When doing AC circuit analysis we take voltage across an inductor to be L(dI/dt) where L and I have standard meaning. Why it's not -L(dI/dt) as the induced EMF is against the driving source? Also why it's magnitude is L(dI/dt) , meaning why the pd across it has to be equal to back EMF.? Please any help.
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Self inductance $L$ is defined as
$$ L = \frac{\phi}{I} $$
where $I$ is the current created by flux $\phi$ through the inductor.
By Faraday's law,
$$ \varepsilon = -\frac{d\phi}{dt} $$
where $\varepsilon$ is the EMF produced by changing flux $\phi$.
Taking the first equation, and differentiating wrt. $t$, we see:
$$ L\dot I = \dot\phi = -\varepsilon . $$
We note that an EMF in a circuit corresponds to an increase in potential. However, we are looking for the potential drop across the inductor (since this is the meaning of the term 'potential difference') and must therefore flip the sign:
$$ -\varepsilon = V = L\dot I . $$
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$\begingroup$ Ok, i understood the first part , thanks. Can you also please tell why pd is always equal to back emf, is there any proof of that for every situation.? $\endgroup$– User 1Commented May 29, 2021 at 13:48
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$\begingroup$ The PD is equal to the negative back emf by definition. Potential difference and EMF are equivalent in all but sign (and even then, that is only by convention). $\endgroup$ Commented May 29, 2021 at 13:53
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$\begingroup$ If there is some voltage source connected to it along with other circuit elements does this holds and why this convention holds ,is there any particular reason. Thank you. $\endgroup$– User 1Commented May 29, 2021 at 13:56
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$\begingroup$ Consider a circuit with a 6V cell and two identical resistors in series. Imagine moving around the circuit. As we cross the cell in the direction of the current, we jump from 0V to 6V. When we cross the first resistor, we jump from 6V to 3V, and crossing the second, from 3V to 0V. We say that the potential difference across one of the resistors is 3V, don't we? This is obtained from $V_{initial} - V_{final} = 6 - 3$. It follows that the potential difference of the cell is 0 - 6 = -6V, even though we know the cell has an EMF of +6V. $\endgroup$ Commented May 29, 2021 at 14:02
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$\begingroup$ (This is called the passive sign convention, by the way.) $\endgroup$ Commented May 29, 2021 at 14:17