Physics Help???

Another very slow link. Just upload the drawing to Yahoo directly, much easier.

It's drawn strange, but you have 4Ω, 48Ω and 16Ω in parallel, those in series with 3Ω and the battery.

1/Rp = 1/4 + 1/48 + 1/16 = 0.25 + 0.0208333 + 0.0625
Rp = 3Ω

total R = 3+3 = 6Ω
I = E/R = 12/6 = 2 amps
which gives you 6 volts across each resistor.

The source sees 3Ω + 1/[1/4+1/48+1/16] = 3 + 1/[16/48] = 3+3 = 6Ω
Thus 12V/6Ω = 2A leaves the source dropping 6V on the first 3Ω leaving 6V across the 4,48 and 16Ω resistors that are in parallel.
i3 = 12v/6Ω = 2A
i4 = 6v/4Ω = 1.5A
i48 = 6v/48Ω = 1/8 A
i16 = 6v/6Ω = 3/8 A
Notice that 1.5+1/8+3/8 = 2A

All the resistors exceopt the 3 ohms are in parallel.
(As billrussell42 says, the diagram is drawn to confuse.)
They have resistance R1, where
1/R1 = 1/4 + 1/48 + 1/16 = 16/48
R1 = 48/16 ohms = 3 ohms
Total resistance across battery = 3 ohms top left in diagram + 3 ohms calculated above = 6 ohms
Battery current = 12v/6 ohms = 2 amps
That current flows in the 3 ohm resistor. <<<
Volts across 3 ohm resistor = 3*2 volts = 6 volts. <<<
That leaves (12 - 6) = 6 volts across the 4, 48 and 16 ohm resistors...all the same- they are in parallel. <<<
Current in 4 ohms = 6/4 = 1.5 amps <<<
Current in 48 ohms = 6/48 = 0,125 amps <<<
Current in 15 ohms = 6/16 = 0,375 amps <<<