C
分析:(1)当开关S闭合时,R
1与R
2串联,电压表V
1测电源的电压,电压表V
2测R
1两端的电压,根据电阻的串联特点和电流特点以及欧姆定律求出R
2的阻值;
(2)当开关S断开时,三电阻串联,根据串联电路的电流特点和P=I
2R表示出R
2和R
3的功率之和是R
1功率的4倍,即可求出R
3的阻值;
(3)根据电阻的串联和欧姆定律分别表示出两种情况下电源的电压,利用电源的电压不变求出两种情况下电路中的电流之比,再根据P=I
2R求出电阻R
2和R
3消耗的功率之比.
解答:(1)当开关S闭合时,R
1与R
2串联,电压表V
1测电源的电压,电压表V
2测R
1两端的电压,
∵串联电路中各处的电流相等,且电路中的总电阻等于各分电阻之和,
∴由欧姆定律可得:
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=
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=
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=
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,
解得:R
2=2R
1=2×10Ω=20Ω,故A不正确;
(2)当开关S断开时,三电阻串联,
∵P=I
2R,且R
2和R
3的功率之和是R
1功率的4倍,
∴
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=
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=
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=
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=
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,
解得:R
3=20Ω,故B不正确;
(3)∵电源的电压不变,
∴
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=
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=
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=
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,
S断开与闭合时,电阻R
2消耗的功率之比:
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=
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=(
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)
2=(
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)
2=
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,故C正确;
S断开与闭合时,电阻R
3消耗的功率之比:
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=
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═(
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)
2=(
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)
2=
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,故D不正确.
故选C.
点评:本题考查了串联电路的特点和欧姆定律、电功率公式的应用,关键是开关闭合、断开时电路串并联的辨别和电压表所测电路元件的判断.