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Re: inductor circuit concept



Ken Fox wrote:

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)

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.

When applying the loop rule you need not know the sign of either I or dI/dt.
However, you do need to assign a positive tangential direction for each
branch of the circuit. (The circuit above has only one branch.) Once that
is done the loop rule is applied. Each time you traverse a resistor or
inductor in the positive tangential direction, the potential drop is IR or
LdI/dt (so the potential difference is -IR or -LdI/dt). The sign of this
potential drop is not known until the equations are solved.

Current is the flux of the current-density vector through a cross-section of
the wire. It is a scalar, so it no direction. For a given
current-density-vector direction, the flux (current) is either positive or
negative, depending on the direction selected for the unit normal on the
integration surface. The convention is that the unit normal is in the
positive tangential direction. The flux, and therefore the current, is
positive if the current density is also in the positive tangential
direction. With this convention for the unit-normal direction, the potential
drop equals IR if you traverse a resistor in the positive tangential
direction.

The back emf of an induction coil is equal to -Nd(phi)/dt (Faraday's Law),
where phi is the magnetic flux through the surface bounded by a single turn
and N is the number of turns. Again, the sign of the flux depends on the
direction selected for the unit normal on the integration surface. The
convention is to curl the fingers of your right hand in the positive
tangential direction along the turn of wire, and your right thumb will point
in the direction of the unit normal. With this convention if you traverse
the coil in the positive tangential direction the potential drop is given by
LdI/dt. To verify the sign of the potential drop, assume that both I and
dI/dt are positive. Then phi and d(phi)/dt are both positive and, in accord
with Faraday's law, the emf is negative. This means that the
nonconservative electric field is in the negative tangential direction
throughout the coil.* If the resistance of the coil is negligible, the net
field driving the current is zero. This means there is a conservative
electric field** within the wire in the positive tangential direction, equal
in magnitude to the nonconservative electric field. A conservative electric
field points in the direction of decreasing electric potential, so the
potential drop is positive if we traverse the coil in the positive
tangential direction. By repeating this argument for the other choices for
the signs of I and dI/dt, it can be shown that the potential drop always has
the same sign as dI/dt.

*It is the line integral of this electric field along the length of the wire
in the positive tangential direction that is negative.

**This field is sourced by surface charges on the wire.

Gene


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* Eugene P. Mosca *
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