A 12.6-V battery is in series with a resistance of 0.330 Ω and an inductor. I got part a right, but I can't seem to figure out the rest. help please. a) After a long time, what is the current in the circuit?I(tinfinity) 38.1 A(b) What is the current after one time constant?I(tτ) AHow is the time constant defined for an RL series circuit? How does the current change as a function of time in such a circuit (c) What's the voltage drop across the inductor at this time?ΔVL VUse Kirchoff's Loop rule to describe the relative sizes of the potential differences across the battery, inductor, and resistor.(d) Find the inductance if the time constant is 0.110 s.L H
b) Current is given by i (E/R)*(1 - e^(-t/tL)) where tL is the inductive time constant. If t tL then i (E/R)*(1 - e^(-1)) You can crank that out. In an RL series circuit, tL L/R c) If the current is i, the voltage across it is i*R. The voltage across the circuit is E. So the voltage across the inductor is E - i*R. d) tL L/R 0.110 s Solve for L. The L H threw me at first -- I think it's telling you that the units are henries.
