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通过R2的电流为I2=0.15A,通过R1和R2的电流之比为I1:I2=2:3,
则可得出通过R1的电流为I1=0.1A,
由欧姆定律和并联电路的电压规律得,
U=U1=I1R1=0.1A×30Ω=3V;
并联电路各支路两端的电压相等,
由欧姆定律得,
R2的阻值:R2=[U
I2=
3V/0.15A]=20Ω.
答:R2的阻值为20Ω;
总电压U为3V.
点评:
本题考点: 欧姆定律的应用.
考点点评: 本题主要考查了欧姆定律及其变形公式的熟练运用,解答时要结合并联电路的电流、电压规律,属常见的题型.
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