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高中课本(80年版)下册P.80有一道例题:如图1所示,R_1为8.0欧姆,电源的电动势(?)为80伏特,内阻r为2.0欧姆,R_2为变阻器。要使变阻器消耗的功率最大,R_2应是多大?这时R_2消耗的功率是多少? 课本解法是,将R_1看成电源内阻的一部分,利用电源输出功率最大的条件求得:当R_2=R_1+r=10.0欧姆时,R_2消耗的功率最大,P_(最大)=(?)~2/4R_2=160瓦特。上述方法实际上是“等效电源法”即戴维南定理的一个应用。按照等效电源法,图1可以改画为图2,虚线框内是等效电源,其电动势(?)′
High School Textbook (80th Edition) The next volume P.80 has one example: As shown in Figure 1, R_1 is 8.0 ohms, the power supply’s electromotive force (?) is 80 volts, the internal resistance r is 2.0 ohms, and R_2 is a varistor. To maximize the power consumed by the rheostat, how large should R_2 be? What is the power consumed by R_2? The textbook solution is to consider R_1 as part of the internal resistance of the power supply, using the condition that the output power of the power supply is the maximum: When R_2=R_1 When +r=10.0 ohms, R2 consumes the largest amount of power, P_(max)=(?)~2/4R_2=160 watts. The above method is actually an “equal power method,” ie, an application of Thevenin’s theorem. According to the equivalent power law, Figure 1 can be redrawn as Figure 2, the dotted line is the equivalent power, the emf (?)’