How do I solve this complex circuit with resistors in series and parallel?

[CL1349]

Homework Statement

Find all resistance, voltage, current, and power values for the circuit.

Homework Equations

I = V/R V = IR R = V/I

Voltage Divider Rule
Current Divider Rule

The Attempt at a Solution

I simplified the circuit down to equivalent values. Sorry for the poor image of the circuit.

I turned the R1 and R2 parallel combo into Req1, = 60 ohms.

Then added the series combo of R6 + R7 in the middle, Req2 = 120 ohms.

The parallel combo of R4 and R5 at the top, Req3 = 160 ohms.

Added the series combo of R3 + Req3, Req4 = 240 ohms.

Solved the parallel combo of Req4 and Req2, Req5 = 80 ohms.

Then I'm left with 3 resistors in series, Req 1 + Req5 + R8, RT= 220 ohms.

I found total current, IT= VT/RT, 660 V/220 ohms, = 3 A

Found total power, PT= IT*VT, 3A*660V = 1980 A

Then I tried to find the current of R1 by using the current divider rule. I came out with 2 A.Now, I am wondering if I did it right and how do I tackle those resistors in the middle, do I just use voltage divers/current dividers?

If anyone can check if I did it right or offer advice, it would be much appreciated. Thanks.

 

[Zryn]

That picture is quite bad, but the numbers look OK.

Remember that current in series is the same everywhere, so after the 3A flows into the first parallel pair, 3A will flow out of it too (since you essentially have 3 groups of resistances in series in this diagram).

Then, just keep in mind that you can reduce the middle clump to a parallel pair (top = 90 + 480||240, bottom = 80 + 40) to find out how the current splits, and then when you know the current in the top path, you can again see how it would split for the top parallel pair.