【答案】
分析:先畫出三種情況的等效電路圖.
(1)根據(jù)P=I
2R分別表示出圖1和圖2中燈泡的電功率結(jié)合燈L的實(shí)際功率為其額定功率的
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得出兩電路圖中電流關(guān)系,再根據(jù)P=I
2R分別表示出圖2和圖3中R
1消耗的功率結(jié)合P
1:P
1′=1:4即可求出兩電路中的電流關(guān)系,進(jìn)一步可得圖1和圖3中的電流關(guān)系,根據(jù)歐姆定律結(jié)合電源的電壓不變即可求出燈泡的電阻;
(2)根據(jù)歐姆定律和電阻的串聯(lián)特點(diǎn)結(jié)合串聯(lián)電路的特點(diǎn)表示出圖1和圖2中電源的電壓即可得出R
2的阻值,根據(jù)電阻的串聯(lián)特點(diǎn)和歐姆定律表示出圖2中的電流,根據(jù)P=I
2R表示出圖2中R
2消耗的功率結(jié)合電阻關(guān)系即可求出電源的電壓;
(3)當(dāng)S
2、S
3都閉合時(shí),電路為R
1的簡單電路,電流表測電路中的電流,此時(shí)電流表的示數(shù)最小,根據(jù)歐姆定律求出其大�。划�(dāng)S
1、S
3都閉合時(shí),燈泡與電阻R
2并聯(lián),電流表測R
2支路的電流,此時(shí)電流表的示數(shù)最大,根據(jù)并聯(lián)電路的電壓特點(diǎn)和歐姆定律求出電流表的示數(shù),進(jìn)一步求出通過電流表的最大電流與最小電流的之比.
解答:解:只閉合開關(guān)S
1時(shí),等效電路圖如圖1所示;只閉合開關(guān)S
3時(shí),等效電路圖如圖2所示; S
2、S
3都閉合時(shí),等效電路圖如圖3所示.
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(1)圖1和圖2中,
∵P=I
2R,且P
L額=
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P
L,
∴
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=
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=16,
解得:
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=
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;
圖2和圖3中,
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=
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=
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,
解得:
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=
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,
∴
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=
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,
∵電源的電壓不變,
∴
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=
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=
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,
解得:R
L=
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R
1=
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×20Ω=10Ω;
(2)圖1和圖2中,
∵電源的電壓不變,
∴
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=
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=
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,即
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=
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,
解得:R
2=10Ω,
圖2中的電流I
2=
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=
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=
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,
R
2消耗的功率:
P
2=I
22R
2=(
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)
2×10Ω=0.9W,
解得:U=12V;
(3)圖3中電流表的示數(shù)最小,最小為:
I
A=
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=
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=0.6A,
當(dāng)S
1、S
3都閉合時(shí),等效電路圖如圖4所示,此時(shí)電流表的示數(shù)最大,
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∵并聯(lián)電路中各支路兩端的電壓相等,
∴電流表的最大示數(shù):
I
A′=
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+
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=
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=1.2A+1.6A=1.8A,
∴在電流表示數(shù)不為零的條件下,通過電流表的最大電流與最小電流的之比:
I
A′:I
A=1.8A:0.6A=3:1.
答:(1)燈的電阻為10Ω;
(2)電源兩端的電壓為12V;
(3)改變?nèi)齻€(gè)開關(guān)的狀態(tài),在電流表示數(shù)不為零的條件下,通過電流表的最大電流與最小電流的之比為3:1.
點(diǎn)評:本題考查了串聯(lián)電路和并聯(lián)電路的特點(diǎn)以及歐姆定律、電功率公式的靈活應(yīng)用,關(guān)鍵是畫出三種情況的等效電路圖和利用好燈泡實(shí)際功率與額定功率、電阻R
1消耗電功率之間的關(guān)系,難點(diǎn)是根據(jù)開關(guān)的閉合、斷開時(shí)判斷出電流表有示數(shù)時(shí)最大與最小電流.