Re: inductor circuit concept

[Rondo JEFFERY <RJEFFERY@WEBER.EDU>]

Here is my take on the RL circuit.

Ken Fox wrote:

> > > kphox@MAIL.CCSD.K12.CO.US 04/27/01 03:00PM >>>
> I have a loop theorem question for an inductive circuit. Maybe someone >can clarify it for me.
>
> 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)

Of course there should be a minus sign in the exponent of this last expression.

> 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 problem here is with the directions.  In the "activating" situation, i and di/dt are both in the same direction, so the signs work out easily.  In the "deactivating" situation, the two are in opposite directions.  That is, i still continues in the direction it was when the switch was moved from connecting to the battery to shorting the inductor and resistor to each other, but di/dt is reversed.  Thus, if you take your loop in the direction of i, then di/dt is in the opposite direction.  The minus sign due to Lenz's law (in the usual expression for the induced emf in the inductor) is there to remind you that the induced emf will always OPPOSE the change in current.  Thus, even though the magnitudes of Ldi/dt and iR are equal, when it comes to writing the differential equation for i, you must put in the minus sign:
 Ldi/dt = - iR
With this you get the final solution i = (E/R)exp(-Rt/L), where E is the battery voltage.

> 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.

Using energy to write the "deactivating" case gives the same result, if you think of what is happening physically.  The energy in the magnetic field is given by (1/2)Li^2.  The rate of energy loss in the resistor is Ri^2.  So, you write this as
    d/dt(.5Li^2) = - Ri^2

The minus sign is because the energy in the inductor is decreasing.  Carry out the differentiation, cancel one i term and you have the same DE as above.

Hope this helps.

> I need help reconciling the algebraic signs here with my concept of
> incresing and decreasing potentials as I traverse a loop.
>
> Thanks
>
> Ken Fox
> AP/IB Physics Teacher
> Smoky Hill High School, CO

Rondo Jeffery
Weber State University
Ogden, UT
rjeffery@weber.edu