解:將滑動(dòng)變阻器的滑片P置于某位置a,閉合開(kāi)關(guān)S
1、斷開(kāi)S
2、S
3時(shí),等效電路如答圖甲所示;
將滑動(dòng)變阻器的滑片P置于最左端,閉合開(kāi)關(guān)S
1、斷開(kāi)S
2、S
3時(shí),等效電路如答圖乙所示;
將滑動(dòng)變阻器的滑片P置于某位置b,閉合開(kāi)關(guān)S
1、斷開(kāi)S
2、S
3時(shí),等效電路如答圖丙所示.

(1)∵P=I
2R,且P
1=0.8W,P
1′=1.8W,
∴

=

------------------①
(2)∵P=UI,且P:P′=2:1,
∴

=

------------------②
∵電源的電壓不變,
∴由甲、乙兩圖可得:U=I
2(R
1+R
2)=I
1(R
1+R
a+R
2)----③
由乙、丙兩圖可得:U=I
2(R
1+R
2)=I
3(R
1+R
b+R
2)------④
由①②③④可得:
2R
a=R
1+R
2------------------------⑤
R
b=R
1+R
2------------------------⑥
∴R
a:R
b=1:2;
(3)因?yàn)镽
a<R
b,所以當(dāng)所有開(kāi)關(guān)均閉合,滑動(dòng)變阻器的滑片P置于位置a時(shí),電路消耗的總功率最大,其等效電路如答圖丁所示.

由①②兩式可知

=

-----------------⑦
∵U=IR,
∴

=

=

---------------⑧
由⑤⑥⑦⑧可得:

=

,

=

;
由圖甲可得:U
1=

U=

U=

,P
1=

=0.8W,
解得

=3.2W,
根據(jù)P
max=

+

+

=

+

+

=22

=22×0.8W=17.6W.
答:(1)I
1與I
2的比值為2:3;
(2)R
a與R
b的比值為1:2;
(3)電路消耗的電功率最大電功率為17.6W.
分析:先畫(huà)出三種情況的等效電路圖:
(1)根據(jù)P=I
2R結(jié)合圖甲和圖乙中電阻R
1消耗的電功率求出I
1與I
2的比值.
(2)根據(jù)P=UI結(jié)合圖乙和圖丙中電路消耗的總功率求出兩者的電流之比,再根據(jù)電源的電壓不變分別根據(jù)歐姆定律和電阻的串聯(lián)特點(diǎn)表示出電源的電壓,利用電流關(guān)系分別得出R
a和R
b與R
1、R
2之間的關(guān)系,進(jìn)一步求出R
a與R
b的比值.
(3)由實(shí)物電路圖可知,因?yàn)镽
a<R
b,所以當(dāng)所有開(kāi)關(guān)均閉合,滑動(dòng)變阻器的滑片P置于位置a時(shí),電路消耗的總功率最大;先根據(jù)以上關(guān)系求出圖甲和圖乙的電流關(guān)系,根據(jù)歐姆定律結(jié)合電壓表的示數(shù)關(guān)系求出R
1、R
2、R
a三電阻之間的關(guān)系;根據(jù)串聯(lián)電路電阻的分壓特點(diǎn)得出圖甲中R
1兩端的電壓與電源電壓之間的關(guān)系,根據(jù)電功率公式表示出電阻R
1消耗的電功率,把以上關(guān)系帶入圖丁即可求出電路消耗的最大功率.
點(diǎn)評(píng):題考查了學(xué)生對(duì)串、并聯(lián)電路的判斷,串聯(lián)電路的特點(diǎn)以及歐姆定律、電功率公式的應(yīng)用.本題難點(diǎn)在于很多同學(xué)無(wú)法將三種狀態(tài)下的功率關(guān)系及電壓關(guān)系聯(lián)系在一起,故無(wú)法找到突破口.解答此類問(wèn)題時(shí),可將每一種情況中的已知量和未知量都找出來(lái),仔細(xì)分析找出各情況中的關(guān)聯(lián),即可求解.