First, we determine what's the equivalent resistance of the 3Ω and 6Ω resistors
Using the quick method, it's
(3×6)÷(3+6) or 2Ω
So the circuit's effectively 3Ω in series with 2Ω, 5Ω in total.
Current according to Ohm's Law is
I=V/R
V = 5V, R = 5Ω so I = 1A So, Ohm's Law again to calculate the voltage drop across the 3//2
combination,
V = IR, I is 1, R is 2, so the answer's 2V.
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