標有“6V、3W”和“6V、6W”的兩盞電燈串聯(lián)后接入電路中,如果讓其中一盞燈正常發(fā)光,另一盞燈發(fā)光較暗,則該電路兩端的電壓應為 V,發(fā)光較暗的燈實際功率為 .
【答案】
分析:由燈的銘牌可知燈泡的額定電壓與額定功率,由電功率的變形公式可以求出燈泡電阻與燈泡正常發(fā)光時的電流;
由串聯(lián)電路特點、歐姆定律及電功率公式計算解題.
解答:解:∵P=
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,
∴電燈的電阻:
R
1=
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=
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=12Ω,
R
2=
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=
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=6Ω;
∵P=UI,
∴燈泡正常發(fā)光時的電流:
I
1=
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=
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=0.5A,
I
2=
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=
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=1A;
兩燈串聯(lián),通過它們的電流相等,
額定電流小的燈L
1正常發(fā)光,實際功率等于額定功率3W,
電路電流I=I
1=0.5A,
∵I=
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,
∴電路電壓U=I(R
1+R
2)=0.5A×(12Ω+6Ω)=9V;
燈L
1的實際功率P
2′=I′
2R
2=(0.5A)
2×6Ω=1.5W;
故答案為:9;1.5W.
點評:本題考查了求電路電壓、燈泡實際功率問題,解題關鍵是:一據(jù)銘牌求燈的電阻和正常發(fā)光電流,二認真審題找出符合條件的電流(選小的電流).