- #1
azi100
- 10
- 0
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, but its not? According to maxwell that path integral is approximately (ignoring fringe fields, etc.) -L dI/dt. The path ingral of E around the curve is only equal to 0 in electrostatics...
I understand that voltage notation is a convenient way of describing circuits, but is not always rigorous (i.e. there doesn't exist a scalar potential in a non-static EB field). But, I do not understand why that formula is correct.
that formula being RI=L dI/dt.
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, but its not? According to maxwell that path integral is approximately (ignoring fringe fields, etc.) -L dI/dt. The path ingral of E around the curve is only equal to 0 in electrostatics...
I understand that voltage notation is a convenient way of describing circuits, but is not always rigorous (i.e. there doesn't exist a scalar potential in a non-static EB field). But, I do not understand why that formula is correct.
that formula being RI=L dI/dt.