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- Thread starter Avichal
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CWatters

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Should that be iR + Ldi/dt = 0

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[tex]\partial_t \vec{B}=-\vec{\nabla} \times \vec{E},[/tex]

and integrate (line integral) along the circuit in direction of the positve current. Then the left-hand side translates into [itex]L \frac{\mathrm{d} i}{\mathrm{d}t}[/itex] for compact circuits, and the right-hand side you can transform into an integral along the surface, translating into [itex]-R i[/itex], where we have made use of Ohm's Law, [itex]\vec{E}=\vec{j}/\sigma[/itex]. From this you get the desired equation,

[tex]L \frac{\mathrm{d} i}{\mathrm{d} t}=-R i.[/tex]

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Sorry I am unaware of some of the things you said. Can you simplify please?

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