PHYSICS HELP??

image won't open. But I suspect ohm's law prevails here...

edit: third try worked.

open, you have two paths, each consisting of 2 resistors in series.
left path, r = 20+60 = 80Ω, I = E/R = 8/80 = 0.1 amp
right path, r = 40+10 = 50Ω, I = E/R = 8/50 = 0.16 amp
total = 0.26 amps
or, use the parallel resistance formula to get total R = 80•50/(80+50) = 30.77 Ω
I = 8/30.77 = 0.26 amp

closed, you have 20 & 40 in parallel, equals 20•40/60 = 40/3 = 13.33 Ω
and 60 & 10 in parallel, = 60•10/70 = 8.57 Ω
the two sets in series add to 21.9 Ω
I = 8/21.9 = 0.365 amps

open swt, 20+60//40+10 = 80//50=30.77Ω. I=260 mA
closed 20//40+60//10=13.33Ω+8.57Ω =21.9Ω. I= 8 / 21.9Ω = 365 mA.

http://falstad.com/circuit/
circuit simulator for confirmation

Switch open:
Two parallel paths:
20 + 60 ohms = 80 ohms
40 + 10 ohms = 50 ohms
Current in first = 8/80 amps = 0.1 amps
Current in second = 8/50 amps = 0.16 amps
Total current = 0.26 amps <<<

Switch closed:
Two sets of parallel resistors in series.
Upper pair: 20 ohms in parallel with 40 ohms.
1/R1 = 1/20 + 1/40 = 3/40
R1 = 40/3 ohms
Lower pair: 20 ohms in parallel with 40 ohms.
1/R2 = 1/60 + 1/10 = 7/60 ohms
R2 = 60/7 ohms
Total resistance = R1 + R2 = (40/3) + (60/7)
=(1/21)*(280 + 180)
= 460/21 ohms
Current = 8/(460/21) amps
= 8*21/460 = 168/460 = 42/115 amps <<<