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標題: 電阻降壓問題 [打印本頁]

作者: =0=......haha! 時間: 2009-10-28 22:16 標題: 電阻降壓問題

我想問下如果我計電阻降壓
咁by V=IR

R1既current係咪I1+I2
而計R2既current係咪只係計I2就得?

[ 本帖最後由 =0=......haha! 於 2009-10-28 22:19 編輯 ]

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作者: lok418 時間: 2009-10-29 15:53

superposition, Thévenin’s

IR1 = I1' +I"

IR2=I2"-I2'

[ 本帖最後由 lok418 於 2009-10-29 15:59 編輯 ]

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作者: =0=......haha! 時間: 2009-10-29 18:04 標題: 回覆 2# 的帖子

sorry...唔係好明
請問點解會有兩個case既?

作者: lok418 時間: 2009-10-29 18:56

原帖由 =0=......haha! 於 2009-10-29 18:04 發表
sorry...唔係好明
請問點解會有兩個case既?

本書的一少部份

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作者: kk825chan 時間: 2009-10-29 21:56

你要給完整電路才能幫你

作者: cmhd 時間: 2009-10-30 00:41 標題: 回覆 1# 的帖子

in, out, 地?

作者: FireBlack 時間: 2009-10-30 01:55

The voltage drop across R2 is I2 x R2. It is very clear.

The voltage drop across R1 is can be (I2 + I1) x R1 or (I2 - I1) x R1, depending on the directions of I2 and I1.

作者: =0=......haha! 時間: 2009-10-30 07:29

原帖由 FireBlack 於 2009-10-30 01:55 發表
The voltage drop across R2 is I2 x R2. It is very clear.

The voltage drop across R1 is can be (I2 + I1) x R1 or (I2 - I1) x R1, depending on the directions of I2 and I1.

thanks~
好在仲有人睇得明

作者: =0=......haha! 時間: 2009-10-30 07:32

原帖由 FireBlack 於 2009-10-30 01:55 發表
The voltage drop across R2 is I2 x R2. It is very clear.

The voltage drop across R1 is can be (I2 + I1) x R1 or (I2 - I1) x R1, depending on the directions of I2 and I1.

咁即係direction of current is from R1 to R2(left to right)
R1=I1 + I2
R2=I2
thats right?
thanks~

作者: FireBlack 時間: 2009-10-30 23:26

If both I1 and I2 flows towards the central point, or both flows away from the central point, then I1 + I2 should be used for R1. Otherwise, I1 - I2 should be used for R1.