解:
(l)R= U额2/P额= (6 V)2/3 W=12 Ω.
(2)当S闭合,S1、S2都断开,L与R1串联,
P位于中点L正常发光时 U1 =U-UL =24 V-6 V=18 V,
I=PL/UL =3 W/6 V =0.5 A,
R1中=U1/I=18V/0.5 A=36 Ω,
R1最大= 2R1中=2×36Ω=72Ω.
(3)当S﹑S1 ﹑S2都闭合,
R1、R2并联且滑片置于最右端时,电路消耗功率最小,
R总= R1R2/(R1+R2)=24Ω×72Ω/(24Ω+72Ω)=18Ω,
P最小= U2/R总=(24V) 2/18Ω=32 W.
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