Q1 http://photos-c.ak.fbcdn.net/hphotos-ak-snc3/hs131.snc3/17861_106839569328759_100000081500430_180641_1568426_n.jpg
Answers is 0.62A,0.75A,1.37A in (a) part ,but i can't calculate these anwsers.
Q2 http://photos-a.ak.fbcdn.net/hphotos-ak-snc3/hs131.snc3/17861_106839572662092_100000081500430_180642_5389158_n.jpg
In b part, my answer is Vc=6(1-e^(-6t))V, but the answer is Vc=6(1+e^(-6t)),
what's my wrong?
My steps:
8=I(32)
I=0.25A
Vc=E(1-e^(-t/RC))
=(8-8(0.25))(1-e^(-t/(1/36)(6)))
=6(1-e^(-6t))V
2 electrical questions
2010-01-02 9:45 pm
回答 (2)
2010-01-03 1:06 am
✔ 最佳答案
Q1:The equations are:I1 + I2 = I3
Left loop: 10-5 =-10(.I1) + 15(I2)
i.e. 1 = =-2(.I1) + 3(I2) ------------------ (1)
Right loop: 10+8 = 15(I2) + 5(I3)
i.e. 18 = 15(I2) + 5[I1+I2] = 5.(I1) + 20(I2) ------------------- (2)
From (1) I2 = [1+2(I1)]/3
substitute into (2)
18 = 5(I1) + 20[1+2(I1)]/3
54 = 15(I1) + 20 + 40(I1)
I1 = 34/55 A = 0.62 A
Hence, I2 = [1+2x0.62]/3 A = 0.75 A
I3 = (0.62 + 0.75) A = 1.37 A
-------------------------------------------------
Q2: When the switch is at A, voltage across the 24-ohm resistor
= 20 x 24/(16+24) v = 12 v
hence, voltage across capacitor = 12 v
When switch is at B, voltage across 24-ohm resistor
= 8 x 24/(8+24) v = 6 v
Therefore, the voltage across the capacitor discharges from 12 v exponentially down to 6 v,
V(c) = [(12-6)exp(-t/(24/36)] + 6 v
Vc = 6[1+exp(-1.5t)]
2010-01-03 1:31 am
it should be
Vc = 6[1+exp(-1.5t)]
Vc=E(1-e^(-t/RC))
can be used if the capacitor is discharges through a pure resistor
in this question V does not drop to zero
Vc = 6[1+exp(-1.5t)]
Vc=E(1-e^(-t/RC))
can be used if the capacitor is discharges through a pure resistor
in this question V does not drop to zero
收錄日期: 2021-04-29 17:33:25
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100102000051KK00680