题目内容
若
=
+
(r1+r2≠0),则R等于
.
| 1 |
| R |
| 1 |
| r1 |
| 1 |
| r2 |
| r1r2 |
| r1+r2 |
| r1r2 |
| r1+r2 |
分析:已知等式右边通分并利用同分母分式的加法法则计算,变形即可表示出R.
解答:解:∵
=
+
=
,
∴R=
.
故答案为:
| 1 |
| R |
| 1 |
| r1 |
| 1 |
| r2 |
| r1+r2 |
| r1r2 |
∴R=
| r1r2 |
| r1+r2 |
故答案为:
| r1r2 |
| r1+r2 |
点评:此题考查了分式的加减法,熟练掌握运算法则是解本题的关键.
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已知总电阻R与R1、R2关系式是
=
+
,若R=6欧姆,R1=3R2,则R1、R2的值分别是( )
| 1 |
| R |
| 1 |
| R1 |
| 1 |
| R2 |
| A、R1=45欧姆,R2=15欧姆 | ||||
| B、R1=24欧姆,R2=8欧姆 | ||||
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D、R1=
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