解:
(1)当K1闭合,单刀双掷开关S2接a时,L与R并联
∵灯泡L正常发光
∴U=6V;
(2)当S1断开,单刀双掷开关S2接b时,L与R串联,电压表测量电阻R两端的电压;
灯L的阻值:RL=
U额2/P额
=(6V)2/3W
=12Ω;
灯L两端的电压:UL=U-UR=6V-3V=3V,
干路电流:I=U/R
=3V/12Ω
=1/4A,
R的阻值:R=U/I
=3V/(1/4)A
=12Ω.
(3)由(2)知:I=1/4 A,UL=3V,t=1min=60s;
∴W=ULIt=3V×(1/4 )A×60s=45J.
答:(1)电源电压6V;(2)R阻值12Ω;(3)消耗电能45J
求采纳
S2至b端短路
至a端灯泡、电阻、电压表并联
http://www.jyeoo.com/physics/ques/detail/a9a22e6a-205b-46cc-96ae-ee6934e9547a