感激……A battery produces 40.8 V when 7.20 A is drawn from it and 47.0 V when 2.50 A is drawn.

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感激……A battery produces 40.8 V when 7.20 A is drawn from it and 47.0 V when 2.50 A is drawn.

感激……A battery produces 40.8 V when 7.20 A is drawn from it and 47.0 V when 2.50 A is drawn.
感激……
A battery produces 40.8 V when 7.20 A is drawn from it and 47.0 V when 2.50 A is drawn.

感激……A battery produces 40.8 V when 7.20 A is drawn from it and 47.0 V when 2.50 A is drawn.
楼上正解,但这种方法层次高了些,我给你讲得详细些
该问题可用闭合电路欧姆定律求解
E=U+Ir
其中E为电源电动势,U为外电压（即电源两极间电压）,I为总电流
式子中有E和r两个未知数,但题目恰好提供了两个情况,正好联立方程组求解
E=40.8+7.2r
E=47+2.5r
联立得r=1.32欧
实际上上面两式相减正好得到楼上兄弟的计算式
当然楼上兄弟的计算式是直接列出的,可以这样理
因为电源电动势不变,而第一次外电压为40.8V,第二次为47V,说明第一次内阻上的电压比第二次大△U=6.2V
又因为内阻是不变的,所以内电压大单纯是由于电流大造成的,应有△Ir=△U
所以r=△U△I /=1.32V

用变大的电压值除以变小的电流值即可
内阻＝（47.0 － 40.8）V／（7.20－2.50）A

一个电池生产40.8 V时7.20取自它和47.0 V时2.50的结论。？

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