As we know,the Kirchhoff circuit laws are applicable for conservative electric fields. Now it is applicable for circuits where inductors are present but the field there is not conservative.
So how are the Kirchhoff's laws applicable there?
Answer
Typically, with inductors, we use the complex impedance, $Z=i\omega L$ with current frequency $\omega$ and inductance $L$, for the voltage: $$ V_{ind}=IZ,\quad V_{rms}=I_{rms}|Z| $$ where the left equation is the voltage of the inductor and the right equation the root-mean-square.
Surely, however, what goes on inside the inductor doesn't matter, it only is only the voltage drop across the inductor that actually matters with Kirchoff's voltage law. Kirchoff's current law, on the other hand, states that the current across a complete circuit is zero because otherwise a charge builds up, so again the non-conservative field is irrelevant.
Professor Lewin (MIT) explains that if we have a magnetic potential driving a circuit, rather than a constant voltage source, the Kirchoff voltage law does not come up with the same answer. This is based on a different definition of Kirchoff's law than what is normally taught. What Lewin means, of course, is that $$ \oint\mathbf E\cdot d\mathbf l=0\tag{1} $$ He writes this on the board integrating from points $D$ to $A$ on a circuit board going to the left & going to the right, the two directions are not equal. However, most textbooks use the formulation,
sum of voltage drops around a loop is equal to the sum of emf around a loop
which is more equivalent to $$ \oint\mathbf E\cdot d\mathbf l=-\frac{d\phi}{dt}\tag{2} $$ which is really Faraday's law that always works.
With definition (1) & using Professor Lewin's example, we have that the net voltage is 1V (as he described) being driven by an emf of 1 V; all is well with Kirchoff's laws indeed.
No comments:
Post a Comment