解:(1)只閉合開關S
l時,等效電路如下圖
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電源的電壓為U=
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=
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=4V;
(2)只閉合開關S
2時:等效電路如下圖
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思路一:R
1=R-R
3→R=
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→I=I
3=
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,
思路二:R
1=
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電路中電流為I
1=I
3=
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=
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=0.4A,
此時R
3兩端的電壓為U
3=
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=
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=1.6V,
電阻R
1兩端的電壓為U
1=U-U
3=4V-1.6V=2.4V;
電阻R
1的阻值為R
1=
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=
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=6Ω;
(3)只閉合開關S
3時,等效電路如下圖
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①滑片左移時,燈泡正常發(fā)光時的電流為0.4A,所以在不損壞電流表和電燈的情況下有:
電路中的最大電流為I=0.4A,
此時電路中的總電阻為R=
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=
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=10Ω,
滑動變阻器接入電路的最小值為R
2=R-R
1=10Ω-6Ω=4Ω;
②當滑片右移時,在不損壞電壓表的情況下有:
電壓表的示數(shù)U
2=3V,
R
1兩端的電壓為U
1′=U-U
2=4V-3V=1V,
此時通過滑動變阻器的電流為I
2=I
1′=
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=
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=
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A,
滑動變阻器接入電路的最大阻值為R
2=
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=
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=18Ω;
∴變阻器R
2的阻值范圍是4~18Ω.
答:(1)電源電壓為4V;(2)燈泡的電阻為6Ω;(3)變阻器R
2的取值范圍為4~18Ω.
分析:(1)只閉合開關S
l時,電路為R
3的簡單電路,根據(jù)U=
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求出電源的電壓;
(2)只閉合開關S
2時,燈泡R
1、R
3串聯(lián),先根據(jù)串聯(lián)電路的電流特點求出電路中電流,根據(jù)U=
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求出R
3兩端的電壓,再根據(jù)串聯(lián)電路的電壓特點求出電阻R
1兩端的電壓,最后根據(jù)歐姆定律求出燈泡的電阻R
1的阻值;
(3)只閉合開關S
3時,滑動變阻器和燈泡串聯(lián),根據(jù)電壓表的量程和燈泡的額定電流確定電路中的最大電流,先根據(jù)歐姆定律求出電路的總電流,再根據(jù)串聯(lián)電路的電阻特點求出滑動變阻器接入電路的最小阻值;根據(jù)串聯(lián)電路的電阻分壓特點可知,滑動變阻器兩端的電壓為3V時接入電路的電阻最大,根據(jù)串聯(lián)電路的電壓特點和歐姆定律求出電路中的電流,進一步根據(jù)歐姆定律求出其最大值,從而得出變阻器R
2的取值范圍.
點評:本題要求學生能通過電路中開關的通斷得出正確的電路圖,并能靈活應用串聯(lián)電路的規(guī)律及歐姆定律求解.