题目内容
| A、R1=30Ω,R2=15Ω | ||||
B、R1=
| ||||
| C、R1=15Ω,R2=30Ω | ||||
D、R1=
|
分析:本题属于并联电路,等量关系为:
=
+
,把R1=2R2代入
=
+
,得
=
+
,解这个分式方程即可.
| 1 |
| R |
| 1 |
| R1 |
| 1 |
| R2 |
| 1 |
| R |
| 1 |
| R1 |
| 1 |
| R2 |
| 1 |
| 10 |
| 1 |
| 2R2 |
| 1 |
| R2 |
解答:解:∵
=
+
,R1=2R2
∴
=
+
,
解得R2=15
∴R1=2R2=30.
故选A.
| 1 |
| R |
| 1 |
| R1 |
| 1 |
| R2 |
∴
| 1 |
| 10 |
| 1 |
| 2R2 |
| 1 |
| R2 |
解得R2=15
∴R1=2R2=30.
故选A.
点评:本题考查分式方程的应用,分析题意,找到关键描述语,找到合适的等量关系是解决问题的关键.
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| 1 |
| R |
| 1 |
| R1 |
| 1 |
| R2 |
( )
| A、30Ω,15Ω | ||||
B、
| ||||
| C、15Ω,30Ω | ||||
D、
|