请推证:将两电阻R1R2并联时大神们帮帮忙
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(1)设流过R1和R2的总电流为L,根据分流定理,则L1=L*R2/(R1+R2) L2=L*R1/(R1+R2) 所以L1:L2=L*R2/(R1+R2) :L*R1/(R1+R2)=R2:R1 (2)电阻消耗的电功率与电流平方成正比,即P1=L1^2*R1= (L*R2/(R1+R2))^2*R1 P2=L2^2*R2=(L*R1/(R1+R2))^2*R2 P1:P2=(L*R2/(R1+R2))^2*R1 :(L*R1/(R1+R2))^2*R2 =R2^2*R1:R1^2*R2 =R2:R1 由于Q=P*t ,所以Q1:Q2=P1*t:P2*t=P1:P2 所以有Q1:Q2=P1:P2=R2:R1