如下图所示，A、B两灯泡额定电压都为110V，额定功率PA=100W、PB=40W，接在电压恒为220V电路上，欲使两灯泡均正常发光，且整个电路消耗的电功率最小，则下面四种电路连接方式中满足要求的

问题描述：

如下图所示，A、B两灯泡额定电压都为110V，额定功率P_A=100W、P_B=40W，接在电压恒为220V电路上，欲使两灯泡均正常发光，且整个电路消耗的电功率最小，则下面四种电路连接方式中满足要求的是（　　）A．B．C．D．

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最佳答案：

根据公式P=

U2

R

可知，灯泡A的电阻RA=

UA2

PA

=

(110V)2

100W

=121Ω，灯泡B的电阻RB=

UB2

PB

=

(110V)2

40W

=302.5Ω．A、图是灯泡A和灯泡B串联，然后接到220V的电源上，根据串联电路的分压特点可知灯泡B两端的电压大于110V，所以不能正常工作，故A不正确；B、图是灯泡A和可变电阻R并联后又和灯泡B串联，灯泡要想正常工作，必须满足灯泡A与可变电阻R并联后和灯泡B的电阻相等；但并联电路中，电阻越并越小，小于任何一个分电阻，所以此电路中灯泡A和灯泡B也不能正常工作，故B不正确；C、图是灯泡B和可变电阻R并联后又和灯泡A串联，灯泡要想正常工作，必须满足灯泡B与可变电阻R并联后和灯泡A的电阻相等；串联电路，电阻越串越大；可以使灯泡A和灯泡B正常工作；所以R总=2RA=2×121Ω=242Ω；D、图是灯泡A和灯泡B并联后又与可变电阻R串联，灯泡要想正常工作，必须满足灯泡A与灯泡B并联后和可变电阻R的阻值相等；由公式

1

R

=

1

R1

+

1

R2

得灯泡B和灯泡A并联后的电阻R≈86.4Ω，故D图的总电阻为R总=2R=2×86.4Ω=172.8Ω；根据公式P=

U2

R

可知，电路电阻越大．消耗的功率就越小，比较C图和D图可知，C图的总电阻大，消耗的功率最小．故选C．

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