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Voltage drop across a thin spherical layer
Posted 12 août 2012, 17:49 UTC−4 Low-Frequency Electromagnetics Version 4.2 2 Replies
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Hello,
I’m modeling the voltage distribution for two mediums separated by a thin spherical layer. I’m using a contact impedance boundary condition to simulate the thin spherical layer that separates the two mediums (the inside and the outside of the sphere), and it seems to be working just fine, I mean the result is what I was expecting. The problem is that I also need to calculate the voltage drop across the thin layer, or in this case across the boundary, and I have no idea how to do that.
I only need the value of that voltage difference on the circumference defined by the sphere and the xy plane (or more exactly I need to plot it against the arc length), and not on the whole sphere. I don’t know if that is important.
I’m a fairly new Comsol user, and well I’ve checked several threads here in the forum dealing with similar topics, but I’m still completely lost, so I would really, really appreciate if somebody could help with this.
I’m modeling the voltage distribution for two mediums separated by a thin spherical layer. I’m using a contact impedance boundary condition to simulate the thin spherical layer that separates the two mediums (the inside and the outside of the sphere), and it seems to be working just fine, I mean the result is what I was expecting. The problem is that I also need to calculate the voltage drop across the thin layer, or in this case across the boundary, and I have no idea how to do that.
I only need the value of that voltage difference on the circumference defined by the sphere and the xy plane (or more exactly I need to plot it against the arc length), and not on the whole sphere. I don’t know if that is important.
I’m a fairly new Comsol user, and well I’ve checked several threads here in the forum dealing with similar topics, but I’m still completely lost, so I would really, really appreciate if somebody could help with this.
2 Replies Last Post 13 août 2012, 16:29 UTC−4