【答案】
分析:(1)當開關S
1閉合,S
2、S
3均斷開時,電阻R
1與R
2串聯;當開關S
2、S
3閉合,S
1斷開時,電阻R
1與R
2并聯,據此畫出等效電路圖;
(2)根據P=I
2R表示出兩圖中電阻R
1消耗的電功率得出圖甲、圖乙中R
1支路的電流關系,再根據電阻的串聯和歐姆定律表示出電源的電壓結合電流關系即可求出兩定值電阻的阻值之比,根據并聯電路的電壓特點和P=
![](http://thumb.zyjl.cn/pic6/res/czwl/web/STSource/20131024153642221320759/SYS201310241536422213207037_DA/0.png)
以及電阻關系得出電源的電壓,利用電阻關系得出兩定值電阻的阻值.
解答:解:(1)當開關S
1閉合,S
2、S
3均斷開時,電阻R
1與R
2串聯,等效電路圖如圖甲所示;
當開關S
2、S
3閉合,S
1斷開時,電阻R
1與R
2并聯,等效電路圖如圖乙所示.
![](http://thumb.zyjl.cn/pic6/res/czwl/web/STSource/20131024153642221320759/SYS201310241536422213207037_DA/images1.png)
(2)圖甲和圖乙中,電阻R
1消耗的電功率:
![](http://thumb.zyjl.cn/pic6/res/czwl/web/STSource/20131024153642221320759/SYS201310241536422213207037_DA/1.png)
=
![](http://thumb.zyjl.cn/pic6/res/czwl/web/STSource/20131024153642221320759/SYS201310241536422213207037_DA/2.png)
=
![](http://thumb.zyjl.cn/pic6/res/czwl/web/STSource/20131024153642221320759/SYS201310241536422213207037_DA/3.png)
=
![](http://thumb.zyjl.cn/pic6/res/czwl/web/STSource/20131024153642221320759/SYS201310241536422213207037_DA/4.png)
=
![](http://thumb.zyjl.cn/pic6/res/czwl/web/STSource/20131024153642221320759/SYS201310241536422213207037_DA/5.png)
,
解得:
![](http://thumb.zyjl.cn/pic6/res/czwl/web/STSource/20131024153642221320759/SYS201310241536422213207037_DA/6.png)
=
![](http://thumb.zyjl.cn/pic6/res/czwl/web/STSource/20131024153642221320759/SYS201310241536422213207037_DA/7.png)
,
∵電源的電壓不變,
∴
![](http://thumb.zyjl.cn/pic6/res/czwl/web/STSource/20131024153642221320759/SYS201310241536422213207037_DA/8.png)
=
![](http://thumb.zyjl.cn/pic6/res/czwl/web/STSource/20131024153642221320759/SYS201310241536422213207037_DA/9.png)
=
![](http://thumb.zyjl.cn/pic6/res/czwl/web/STSource/20131024153642221320759/SYS201310241536422213207037_DA/10.png)
,
解得:
![](http://thumb.zyjl.cn/pic6/res/czwl/web/STSource/20131024153642221320759/SYS201310241536422213207037_DA/11.png)
=
![](http://thumb.zyjl.cn/pic6/res/czwl/web/STSource/20131024153642221320759/SYS201310241536422213207037_DA/12.png)
,
圖乙中:
P
1=
![](http://thumb.zyjl.cn/pic6/res/czwl/web/STSource/20131024153642221320759/SYS201310241536422213207037_DA/13.png)
=
![](http://thumb.zyjl.cn/pic6/res/czwl/web/STSource/20131024153642221320759/SYS201310241536422213207037_DA/14.png)
=
![](http://thumb.zyjl.cn/pic6/res/czwl/web/STSource/20131024153642221320759/SYS201310241536422213207037_DA/15.png)
×
![](http://thumb.zyjl.cn/pic6/res/czwl/web/STSource/20131024153642221320759/SYS201310241536422213207037_DA/16.png)
=
![](http://thumb.zyjl.cn/pic6/res/czwl/web/STSource/20131024153642221320759/SYS201310241536422213207037_DA/17.png)
×4A=16W,
解得:U=12V,
R
2=
![](http://thumb.zyjl.cn/pic6/res/czwl/web/STSource/20131024153642221320759/SYS201310241536422213207037_DA/18.png)
=
![](http://thumb.zyjl.cn/pic6/res/czwl/web/STSource/20131024153642221320759/SYS201310241536422213207037_DA/19.png)
=3Ω,
R
1=3R
2=3×3Ω=9Ω.
答:(1)等效電路圖如上圖所示;
(2)R
1、R
2的阻值分別為9Ω和3Ω.
點評:本題考查了串聯電路的特點和并聯電路的特點以及歐姆定律、電功率公式的靈活應用,利用好電源的電壓不變是關鍵.