# Induced emf and self-inductance: clarification about signs [closed]

I'm a little confused.. When I write the equation of the circuit, do I have to keep the signs of the induced emf and self-inductance opposite?

I'll try to explain better my doubt: Consider a coil that has resistance R and inductance L. Running a current will give rise to both an induced emf and self-inductance, and the situation is:

induced emf= -k

self-inductance= $-L \frac{dI}{dt}$

Now, my question is: the equation of the circuit is given by:

$-k -L \frac{dI}{dt}=RI$

or by

$k -L \frac{dI}{dt}=RI$ ?

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## closed as off-topic by Emilio Pisanty, tpg2114, Dimensio1n0, Waffle's Crazy Peanut, Qmechanic♦Nov 6 '13 at 12:16

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Refer to Lenz's Law: Induced current has a direction such that its magnetic field opposes the change in magnetic field induced in the current. –  Signus Sep 24 '13 at 20:06
@Signus and so..? –  sunrise Sep 24 '13 at 20:07
@Signus I think that the correct expression is the second, because the self inducted emf opposes to the inducted emf.. do you agree? –  sunrise Sep 24 '13 at 20:12
I do believe the second equation is correct as the self-inductance does oppose the induced current already on the wire/coil. That is why I mentioned Lenz's Law. I always think to myself "induction is always in the opposite direction of what my brain thinks." –  Signus Sep 24 '13 at 20:25
Doesn't the term $-LdI/dt$ already account for the induced EMF $E$ so there must be only one of these in the equation? –  Satwik Pasani Sep 25 '13 at 1:25