things into a calculator or computer incorrectly. You WILL mess up parens when entering formulas into spreadsheets and program statements. You WILL! We all DO! The calculator and the computer do not care and they will do exactly what you told them to do. It is up to YOU to check your work. Get in the habit -- NOW!; do not wait until you are on trial for a wrongful death because you couldn't be bothered to perform reasonable due diligence. Sorry it that sounds harsh, but better that than the potential alternatives. You now KNOW that you have particularly weak basic math skills (again, very common today and we all suffer from it to greater and lesser degrees in this world of machines that we have at our beck and call) so you have a particular responsibility to employ proper safeguards.

 

Simply find equivalent resistance seen by looking into A-B terminals
View attachment 75613
This is nothing more than a resistor combinations (series and parallel).

ok so i have come out with Vth= 2.7v and Rth= 2.15 ohms.

What would be the next step to finding the current passing through the 5 ohm resistor?

 

Any ideas?

 

Any ideas?

I have been trying to figure it out for the last 2 days and i havent had a clue. haha
I think my brain has melted!

 

I have been trying to figure it out for the last 2 days and i havent had a clue. haha
I think my brain has melted!

Well simply, you find Vth and Rth so now you can replace this circuit
View attachment 75612

With this one

 

I have been trying to figure it out for the last 2 days and i havent had a clue. haha
I think my brain has melted!

Remember what the whole point behind a Thevenin or Norton equivalent circuit is:

You have a complicated circuit that has a bunch of stuff including one component (which may, of course, be a group of components) that you are interested in knowing the voltage and current for. The points at which the components you are interested in joins the rest of the circuit are the "ports". If there are only two ports, then you break the circuit into two pieces at the ports -- the piece you want to know about and everything else. You then analyze just the "everything else" part of the circuit and device a very simple equivalent circuit that has the same voltage/current behavior at the ports as the original part of the circuit that had "everything else". That means that you can replace "everything else" with the simple equivalent circuit and the component that you are interested in can't tell the difference.

With that in mind, now that you have a Thevenin equivalent circuit with a known voltage and resistance, what should you do next?

 

i DID I= Vth/ Rth + 5 = 2.7/ 7.15= 0.37a

Does this look right??

 

i DID I= Vth/ Rth + 5 = 2.7/ 7.15= 0.37a

Does this look right??

Always, always, ALWAYS track your units. Do NOT just throw whatever units you WANT the answer to have onto whatever number happens to end up at the end.

What you wrote and what you did are two different things. What you wrote:

I = (Vth/Rth) + 5Ω [Remember: Multiplication/division before addition/subtraction]

What you did:

I = Vth/(Rth + 5Ω)

The latter is what you wanted, so be sure that that is what you write. This is the kind of mistake that is VERY easy to make when you enter formulas into programs and spreadsheets and they are very difficult to catch because when YOU look at it your sloppiness has trained you to see what you WANT to see and not what you ACTUALLY wrote. The computer only cares about what you actually wrote.

Following through with units, you should have proceeded as:

I = 2.7V/(2.15Ω + 5Ω) = 2.7V/7.15Ω = 0.378 A

As a general rule when nothing else argues differently, answers should be given to three sig figs. Intermediate results should be carried with four or five sig figs to minimize cumulative round off errors. So what you actually should have ended up with was:

I = 2.717V/(2.151Ω + 5Ω) = 0.380 A

While it appears like a minor difference (and by and large it is), even this mild amount of round off error represents an error of 2.6% after just these few steps on this simple of a circuit. Hopefully you can appreciate how quickly round-off errors can accumulate to throw answers off by significant amounts.