Re: inductor circuit concept

[John Mallinckrodt <ajmallinckro@CSUPOMONA.EDU>]

Rondo Jeffery has answered this question, but it may be worth
emphasizing how never to go wrong in this perennially confusing
situation.

When analyzing a circuit using the loop rule, one generally
chooses (rightly or wrongly) a direction for the current flow in
each leg.  Having done that, the potential difference across a
resistor is *always* -iR when one "goes with the flow" and +iR
when one "goes against the flow."  This seems natural because of
the obvious and proper analogy with a stream as implied by the
words.

In exactly the same way, the potential difference across an
inductor is *always* -Ldi/dt when one "goes with the flow" and
+Ldi/dt when one "goes against the flow."  There is absolutely no
need to make any prior "choice of direction" for di/dt or to
concern yourself in any way whatsoever with the likely sign of
di/dt itself.  It works *itself* out in the ensuing algebra in
exactly the same way that the correct sign of i works itself out.
If I have the sign of i wrong, then I have the sign of di/dt wrong
as well but *both* will be fixed automatically by the math.

To remember the correct sign to use in the loop rule, it helps to
carry around a single concrete example.  I always refer to the
fact that to make a *positive* current through an inductor
*positive* rate of change, I will need to raise the potential of
the input side relative to the output side thereby creating a
voltage *drop* when I go "with the flow."

John Mallinckrodt      mailto:ajm@csupomona.edu
Cal Poly Pomona          http://www.csupomona.edu/~ajm

On Fri, 27 Apr 2001, SSHS KPHOX wrote:

> When activating a RL circuit with battery and switch loop
> theorem gives:
>
> emf - iR - L(di/dt) = 0  form this we get i = (emf/R)(1-e^t/timeconstant)
>
> when allowing to decay, I want to write the loop theorem as:
> -iR + L(di/dt) = 0 as there is a potential drop across the
> resistor and gain in the inductor. This makes it hard to get a
> negative exponent in the current equation. The text says we
> can use the original equation and set emf to zero. This works
> algebraically but does not set well with my concept of energy
> conservation. I know that di/dt is negative which will make
> the term positive but .....I also know that the self induced
> emf is -Ldi/dt which could make a problem in the first
> equation.
>
> I need help reconciling the algebraic signs here with my concept of
> incresing and decreasing potentials as I traverse a loop.