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/ee/ - Electrical Engineering

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25 Dec 2021Mathchan is launched into public
Catalog Archive

File: tesla coil.jpg ( 58.38 KB , 737x343 , 1641826301707.jpg )

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  No. 6
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This board is for the discussion of electrical engineering relating to topics such as electronic circuits, control systems, telecommunication, power distribution and more.

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  No. 10

File: circuitikzmanual.pdf ( 255.79 KB , 1641826905597.pdf )

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A list of components can be found in the attached PDF.

File: RC circuit captcha.png ( 155.04 KB , 775x933 , 1662187429613.png )

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  No. 13
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>>16
What happened to the RC circuit captcha?
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  No. 14
It's a draft.
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  No. 16
>>13
That puzzle is one of the easiest on the site, I now feel stupid for thinking it was more complex than it actually is. The delta symbol and the way it is redacted made me think it was asking for something slightly more complicated.
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  No. 19
The answer is 100ohm?

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  No. 17
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>>18
yay I'm studying ee in school top 300 uni
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  No. 18
>>17
where the fuck did that "school" come from?

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  No. 12
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Hullo anons,
I am on my way of self studying some topics that is not offered in my University, but am having difficulty finding topics in regards of RF designs and implementations, any pointers to help with that? I am familiar with analog circuits, but not in how things change when they're high frequency

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  No. 4
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Gauss's law

∯∂ΩE⋅dS=1ε0∭Ωρ dV\qquad \oiint_{\partial\Omega}\textbf{E}\cdot\textrm{d}\textbf{S} = \frac{1}{\varepsilon_0}\iiint_\Omega\rho\,\mathrm{d}V


Gauss's law for magnetism
∯∂ΩB⋅dS=0\qquad \oiint_{\partial\Omega}\textbf{B}\cdot\textrm{d}\textbf{S} = 0


Faraday's law

∮∂ΣE⋅dℓ=−ddt∬ΣB⋅dS\qquad \oint_{\partial\Sigma}\textbf{E}\cdot\textrm{d}\ell = -\frac{\textrm{d}}{\textrm{d}t}\iint_\Sigma\textbf{B}\cdot\textrm{d}\textbf{S}


Ampere's law

∮∂ΣB⋅dℓ=μ0(∬ΣJ⋅dS+ε0ddt∬ΣE⋅dS)\qquad \oint_{\partial\Sigma}\textbf{B}\cdot\textrm{d}\ell = \mu_0\left(\iint_\Sigma\textbf{J}\cdot\textrm{d}\textbf{S} + \varepsilon_0\frac{\textrm{d}}{\textrm{d}t}\iint_\Sigma\textbf{E}\cdot\textrm{d}\textbf{S}\right)
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  No. 5
In differential form:

∇⋅E=ρε0∇⋅B=0∇×E=−∂B∂t∇×B=μ0(J+ε0∂E∂t) \begin{aligned} \nabla\cdot\textbf{E} &= {\rho \over \varepsilon_0}\\ \nabla\cdot\textbf{B} &= 0\\ \nabla\times\textbf{E} &= -\frac{\partial\textbf{B}}{\partial t}\\ \nabla\times\textbf{B}&=\mu_0\left(\textbf{J} + \varepsilon_0\frac{\partial\textbf{E}}{\partial t}\right) \end{aligned}

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