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一、问题的提出例1:如图所示,电源电动势E=6V,内电阻r=2Ω,定值电阻R_1=4Ω,可变电阻R_2为0~10Ω,求R_2消耗的最大功率。方法 1:R_2的电功率 P_2=(E/(R_1+R_2+r))~2R_2=E~2R~2/{[R_2-(R_1+r)]~2+4(R_1+r)R_2}=E~2/{[R_2-(R_1+r)]~2/R_2+4(R_1+r)}…………1当 R_2= R_1+r 即 R_2=6Ω时,P_(2max)=E~2/[4(R_1+r)]=1.5w 最大,“方法 2:通过1式看出,可以将虚线框内部分视为R2的等效供电电源,等效电源的电动势为E,等效电源的内电阻为R1+r,则R2为等效电源的外电阻。根据电功率计算的结论可知,当内、外电阻相等
First, the problem raised Example 1: As shown, the power electromotive force E = 6V, the internal resistance r = 2Ω, the value of the resistor R_1 = 4Ω, variable resistor R_2 is 0 ~ 10Ω, find the maximum power consumption R_2. Method 1: The electrical power of R 2 P 2 = (E / (R_1 + R_2 + r)) ~ 2R_2 = E ~ 2R ~ 2 / {[R_2- (R_1 + r)] ~ 2 + 4 (R_1 + r) When R_2 = R_1 + r is R_2 = 6Ω, P_ (2max) = E ~ 2 / {[R_2- (R_1 + r)] ~ 2 / R_2 + 4 (R_1 + r) 2 / [4 (R_1 + r)] = 1.5w maximum, ”Method 2: Through the 1-type can be seen that the part of the dashed box can be regarded as equivalent power supply R2, the equivalent power of the electromotive force E, etc. The internal resistance of the power supply is R1 + r, then R2 is the external resistance of the equivalent power supply.According to the conclusion of the calculation of electric power, when the internal resistance and the external resistance are equal