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题目有一阻值8Ω的电阻 R_1与一最大阻值为24Ω的滑动变阻器 R_2串联,接在电压 U=4V 的电源上,问当 R_2为多少时,R_2消耗的电功率最大?是多少?"常规解法由于问 R_2为多少时,P_2有最大值?最大值是多少?所以常规思路是选 R_2为自变量列出二次函数式为 P_2=I~2R_2=(U/R_1+R_2)~2R_2=16R_2/(8+R_2)~2.此式列出后大部分学生无从着手.因为要得出答案要有深厚的数学知识,且会灵活运用.它只有通过变形后
A resistor R_1 with a resistance of 8Ω is connected in series with a sliding rheostat R_2 with a maximum resistance of 24Ω and is connected to a power supply with voltage U=4V. When R_2 is the maximum, what is the maximum electric power consumed by R_2? The solution is to ask how much R_2, P_2 has a maximum value? What is the maximum value? So the conventional idea is to choose R_2 as an independent variable to list the quadratic function formula: P_2=I~2R_2=(U/R_1+R_2)~2R_2= 16R_2/(8+R_2)~2. After this formula is listed, most of the students are unable to start. Because the answer to the answer is to have a deep knowledge of mathematics, and will be flexible. It can only be transformed