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p486ckl3 2015-7-16 08:18 PM

求教學計算電流

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[url=http://98745612345678.blogspot.hk/2015/07/blog-post.html]求教學計算電流[/url]A^4 AND A^5

tangents 2015-7-18 03:38 PM

Iac = 4 mA

Vab = 6V, Vac= 4V

so Vcb = 2V, Icb = 1mA

A4 = 7mA,  A5 = 3mA

p486ckl3 2015-7-19 06:08 PM

我都係睇五明。。。



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kiwakwok 2015-8-13 09:50 AM

Current source = 10 mA.
since Iab = 6mA so Iac = 10 - 6 = 4 mA.
Suppose the potential at a is 10 V (arbitrary), then potentials at b and c are, respectively, 4 V and 6 V.
Therefore, current flows from c to b, Ibc = 2 V / 2 kΩ = 1 mA.
Hence, A4 = 7 mA, A5 = 3 mA.
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