AB
分析:(1)開關(guān)S
1、S
2都閉合時,電阻R
1與R
2串聯(lián),電壓表V
1測電阻R
1與R
2的串聯(lián)電壓,電壓表V
2測電阻R
2的電壓,由歐姆定律可求出電阻R
1與R
2的阻值關(guān)系、電路電流,由功率公式可求出電阻R
1消耗的功率.
(2)S
1閉合、S
2斷開時,電阻R
1、R
2、R
3串聯(lián),電壓表V
1測電阻R
1、R
2、R
3的串聯(lián)電壓,電壓表V
2測電阻R
2與R
3的串聯(lián)電壓,由歐姆定律可求出電路電流、電壓V
1、V
2的表示數(shù)之比,由功率公式求出電阻R
1的功率.
(3)根據(jù)(1)與(2)的分析進一步判斷各選項的正誤.
解答:(1)開關(guān)S
1、S
2都閉合時,電阻R
1與R
2串聯(lián),電壓表V
1測電阻R
1與R
2的串聯(lián)電壓,電壓表V
2測電阻R
2的電壓,
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=
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=
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=
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,則
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=
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,故A正確;R
2=
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R
1,電路電流I=
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=
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=
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①,
電阻R
1的功率P
1=I
2R
1=(
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)
2R
1=
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③.
(2)S
1閉合、S
2斷開時,電阻R
1、R
2、R
3串聯(lián),電壓表V
1測電阻R
1、R
2、R
3的串聯(lián)電壓,電壓表V
2測電阻R
2與R
3的串聯(lián)電壓,電壓表V
1、V
2的示數(shù)之比為
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=
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=
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=
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,故B正確;
電路電流I′=
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=
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=
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②;
電阻R
1功率P′
1=I′
2R
1=(
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)
2R
1=
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④.
(3)由①②得:
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=
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=
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,故C錯誤;由③④得:
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=
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=
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,故D錯誤.
故選AB.
點評:靈活應用串聯(lián)電路的特點、歐姆定律、電功率公式是解題的關(guān)鍵,解題時注意比值法的應用.