Exercise:
The making current for an RL circuit with resistance RO is graphed in the diagram below. center includegraphicswidthcm#image_path:making-current# center abcliste abc Read the half-life for the making current from the graph and calculate the inductance of the circuit and the applied voltage. abc Calculate the self-induced emf immediately after switching on the voltage supply. Discuss the result. abc Sketch the graph for the braking current of the same circuit. abcliste

Solution:
abcliste abc The stationary current is I_IO. The half-life corresponds to the time where the current has dropped to half of the stationary value see figure. This corresponds to a value of T_/ ThO center includegraphicswidthcm#image_path:making-current-sol# center Since the half-life of an RL circuit is given by T_/ tau ln fracLR ln we find for the inductance L LF fracThtimes Rln L approx resultLP- The applied voltage is given by V_ VF Rtimes I resultVP abc The making current is given by It I_left-e^-t/tauright It follows for the self-induced emf mathcalEt -L dotIt -L I_ left-e^-t/taurightleft-fractauright -L I_ e^-t/tauleftfracRLright -R I_ e^-t/tau For t the exponential term is equal to and we find that the self-induced emf is equal to V_ ViF -RtimesI resultViP This is equal in magnitude to the applied emf voltage supply but with the opposite polarity. The result makes sense since immediately after switching on the voltage supply the current is still close to zero so the total emf has to be close to zero. abc The breaking current is given by the expression It I_ e^-t/tau The time constant tau and therefore the half-life is the same as for the making current. The graph below ases that the voltage is supplied is switched off after the current has reached the stationary value I_V_/R. It corresponds to the graph of the making current mirrored about a horizontal axis going through I_/. center includegraphicswidthcm#image_path:making-and-breaking-current# center abcliste