如图所示的电路中,设电源电压不变,灯泡L上标有“6V 3W”字样,当S1闭合,单刀双掷开关S2

2025-05-10 13:19:21
推荐回答(3个)
回答1:

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

回答2:

S2至b端短路

至a端灯泡、电阻、电压表并联

回答3:

http://www.jyeoo.com/physics/ques/detail/a9a22e6a-205b-46cc-96ae-ee6934e9547a