Electronics > Beginners
Why does inc R of a RL circuit increase dVL/dt. Here I have circuits and details
renzoms:
Assuming the current supplied by the voltage source produces the current 25 mA, from 0 A, ideally, for practical reasons, what is happening to the current through the resistor? I imagine the resistor conducts 25mA immediately and the measured current that appears on the graph, above it, is the measurement of the current passing through the inductor around the circuit in a loop. That current increases exponentially because of the self-induced voltage.
I believe the current change 0 A to 25 mA created the exponentially changing back emf, that affects the current 0 A to 25 mA to change exponentially, through the inductor. Through the resistor it went 0 A to 25 mA. Can you please explain what I've got wrong and misunderstand?
Edit: had to make a fix in the second to last sentence. I meant resistor I wrote inductor, previously.
renzoms:
I'm going to restudy inductors using my book from the library and the help on the internet. Idk why I've done mental gymnastics leading to my misunderstanding of current in this circuit, among other things. For example I imagine current across the resistor is instantly 25 mA and the measured current on the graph is only the current that passes around the circuit back to the resistor. I imagined this because I believed the voltage drop/back emf/self-induced voltage was created by the rate of change of current 0 A to 25 mA created by the voltage source from off 0 V to on 5 V. Def wrong 0.0
rstofer:
As my plot shows, the current just after t0 is 0 mA because the inductor voltage rose instantly to the supply voltage. With no voltage difference across the resistor, there is no current flow through the resistor. The current flow is everywhere the same so what flows through the resistor flows through the inductor.
renzoms:
Good explanation and I'm happy to have this help.
How did the voltage across the inductor rise to the supply voltage?
renzoms:
The voltage across the inductor rose to the voltage supply because the voltage across the voltage supply V, provoked the self-induced emf and voltage drop of the inductor, by the potential difference between the voltage across the voltage source and the resistor and the inductor.
The voltage source changed from 0 V to 5 V, practically ideally, and the rate of change of current through the circuit began at 0 and increased to V/R, because of the self-induced emf.
The rate of change of current is affected by the resistance because of the voltage drop across the resistor, causing less voltage seen/available to the inductor, abiding by the laws of physics (kvl), (i am being dramatic at the end)
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