有一盏“220V40W的灯泡L1,问(1)正常发光的电流多大

2025-03-29 19:09:57
推荐回答(5个)
回答1:

1、因为P=UI,所以电流 I=40/220=0.182A2、因为P=U^2/R,所以电阻 R=220^2/40=1210欧姆3、因为W=Pt,所以W=40*10*3600=1440000焦耳
4、t=W/P=1000/40=25小时5、n=5/0.182=27盏6、因为灯泡电阻可以看作没变,所以P=U^2/R=110^2/1210=10w7、L1实际功率40W,L2实际功率100W,总功率140W,L2更亮 第一题是I=p/U,第三题是W=pt,也是W=UIt,这两种答案是一样的,因为这里我们不考虑灯泡的电能转化为光能的那一部分,就是说认为灯泡是纯电阻,这样的话,上面两式就是等价的。

回答2:

1:电流*电压=功率电流=40/220=0.18(A)2:电阻R=电流/电压=1210欧姆3:电能W=P*t=400*3600=1440KJ=0.4KW.h4:t=1/0.4=2.5h5:n=5/0.18=27.5 只能接27盏 而其必须并联6:W=U的平方/R=10W7:功率各为40 100总功率140第二个灯更亮第一题是I=p/U 第三题是W=pt还是W=UIt是一样的

回答3:

1、I=P/U=40/220=0.1818A2、R=U/I=U^2/P=12103、w=Pt=40*10*60*60=1440000J4、t=w/p=1000*3600/40=90000s=25h5、n=5/I=276、P=U^2/R=10W7\L1为40w L2为100W 总的为140w L2更亮

回答4:

你说的都是对的
p=UI,所以W=UIt,然而是正常工作,所以用w=Pt就可以了。

回答5:

I=40\220=2\11A R=220�0�5\40=1210ΩW=220*2\11*60*60*101440000Jt=3600000\40=90000sn=5\(2\11)=55\2=22p=110�0�5\1210=10W第七题用电阻算吧,由分压算功率,总功率是相加,功率大的亮是都可

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