Find the current in the other resistors

Consider the group of resistors in the figure below, where R1 = 15.9 Ω and R2 = 8.25 Ω. The current in the 13.8-Ω resistor is 0.795 A. Find the current in the other resistors in the circuit.

Use this diagram: http://www.webassign.net/walker/21-33alt.gif

I(17.2) = 0.638
I(4.11) = 0.887
I(15.0) = ???
R1 = I(15.0) = ???
R2 = 0.887

All of the above answers are correct.

I tried doing this to find I(15.0):
V/(15+15/9)=(15-10.97) = 4.03V
I(15) = I(15.9) = 4.03V/(15+15.9) = 0.130A

0.130A did not work...

Thanks for any help!

R2 & 4.11 Ω are in series
=>R3 = 8.25 + 4.11 = 12.36 Ω
As now R3, 17.2 & 13.8 are in parallel
=>By V = i1R1 = i2R2 = i3R3
=>V = 13.8 x 0.795 = i3 x 12.36 = 17.2 x i2
=>i2 = 0.638 amp (in 17.2Ω)
=>i3 = 0.888 amp (in 8.25 Ω & 4.11 Ω )
Current in R1 & 15Ω => i = i1+i2+i3 = 0.795 + 0.638 + 0.888 = 2.321 amp

It's a shame that you didn't number the nodes to make this easier to describe in words.

If you know the current in the 13.8Ω resistor, you also know the voltage across it:  V=IR

That voltage allows you to calculate the current in the 17.2Ω resistor, as well as the series pair to the right.

Summing those 3 currents gives the total current drawn from the source (there are no other paths).  R1 and the 15Ω carry the total current.

Doesn't the current divide itself through the three branches (right side of diagram) in INVERSE proportion to the resistance of each branch?

In other words,
the three branches, respectfully, have resistances = 13.8, 17.2, 12.4 ohms.
So the most current flows thru 12.4 Ω and the least thru 17.2 Ω
The three current proportions are:
13.8/43.4 = 0.32
17.2/43.4 = 0.40
12.4/43.4 = 0.28
with the one between the other two being that thru the 13.8 Ω resistor = 0.795 A
so
0.32(I) = 0.795 A {where I = total current from battery}
I = 2.48 A
current thru 12.4Ω = (0.40)(2.48) = 0.99 A
current thru 17.2Ω = (0.28)(2.48) = 0.69 A