將兩燈 L 1 和 L 2 并聯接在U=24V 的電源上 , 若通過L 1 的電流為 2A , 且兩燈電阻
R 1 : R 2 = 1:3 , 求電燈 L 2的電阻和主路電流 ?
將一段阻值原為 1 歐姆的導線均勻拉長 1 倍後接回原電路中使用 , 此時其阻值應變為多少?
請幫忙解答這問題
2007-03-31 8:09 am
回答 (2)
2007-04-02 3:26 am
✔ 最佳答案
Q2Let R be the original resistance of the wire, A be its cross-sectional area and L its length
then R = (ro)L/A
where (ro) is the resistivity
When the length of the wire is doubled, its cross-sectional area is halved, thus the new resistance R(new) is
R(new) = (ro)(2L)/(A/2) = 4.(ro)L/A = 4R
the resistance is four times the original value.
2007-03-31 8:25 am
1)
Let the resistance of L1 be x, then the resistance of L2 is 3x
Voltage across L1=24V
By V=IR,
24=(2)(x)
x=12
So the resistance os L1 is 12Ω,
and the resistance of L2 is 36Ω.
Current in main circult, by V=IR
24=I[(12×36)/(12+36)]
I=2.67A
2)
Resistance of a wire directly proporional to it's length.
Double the length of wire will double it's resistance.
New resistance=2Ω
Let the resistance of L1 be x, then the resistance of L2 is 3x
Voltage across L1=24V
By V=IR,
24=(2)(x)
x=12
So the resistance os L1 is 12Ω,
and the resistance of L2 is 36Ω.
Current in main circult, by V=IR
24=I[(12×36)/(12+36)]
I=2.67A
2)
Resistance of a wire directly proporional to it's length.
Double the length of wire will double it's resistance.
New resistance=2Ω
收錄日期: 2021-04-12 22:55:39
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070331000051KK00059