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Inductors Jump to: navigation, search   Inductors in Series.svg
Consider n inductors in series, as above. The voltage across the
whole arrangement (i.e. across the two terminals) must be equal
to the sum of the voltages across the individual inductors:
v_{tot}=v_1 + v_2 + \cdots + v_n.  from our definition of
inductance, where the voltage is the inductance multiplied by
the rate of change of current, we get  v_{tot}=L_1
\frac{di_1}{dt}+L_2 \frac{di_2}{dt}+ \cdots + L_n
\frac{di_n}{dt},  where i1 is the current in element 1, and so
on. Since the current in every element the series must always be
the same (by Kirchhoff's Current Law), we can see that
v_{tot}=L_1 \frac{di}{dt}+L_2 \frac{di}{dt}+ \cdots + L_n
\frac{di}{dt},  where i is the current in the network.
Factorising, we get:  v_{tot}=\left(L_1+L_2+\cdots+L_n
\right)\frac{di}{dt}.  If we now call all the element in the
series a single, equivalent inductance, Leq, we see that
v_{tot}=L_{eq}