(20 14? In the comprehensive practical activity class, Xiao Ming designed the analog dimming lamp circuit as shown in the figure, in which the power supply voltage is constant and the light bulb is L-shaped. (1) The resistance of the light bulb when it emits light normally: RL = U2LPL = (6V) 23W =12ω;

(2) when p is at a certain position, U2= 1V, I=0.5A, and the current when the bulb normally emits light: IL = Plul = 36V = 0.5A, so the bulb normally emits light at this time, and the voltage at both ends of the bulb is 6V;

According to I=UR, the voltage across the resistor R 1: u1= IR1= 0.5a× 6ω = 3v;

Power supply voltage: u = ul+u1+U2 = 6 v+3 v+1v =1v;

When P is at the far right, U2 ′ = 6V, and I ′ = 0.3A..

The voltage across the resistor R 1: u1'= I' r1= 0.3a× 6Ω =1.8v,

The actual voltage across the bulb: u1'= u-u1'-U2' =10v-1.8v-6v = 2.2v,

Actual power of bulb: PL ′ = UL ′ IL ′ = 2.2V× 0.3A = 0.66W;

(3) According to the meaning of the question, when the slider P is at the leftmost position, the light bulb normally emits light. At this time, the total resistance in the circuit is: r = uil =10v0.5a = 20ω; Therefore, the resistance value of a constant resistor needs to be connected in series in the circuit: r' = r-rl-r1= 20ω-12ω-6ω = 2ω.

Answer: (1) The resistance of the light bulb L when it emits light normally is12Ω;

(2) When the slider P of the sliding rheostat R2 is at the extreme right end, the switch is closed, and the actual power of the bulb L is 0.66 W；;

(3) Connect a 2 Ω resistor in series in the circuit.