Hi.

I understand that the self-induced emf across an inductor is \[\epsilon=-L\frac{di}{dt}\]. Now with reference to the circuit below, assume that the current is increasing. Then the emf across the inductor will be in the direction show, opposing the increase in current. Why then is the voltage across the inductor \[V_{ab}=L\frac{di}{dt}\] , i.e. in the other direction?

I don't understand why \[V_{ab}\neq\epsilon\] ?

If someone could explain it to me I would be most grateful, as i've been trying to understand it all afternoon!!

Many thanks,

Jon.

I understand that the self-induced emf across an inductor is \[\epsilon=-L\frac{di}{dt}\]. Now with reference to the circuit below, assume that the current is increasing. Then the emf across the inductor will be in the direction show, opposing the increase in current. Why then is the voltage across the inductor \[V_{ab}=L\frac{di}{dt}\] , i.e. in the other direction?

I don't understand why \[V_{ab}\neq\epsilon\] ?

If someone could explain it to me I would be most grateful, as i've been trying to understand it all afternoon!!

Many thanks,

Jon.

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