【答案】
分析:(1)有電流和電阻,由P=I
2R可以寫(xiě)出P
1和P
1′的表達(dá)式,根據(jù)P
1:P
1′=9:25.可求出電流之比.
(2)分析電路:當(dāng)開(kāi)關(guān)S
1閉合,S
2斷開(kāi),滑動(dòng)變阻器的滑片P移到B端時(shí),燈泡短路,R
1和滑動(dòng)變阻器串聯(lián),此時(shí)滑動(dòng)變阻器電阻全部接入電路.電阻為R
3,當(dāng)開(kāi)關(guān)S
1斷開(kāi),S
2閉合時(shí),滑動(dòng)變阻器短路,R
1和燈泡串聯(lián).燈泡電阻為R
L,根據(jù)(1)中的電流之比可求出
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之比;
當(dāng)開(kāi)關(guān)S
1、S
2 都斷開(kāi),滑動(dòng)變阻器的滑片P在B點(diǎn)時(shí),R
1、R
L、R
3串聯(lián),電路中電阻最大,功率最�。�
當(dāng)開(kāi)關(guān)S
1、S
2 都閉合時(shí),滑動(dòng)變阻器和燈泡短路,電路中電阻是R
1電阻最小,功率最大.最大電功率與最小電功率之比為3:1.可求出
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之比,然后與上式整理可求得燈的電阻與滑動(dòng)變阻器最大阻值之比;
(3)滑動(dòng)變阻器的滑片P在C點(diǎn)時(shí),U
1:U
2=3:2,可求得
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之比,根據(jù)電阻關(guān)系找出總功率P
總與R
C的電功率的關(guān)系,由R
C的電功率是9W,可求出總功率.
(4)根據(jù)電阻關(guān)系找出燈L的額定功率P
L0與R
C的電功率的關(guān)系,由R
C的電功率是9W,燈L的額定功率P
L0.
解答:解:相關(guān)電路狀態(tài)的等效電路圖如所示:

(1)由題可知,P
1=I
12R
1,P
1′=I
22R
1,因?yàn)�,P
1:P
1′=9:25,所以,I
1:I
2=3:5;
(2)當(dāng)開(kāi)關(guān)S
1閉合,S
2斷開(kāi),滑動(dòng)變阻器的滑片P移到B端時(shí),燈泡短路,R
1和滑動(dòng)變阻器串聯(lián),如圖1所示,
當(dāng)開(kāi)關(guān)S
1斷開(kāi),S
2閉合時(shí),滑動(dòng)變阻器短路,R
1和燈泡串聯(lián).如圖2所示,由I
1:I
2=3:5,
可得,
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=
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…①;
當(dāng)開(kāi)關(guān)S
1、S
2 都斷開(kāi),滑動(dòng)變阻器的滑片P在B點(diǎn)時(shí),R
1、R
L、R
3串聯(lián),如圖4所示,電路中電阻最大,功率最��;
P
最小=
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,
當(dāng)開(kāi)關(guān)S
1、S
2 都閉合時(shí),滑動(dòng)變阻器和燈泡短路,電路中電阻是R
1電阻最小,如圖5所示,功率最大.
P
最大=
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,因?yàn)�,最大電功率與最小電功率之比為3:1,
所以,
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=
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…②,
由①②得,
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=
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,R
1=2R
L.
(3)當(dāng)開(kāi)關(guān)S
1、S
2 都斷開(kāi),滑動(dòng)變阻器的滑片P在C點(diǎn)時(shí),如圖3所示,U
1:U
2=3:2,
所以,得
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=
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…③將R
1=2R
L代入③式,可得R
L=R
C,
又P
C=
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×R
C=9W.
P
總=
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,將
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=
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,R
1=2R
L,R
L=R
C代入兩式整理得,P
總=36W.
(4)當(dāng)開(kāi)關(guān)S
1斷開(kāi),S
2閉合時(shí),滑動(dòng)變阻器短路,R
1和燈泡串聯(lián).如圖2所示,燈泡正常發(fā)光,
P
L=
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×R
L,
又P
C=
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×R
C=9W.將R
1=2R
L,R
L=R
C代入以上兩式整理得,P
L=16W.
答:(1)電流表I
1與I
2的比值是3:5;
(2)燈的電阻與滑動(dòng)變阻器最大阻值之比是1:3;
(3)總功率為P
總是36W;
(4)燈L的額定功率是16W.
點(diǎn)評(píng):正確分析電路畫(huà)出電路圖是解題的關(guān)鍵;求出各電阻之間的關(guān)系是解題的突破口.
在計(jì)算電功率時(shí),若沒(méi)有給出具體的電壓、電流和電阻,就要設(shè)法找出要求的功率與已知功率的關(guān)系,求出要求的功率.