Way #1: The current in the series loop is (12 V) / (total resistance) . (Turns out to be 2 Amperes, but the question isn't asking for that.)
In a series loop, the current is the same at every point, so it's the same current through each resistor.
The power dissipated by a resistor is (current)² · (resistance), and the current is the same everywhere in the circuit, so the smallest resistance will dissipate the least power. That's R1 .
And by the way, it's not "drawing" the most power. It's dissipating it.
Way #2: Another expression for the power dissipated by a resistance is
(voltage across the resistance)² / (resistance) .
In a series loop, the voltage across each resistor is
[ (individual resistance) / (total resistance ] x battery voltage.
So the power dissipated by each resistor is
(individual resistance)² x [(battery voltage) / (total resistance)²]
This expression is smallest for the smallest individual resistance. (The other two quantities are the same for each individual resistor.) So again, the least power is dissipated by the smallest individual resistance. That's R1 .
Way #3: (Einstein's way) If we sat back and relaxed for a minute, stared at the ceiling, let our minds wander, puffed gently on our pipe, and just daydreamed about this question for a minute or two, we might have easily guessed at the answer.
===> When you wire up a battery and a light bulb in series, the part that dissipates power, and gets so hot that it radiates heat and light, is the light bulb (some resistance), not the wire (very small resistance).
