网友的相关建议:
看不到图啊,不过根据你的描述,得知R1的电阻比较低大,所以是保温电热丝。R2的电阻比较小,所以是加热丝。
现在回答你的第一个问题:S闭合,R1被短接,此时电流不经过RI而只经过R2。此时只有R2工作,因为R2电阻小,根据公式U=IR和P=UI知道电阻小电流就大(反比关系),电流大功率就大(正比关系),所以就是加热
第二个问题:已知加热时P=U^2/R2=1000 W 求得保温时的功率P=U^2/R1+R2=100W(因为保温时开关S断开,电流经过R1和R2,就看成一个串联电路因此R1R2要相加)
网友的相关建议:
怎么看不到图??
不过想象的到应该是R1和R2串联,然后R1上面加了个S温控开关,AB就是电源电压…
(1)S闭合,R1被短接,处于加热状态
(2)p=U*I=U平方/R 得到R2=60.5 , R1=9*60.5=544.5,总电阻=60.5+544.5=605 ,然后再利用P=U平方/R =220*220/605=80W
网友的相关建议:
There are 3 people in my family.My mother , my father and I.I love them very much .Especially my father.My father is a worker. He works very hard.Everyday when he comes home ,he feels very tired ,I really want to help him to do something ,bring him a cup of tea or do some housework.But he said ,no,my dear doughter,you just study hard.That's the way to help me.I just want to say ,dad ,I will study hard ,because I love you very much.
网友的相关建议:
图呢??我没看到啊。。。
网友的相关建议:
梵蒂冈
网友的相关建议:
0000