论文部分内容阅读
用一般的方法来解电阻的并联,特别是三个或三个以上电阻的并联问题是比较麻烦的.这里介绍的诺模图可以很方便地解这类问题,计算公式为: 1/R=1/(R_1)+1/(R_2)+1/(R_3)+…+1/(R_n). 串联电容或并联电感的计算也可用此图来解,它们的计算公式分别为: 1/C=1/(C_1)+…+1/(C_n);1/L=1/(L_1)+…+1/(L_n). 图中各尺度所标的数值,可视需要同时乘以10的任意次方,如图中的8(或16)可以是80(或160),也可以是8×10~(-2)(或16×10~(-2))等.
In general, it is troublesome to solve the parallel connection of resistors, especially the parallel connection of three or more resistors. The nomogram introduced here is easy to solve such problems, and the formula is: 1 / R = 1 / (R_1) + 1 / (R_2) + 1 / (R_3) + ... + 1 / (R_n). The calculation of the series capacitor or the shunt inductor can also be solved by using this graph. Their calculation formulas are as follows: 1 / C = 1 / (C_1) + ... + 1 / (C_n); 1 / L = 1 / (L_1) + ... + 1 / (L_n). The values in each scale in the figure can be multiplied by 10 The second party, such as 8 (or 16) in the figure, can be 80 (or 160) or 8 × 10 -2 (or 16 × 10 -2).