What is difference between EMF given as \$\epsilon=-L\frac{dI}{dt}\$ and voltage on inductor \$V=L\frac{dI}{dt}\$? In circuit analysis is \$V\$ used as the voltage on inductor, and why is it equal to \$-\epsilon\$ ?
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2 Answers
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The symbol \$\epsilon\$ is the back-emf produced inside the inductor when voltage \$V\$ is applied. Its formula has a negative sign because "it is seen" to oppose the forward voltage \$V\$.
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What we call "voltage" can usually be described more precisely as "potential difference" (p.d.). The potential difference between any two points is the same regardless of what route one takes between those points. Thus, if there is a circuit consisting of a battery and a resistor, the potential difference (voltage) between the battery terminals is the same as the potential difference across the resistor terminals.
An electro-motive force (emf) is something that "supplies" a potential difference. In the previous example, the battery provides an emf, but the resistor does not.
When current changes in an inductor, the inductor supplies an emf (a "back" emf). This also happens to be the p.d. (voltage) across the inductor. But it is important to note that this is not just a voltage drop like in a resistor, but is also an emf. It "supplies" a p.d..
answered Jan 5, 2021 at 14:11