Non conservative electric field and kirchoff law

[phymatter]

in circuits involving inductors can we use kirchhoff law ?
i have seen this in many books but the electric field inside inductors is non conservative !

 

[sphyics]

if the direction of assumed current is opposite to the direction of motion, the voltage across the inductor rises.

i'm unable to get ur point regarding the relation between KVL and electric field :)

 

[tiny-tim]

Welcome to PF!

Hi phymatter! Welcome to PF!

(two h's in Kirchhoff! )

in circuits involving inductors can we use kirchhoff law ?

Yes, V = IZ, or V_rms = I_rms|Z| where Z is the (complex) impedance of the inductor.

And Z = iωL where ω is the frequency of the current, and L is the inductance, and i = √(-1).

(does it matter what's going on inside the inductor? … you'll only be measuring the voltage drop across it )

 

[willem2]

in circuits involving inductors can we use kirchhoff law ?
i have seen this in many books but the electric field inside inductors is non conservative !

Yes you can. you can use $$V = L \frac {dI}{dt}$$ for the potential across the inductor, or you can use an impedance of $j \omega L$ if you can work with complex impedances and frequencies.

That the field is nonconservative is not a problem. Even if conservation of energy is violated (because of a non-closed system) Kirchhofs laws are still valid.

The current law says that all currents going into a node must sum to 0, because otherwise the electric charge will pile up.

You might think that the voltage law is not valid, because in a non-conservative electric field you can't define a potential. The potential difference if you put an open loop of wire in this field is
still the integral of the electric field along this loop however.
If you connect one side of the loop to a point with a known potential you can compute the
potential of any point of the wire.

Kirchhof voltage law tells us that for a circuit that goes through nodes A,B,C and D:

$$(V_B - V_A) + (V_C - V_B) + (V_D - V_C) + (V_A - V_D) = 0$$ you can prove this
with only arithmetic using no properties of the electric field.