Recent content by azi100

anyone?? the only progress i have made is to assume that kinetic energy is equal to k*I^2 where k is some constant so that d Ke /dt = k*I*I' and from there I can derive the amount of energy lost to heat. Do you think this is the right approach?

Well if the energy put in by the electric field is all lost to heat then no work is done on the charges and so there kinetic energy must remain constant.

Consider a circuit with an AC source (time varying emf) and a resistor. The voltage drop through the resistor is of course equal to the emf. Consider the region which bounds the resistor. The work done on a charge as it moves from one end of the resistor to the other (and in turn the...

thanks guys. i got my answer and I've been thinking about this question for a while.

sections 8.3 and 8.4 are awesome. thanks atyy. i have never seen the scalar magnetic potential and the rest of it in any e+m text

So, elect_eng if the voltage gradient is fully concentrated in the resistor that means that the E-field along the inductor is practically 0 (since its a perfect conductor). I just want to make sure that's what your saying.

Wait. atyy: are you saying that the path integral of E around a closed curve C is approximately 0, if C is a circuit containing an inductor??

To all: I guess what I'm looking for is a translation from circuit theory to reality. So, quite simply, if i give you a resistor, an inductor and i guess an ac source (to generate some current) what will the E field be... 1. along the inductor (which has negligable resistance =Ro and...

Given an inductor (maybe in an RL circuit, whatever) that starts at point A and ends at point B. If the voltage drop across the inductor is indeed -L dI/dt then ohms law should imply that -L dI/dt = RI, where R := the resistance of the inductor. This implies that -L dI/dt=0 which is definitely...

Well... Consider a basic RL circuit. Maxwell tells us that the induced EMF=-L dI/dt. The voltage drop across the resistor is of course RI, but why does this mean that RI =L dI/dt as every physics textbook says. That statement relies on the fact that path integral around the circuit is = 0...