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I thought I understood inductance. It's sort of an inertial property in circuits. Physically, it accounts for the fact that "it takes work to do stuff", meaning that if you put a voltage across a coil of wire, the current through that coil can't jump up to whatever steady state value it will attain instantaneously. That would mean an infinite rate of change of current, which, for non-zero inductance (always the case) could only be provided by an infinite voltage. Inductors oppose changes in the current across them. They produce back emf that tries to oppose the change in magnetic flux across the coil that would result from said change in current. This is all a result of Faraday's law and can be expressed as follows:
[tex] \mathcal{E} = -L\frac{di}{dt} [/tex]
Where the curly E is the back emf being referred to, and the the negative sign indicates that the emf opposes whatever change in flux is occurring. So far so good. Now, if you open up an electrical engineering textbook on basic circuit theory, it will tell you that if the instantaneous current through an inductor is given by i(t), then the instantaneous voltage across it v(t) is given by:
[tex] v(t) = L\frac{di}{dt} [/tex]
I'm confused about the signs. What is the "voltage across the inductor?" How is this different from the back emf produced by the coil when you try to change the current flowing through it?
[tex] \mathcal{E} = -L\frac{di}{dt} [/tex]
Where the curly E is the back emf being referred to, and the the negative sign indicates that the emf opposes whatever change in flux is occurring. So far so good. Now, if you open up an electrical engineering textbook on basic circuit theory, it will tell you that if the instantaneous current through an inductor is given by i(t), then the instantaneous voltage across it v(t) is given by:
[tex] v(t) = L\frac{di}{dt} [/tex]
I'm confused about the signs. What is the "voltage across the inductor?" How is this different from the back emf produced by the coil when you try to change the current flowing through it?
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