As shown in the figure, the specification of the small bulb is "6V & nbsp; 3.6W" (filament resistance does not change), close the switch, when the slide P of the sliding rheostat moves to the a end, the current indication is 0.6A; when the slide P moves to the B end, the voltage indication is 4V, then () A. The filament resistance of the small bulb is 6 Ω B. the power supply voltage is 10vc. The maximum resistance of the sliding rheostat is 10 Ω D. when the slide P is located at the B end, the power consumed by the sliding rheostat is 0.8W

As shown in the figure, the specification of the small bulb is "6V & nbsp; 3.6W" (filament resistance does not change), close the switch, when the slide P of the sliding rheostat moves to the a end, the current indication is 0.6A; when the slide P moves to the B end, the voltage indication is 4V, then () A. The filament resistance of the small bulb is 6 Ω B. the power supply voltage is 10vc. The maximum resistance of the sliding rheostat is 10 Ω D. when the slide P is located at the B end, the power consumed by the sliding rheostat is 0.8W

A. RL = U2, P = (6V) 23.6w = 10 Ω, so a is wrong; B. when slide P moves to a end, there is only lamp L in the circuit, I = 0.6A, power supply voltage U = IR = 0.6A × 10 Ω = 6V, so B is wrong; C. when slide p moves to B end, all slide rheostat is connected as resistance R, lamp L and R are connected in series, ur = 4V, UL = u-ur = 6v-4v = 2V

The specification of small bulb is 12V 6W, the maximum resistance of sliding rheostat is 24 Ω, and the constant power supply voltage is 12V (regardless of the influence of temperature on filament resistance)
Calculate the power consumption of the small bulb when it is powered on for 10s when the light of the small bulb is the darkest

R=U^2/P=(12V)^2/6W=24Ω
When it is darkest, the sliding resistance is the largest
I=U/R=12V/(24Ω+24Ω)=0.25A
W=I^2Rt=(0.25A)^2×24Ω×10s=15J