(3)W2=G S2=500N?90m=4.5?104J ··························· (1分)
P=
W2 ···························································· (1分) t24.5?104J==3?103w ·········································· (1分)
15s(注:其它正确解法,参照计分)
28.(9分) (1)只闭合开关S1,将变阻器滑片P移到中点时,L与1/2R2串联,电流表A1测电路中的电流,此时电压表被短路,示数为0V, ……(1分) 由P=I2R可得,灯泡的电阻: PL实RL=2……(1分) I1.6W ==40Ω……(1分) (0.2A)2因串联电路中总电阻等于各分电阻之和, U所以,由I=可得,电源的电压: R1U=I(RL+R2)……(1分) 21=0.2A×(40Ω+×40Ω)=12V;……(1分) 2(2)同时闭合开关S1、S2,且将变阻器滑片P移到b端时,L与R2串联再与R1并联,电流表A1测干路电流,电流表A2测R1支路的电流,因并联电路中各支路两端的电压相等, 所以,通过L的电流: UIL=……(1分) RL+R212V==0.15A,……(1分) 40Ω+40Ω因并联电路中干路电流等于各支路电流之和, 所以,电流表A1的示数: I′=IL+I1……(1分) =0.15A+0.2A=0.35A,……(1分) (注:其它正确解法,参照计分)
益阳市2018年普通初中毕业学业考试物理样卷(含答案)
(3)W2=GS2=500N?90m=4.5?104J···························(1分)P=W2····························································(1分)t24.5?104J==3?103w·······················
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