Answer:
3 Ω
Explanation:
I find a calculator with a reciprocal key quite useful for calculating parallel resistance.
Req = 1/(1/r1 +1/r2 +... +1/rn))
Of course, resistors in series add.
The far right two branches are 10 Ω in parallel with (8+2) Ω = 10 Ω. Of course, the parallel combination of two equal-value resistors is half the value of either of them. Thus the two right branches resolve to 5 Ω, which is then added to 1 Ω to give an effective middle section that is 12 Ω || 4 Ω || 6 Ω.
That combination has an effective resistance of ...
Req = 1/(1/12 +1/4 +1/6) = 1/(1/12 + 3/12 +2/12)) = 12/6 = 2 . . . ohms
This is effectively in series with the upper left 1 Ω resistor, so the equivalent load on the source is 1+2 = 3 Ω.
