Current through an inductor - Page 1
Forum section: electronics / beginners

author: renzoms
author_rank: Regular Contributor
timestamp: November 27, 2019, 04:28:49 pm
content: Really need some help. I have not been able to define the reason for the current passing through an inductor supplied DC voltage, in series with a resistor (DC supplied RL circuit), exhbiting exponentially delayed increase.In contrast, I understand this behavior (exponentially delayed increase) in a voltage across a charging capacitor in a DC supplied RC circuit. The capacitor experiences the largest change in voltage across it (100 - 0) and charges up to 38 because of the resistance and capacitance in the circuit or quickly said the RC constant. Then the capacitor experiences a slightly smaller change in voltage across it (100 - 38) and charges up to 68.7 due to the resistance and capacitance in the circuit. Then 27, 16, 9, 3, 1... 0.I'm trying to extrapolate this onto inductors and I've been wrong before so I appreciate help.I'm thinkin:The inductor exhibits exponentially delays increase in current through the inductor in a DC supplied RL circuit. The inductor experiences the largest change in current across it (5 - 0) and passes 1.9 because of the resistance and inductance in the circuit or quickly said the L/R constant. Then the inductor experiences a slightly smaller change in current across it (5 - 1.9) and passes 3.078 due to the resistance and capacitance in the circuit. Then 4.108, 4.8, 4.9 .... 5. I made those last numbers up.In summary, understanding the exponentially delayed behavior of capacitors charging and discharging helped me understand the response of an RC circuit supplied square wave, sine wave, sawtooth wave, and triangular wave, so I want to understand why the exponentially delayed increasing current through an inductor occurs before I move onto learning about inductors and resistors supplied AC wave forms.Thanks!

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 27, 2019, 08:24:30 pm
content: There is no delay. That would imply a time shift. (You can define delay in a certain way, e.g. the 50%-50% rise time, etc., but such a definition is arbitrary.)It's simply that, for a constant voltage applied to an R+C, initially all the voltage (Vin - Vc(0)) appears across the resistor, so the capacitor current is maximal; in the next instant dt, the voltage approaches Vin an incremental amount dV = I/C dt, and so on. We integrate (substituting I = (Vin - Vc) / R), rearranging as needed to make it integrable, and obtain the exponential function.There are no steps like 38%, 27, 16, etc., the process is completely continuous; we can define thresholds, which are proportional to the time constant (i.e., a ratio of ln(2) for example, where the proportion is 1), but which one we choose is arbitrary.We also cannot consider the problem wholly stepwise, as we will get some kind of error if we do. If we perform the  sum of ΔV = I/C Δt, what Δt shall we choose? Again it is arbitrary (the correct answer is lim Δt --> 0, and the sum becomes integral dt)....This discussion may not be very helpful if you haven't taken calculus yet. May I recommend 3blue1brown's series on the subject?Okay. Why emphasize the capacitor? Because the inductor is the same.Everything an inductor does with current, a capacitor does with voltage.Capacitor: I = C dV/dtInductor: V = L dI/dtJust swap V and I.We phrase it:For a constant current applied to an R||L, initially all the current (Iin - IL(0)) drops through the resistor, so the inductor voltage is maximal; in the next instant dt, the current approaches Iin an incremental amount dI = V/L dt, and so on. We integrate (substituting V = (Iin - IL) * R), ...Note that I used a series-parallel transformation here -- all voltages become currents and vice versa, so the source is actually a current source now, and the circuit is a current divider (everything in parallel), instead of a voltage divider (everything in series). The topology change may not be obvious, but it is easy to show that they are equivalent.Interestingly (but by no accident), the result is precisely the same in the series circuit -- simply convert the parallel current source and resistor from Norton to Thevenin equivalent.Tim