初中物理电学两道填空题~求详细解题思路~~~~好的会追加分数~
可以传图片来~~
推荐回答(4个)
分析:
4、第一条填空的考虑:∵R与R变是串联,电路中电流在增大,∴电路总电阻必然变小,既然R是定值电阻,那只能是变阻器电阻“变小”;
第二条填空的考虑:这类题目通常都从“电源电压不变”这一点着手,建立方程组来求解:
由题意可知电源电压 U=IR+U变,代入相应条件,即可得:
U=0.1R+U变……①,及 U=0.3R+U'变……②
①-②,可得:0.2R=(U变-U'变)=4,∴R=20(Ω)
∴定值电阻R电功率的变化量:
△P=P'R-PR=I'^2R-I^2R=(0.3^2-0.I^2)×20=1.6(W)
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5、根据题意,要保证电路中电路元件安全,只须考虑三个元件:电灯、滑动变阻器、电压表
①、先考虑电灯:小灯正常工作时的电流:I额=P额/U额=0.9/3=0.3(A),
显然这一电流值应为最大,即: I最大=I额
∴R变最小=(U-U额)/I最大=(4.5-3)/0.3=5(Ω);
②、再考虑滑动变阻器:由于I最大<I变最大=1A,∴变阻器始终处于安全状态;
③、最后考虑电压表:∵U表最大=3V,即变阻器两端电压最大值为3V,
∴ U灯最小=U-U表最大=4.5-3=1.5V
由于R灯=U额^2/P额=3^2/0.9=10(Ω),
由分压原理可得:U表最大/U灯最小=R'变/R灯,即 3/1.5=R'变/10
∴R'变=20(Ω)<R最大=50Ω,
综合考虑①~③,得结论:
滑动变阻器允许接入电路的阻值范围是:5Ω~20Ω
我自己想的,思路 答案有可能不对,主要是做个参考~~如有错误请见谅
第4题.因为电源电压保持不变,而滑动滑动变阻器的电阻值变小,电路电流增大,则定值电阻的电压值就会随之增大,这里的电压表改变了4伏(即滑动变阻器的电压值变小4V),电流表从0.1A变成了0.3A(即定值电阻的电流改变了0.2A),则电功率改变了P=UI=4V×0.2A=0.8W
第5题.首先这里要考虑到各元件最高承受范围(这里可理解为额定的范围,一般考虑电灯泡 电压表的最高承受范围)即电压表最高承受范围(量程)3V 电灯泡的最高承受范围(额定电压)3V,当电路电流值达I=P/U=0.9W/3V=0.3A时为电灯泡的最高值,则滑动变阻器的电压值为U=U电源 - U灯=4.5V - 3V=1.5V,滑动变阻器的电阻值应为R=U/I=1.5V/0.3V=5欧;当电压表达最高值3V时,电灯泡的电压值为U灯=U电源 - U电阻=4.5V - 3V=1.5V,则电路电流值为I=U/R=1.5V/10欧=0.15A,则滑动变阻器的电阻值为R=U/I=3V/0.15A=20欧,所以滑动变阻器的取值范围应为5~20欧
如有不懂,请追问...
可知,电流表示数从0.1A变为0.3A时,电路中的电阻变小,即滑动变阻器接入电路的电阻变小;
当电路中的电流I=0.1A时,电源的电压U=IR+U滑=0.1R+U滑;
当电路中的电流I′=0.3A时,电源的电压U=I′R+U滑′=0.3R+U滑′;
因电压表示数改变了4V,
所以U滑-U滑′=(U-0.1R)-(U-0.3R)=4V,
解得:R=20Ω;
定值电阻R电功率的变化量:
△P=(I′)2R-I2R=(0.3A)2×20Ω-(0.1A)2×20Ω=1.6W.
故答案为:变小;1.6.
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5
(1)灯泡L正常工作时的电流为:
I= = =0.3A;
(2)灯泡的电阻为RL= = =10Ω,
滑动变阻器两端的最小电压为U滑=U总﹣U=4.5V﹣3V=1.5V,
滑动变阻器连入电路中的最小阻值R= = =5Ω;
当电压表的示数最大U滑′=3V时,滑动变阻器接入电路的电阻最大,
此时灯泡两端的电压为U′=U总﹣U滑′=4.5V﹣3V=1.5V,
此时电路的电流为I′= = =0.15A,
滑动变阻器接入电路的电阻值为R′= = =20Ω.
故答案为:0.3;5~20Ω
希望可以帮你,我也在备考中
4、R=4/0.2=20欧 变化的功率=0.3*0.3*20--0.1*0.1*20=1.6w
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