初二物理关于电功率问题

2024-12-14 00:17:48
推荐回答(6个)
回答1:

1、电路中只有电视机一个用电器
2、当电视机工作时,用秒表测量电能表转一转的时间
3、得出时间后再计算电能表一转所耗电能,即3.6*10的六次方/3000,一转为1.2*10的三次方焦
4、根据公式P=W/T得出功率 *注意单位

回答2:

将电能表接入电视机电路中,

打开电视,开启秒表计时一分钟,

数电能表在一分钟内转了n圈,

即消耗了n/3000kWh,

除以时间是1分钟(六十分之一小时)就得到了电视机的实际功率了。

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回答3:

这是电功率知识点中重难之处:
1清楚从能量的转化上理解,热量是电功转化来,单位时间内电功即分析电功率的问题;
2理解电学中P=W/t=UI的意义,对于纯电阻的导体由欧姆定律带出电功率的另外两个常用的式子P=U2/R
,P=I2R
3在处理你提到的两种情况应重视P=UI的分析,分析在不同的电路(串联或并联)条件,电阻的大小对电流或电压大小的联系,对两个说法的认同注意这个转换
当串联时,电流相等,两个不同的电阻比较,对电压分析,从而得出电阻大的,电功率大.
当并联时,电压相同,两个不同的电阻比较,对电流的影响,从而得出电阻大的,电功率应小.
楼上那位说的最后一句,应启好学好这节的知识.加油吧!

回答4:

1.小灯泡正常发光,那么流过电流是0.2A,它和变阻器串联,所以流过变阻器电流也是0.2A,那么变阻器两段电压是U×0.2=W=1.04,U=5.2v,小灯泡两段电压是9-5.2=3.8v,所以小灯泡电阻是3.8/0.2=19欧姆

回答5:

(1)变阻器上电压为1.04除以0.2等于5.2,小灯泡电压为3.8,电阻为19。

回答6:

(1)因为灯泡正常发光,所以串联电路电流I=0.2A
P滑=U滑I
U滑=P滑/I=1.04W/0.2=5.2V
U灯=U-U滑=9V-5.2V=3.8V
R灯=U灯/I=3.8V/0.2A=19欧
(2)P滑=I^2*R滑
R滑=P滑/I^2=1.04W/(0.2A)^2=26欧
(3)题目有错,不可能达到1A的电流,电阻没有那么小,而且超过灯泡的额定电流,会烧坏灯泡
题目应该是I实=0.1A的吧
P灯实=I实^2*R灯=(0.1A)^2*19欧=0.19W
U灯实=I实R灯=0.1A*19欧=1.9V
U滑=U-U灯实=9V-1.9V=7.1V
R灯:R滑实=U灯实:U滑实、
19欧:R滑实=1.9V:7.1V
R滑实=71欧

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