Simplifying a series / parallel circuit and calculating unknown values

[jimmystevens123]

Homework Statement

I'm struggling with interpreting the circuit correctly, and checking my work afterwards yields confusing results.

Relevant Equations

Kirchhoff's Current Law
Ohms Law

Given the circuit above, I have to solve for the labelled currents, find V total and R total accordingly. 1A is flowing through the 5Ω resistor as shown. Assuming electron flow (negative terminal to positive) for circuit.

The connector in the middle was somewhat confusing. Without it, this would be much simpler. At any rate, I redrew the circuit (perhaps incorrectly) as follows:

If I have a 1A going through one identical parallel resistor, I must have 1A going through the other.

Adding the equivalent parallel resistor values, I get 4.5 + 2.5 = 7Ω.

2A across 2.5Ω = 5V dropped.
2A across 4.5Ω = 9V dropped.

R total = 7Ω
I total= 2A
V total = 14V

Using Kirchoffs Current Law - (i2 = It * R1 / R1+R2) 2A * 6/6+18 = 0.5A?
i1 = 2A * 18/6+18 = 1.5A?

I feel like the redraw was a mistake, because without that junction, the values don't make any sense in the original circuit.

My question is, how do I interpret this circuit correctly? How do I know if any current is going to flow through the middle connector, and if so, how much? It's not intuitive to me at all. Apologies for the length.

 

[scottdave]

The thing to think about the tie connection in the middle - the voltage is the same at that point.

If it is an ideal conductor (0 ohms) then regardless of the amount of current through it, there is zero voltage drop. This makes the two bottom resistors in parallel, and the two top ones in parallel.

The way you redrew it is electrically the same as the original drawing.

Your calculations look correct - I would add some parentheses to make it a little more clear.

So if 1.5 A flows through R1, and 0.5 A flows through R2, and you have the currents in the bottom two resistors, how much flows through the tie connector?

 

[Averagesupernova]

My 2¢ is simply that you are over thinking it. Remember that no matter what, the top two resistors are in parallel. Of course it's the same case for the bottom two. If you are given a current through one resistor you know the voltage across it. It may help you to redraw with the top and bottom each having one equivalent resistor. You may find you go back and forth between the two schematics to help in solving.

 

[jimmystevens123]

Thanks very much for your replies gentleman, I can see now where I was confused.