(20 13? Xiao Gang designed a circuit model of a chicken incubator, as shown in the figure. Under normal circumstances, only lighting is needed, when the temperature is low.

∴ The current of bulb L in normal operation is: IL = PLUL = 3W12V = 0.25a. 。

(2) Close S and disconnect S 1 and S2. The bulb l is connected in series with the resistor R 1, and the bulb works normally at this time.

Then the voltage across the bulb is UL= 12V, and the power supply voltage is U=72V.

The voltage across the resistor R 1 is: u1= u-ul = 72v-12v = 60v.

When the bulb L works normally, the current of the bulb L is 0.25A, and the current in the circuit is 0.25A at this time.

∴ the resistance value of the resistor R 1 is r1= u1I = 60v 0.25a = 240ω.

(3) When S, S 1 and S2 are all closed, the resistor R 1 is connected in parallel with the sliding rheostat R2, and the power supply voltage is U=72V.

Then U 1=U2=U=72V,

The maximum resistance of the sliding rheostat is r = 200 ω,

The minimum power consumption of R2 is: p small =U2R large = (72v) 2200Ω = 25.92w. 。

The resistance of R 1 is 240Ω, while U 1=U2=U=72V.

∴ At this time, the current through R 1 is: I1= UR1= 72V240Ω = 0.3A.

The ammeter A 1 measures the current flowing through the resistor R 1, and the ammeter A2 measures the total current in the main circuit. The range of ammeter A 1 is 0 ~ 0.6a, and the range of ammeter A2 is 0 ~ 3a.

The maximum current through R2 is I2=I-I 1=3A-0.3A=2.7A,

The maximum power consumption of R2 is: p = ui2 = 72v× 2.7a =194.4w. 。

Answer: (1) The current of the bulb is 0.25 A when it works normally.

(2) Close S, disconnect S 1 and S2, the bulb works normally, and the resistance value of resistor R 1 is 240Ω.

(3) When S, S 1 and S2 are all turned off, the minimum value of electric power consumed on R2 is 25.92W, and the maximum value is194.4 W. 。