② Before closing the switch, the sliding blade of the sliding rheostat is at the maximum resistance. According to the law of series resistance voltage division, the voltage at both ends of the small bulb is the smallest at this time. When the light bulb normally emits light, the voltage indicator changes by 0.5V, that is, the voltage at both ends of the light bulb increases by 0.5 V, so: u = 2 V+0.5 V = 2.5 V; The small light bulb is marked with the word "0.2A", that is, the current during normal operation: I quantity = 0.2A.
Rated power of small bulb: P=U, I = 2.5V× 0.2A = 0.5W 。
So, the answer is: ① P = UI; ② Solution: U = 2 V+0.5 V = 2.5 V; Current when the small bulb works normally: I quantity = 0.2a
Rated power of small bulb: P=U, I = 2.5V× 0.2A = 0.5W 。