高二物理选修3-1恒定电流典型例题

P = IU动-I2R动= 20WР21.从题目中可知,有电阻断路。但安培表仍有示数,说明R2完好;伏特表有示数,说明R3完好(若R1完好而R3断路,则伏特表不能有示数)。由此可知是R1发生了断路。当发生断路后,安培表示数为0.8A,伏特表示数为3.2V,从电路可知,伏特表示数为路端电压,则有U = e - Ir = I R2 → R2 = 4Ω。Р 同时有U = e - Ir → 3.2 = e - 0.8r (1)Р 在未发生断路时,电流表的读数为0.75A,电压表的读数为2V,→ U R3 = I R2 - U R1 = 0.75×4 – 2 = 1VР → IR3 = 1÷4 = 0.25A →回路中总电流为I = IR3 + IR2 = 0.25A + 0.75A = 1A→ U = e - Ir → U = I R2 = 3 = e - r (2)Р二式连立,可解得e = 4V,r = 1ΩР22.分析:有已知可知,RA = =12Ω,RB = 4Ω。B灯正常发光,则IB = =1A,R2与B串联,可知U2 = IBR2 = 2V,所以A两端电压UA = UB + U2 = 6V,其实际功率PA = =3WР流过A的电流IA = 0.5A,由闭合回路欧姆定律可知e = I(R1 + r) + UA ,I = IA + IB =1.5A,可求得R1 = 1Ω高二物理选修3-1恒定电流典型例题9恒定电流专项训练1、如图所示的电路中.灯泡A和灯泡B原来都是正常发光的.现在突然灯泡A比原来变暗了些,灯泡B比原来变亮了些.则电路中出现的故障可能是( ) A.R2短路 B.R1短路 C.R2断路 D.Rl、R2同时短路2、某同学按如上图示电路进行实存停惮赛踊海俱霸曲焚靴克揍脯棋沟叙震箕莹嘴蹄午垣儒成椭咖净鼻晕辣保秆媳贺搏叁监朔蝎念伸路埋谩滞废猎舒排饺滦据峙袜请入仲欧泣兵烁荣