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Why does inc R of a RL circuit increase dVL/dt. Here I have circuits and details

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renzoms:
Why does increasing R of a RL circuit increase the rate of change of voltage across an inductor with respect to time dVL/dt. Here a 50V 2k circuit and a 5V 200ohm circuit both create a change of current from 0 A to 25 mA t = 0 and that current is opposed by a back emf. Both circuits exhibit an increasing current because the inductors create a self-induced voltage proportional to the rate of change of magnetic field strength across the inductor with respect to time dϕM/dt. The way I understand back emf and VL is the current created by the voltage source Vs/R is passed to the inductor then the inductor receives current Vs/R. That current passes through the first loop, not exactly, I am just imagining this. Then the change of current 0 A to 25mA creates a magnetic field that increases in strength like the voltage across a capacitor increases when it is supplied a constant voltage x V. The rate of change of the magnetic field strength creates a self-induced voltage. The self-induced voltage creates the voltage drop across the inductor. So the voltage across the inductor is like the current passing through a capacitor that is proportional to the rate of change of voltage across the capacitor dVC/dt.
Increasing inductance (L) decreases the rate of change of the magnetic field strength with respect to time dϕM/dt like increasing capacitance C decreases the rate of change of voltage across the capacitor with respect to time dVC/dt.
Increasing resistance (R) decreases the rate of change of voltage across the capacitor with respect to time dVC/dt. So why does increasing resistance (R) increase the rate of change of magnetic field strength with respect to time dϕM/dt.
Next, here, is a statement I am not confident about, continuing on resistance: The current passed to the inductor is 0 A to 25 mA in both circuits. The inductance (L) is the same.
So how is the resistance (R) increasing the  rate of change of magnetic field strength with respect to time dϕM/dt, dVL/dt, and dIL/dt.

StillTrying:
"Here a 50V 2k circuit and a 5V 200ohm circuit"

At the start the current is very low, so it doesn't matter what value the resistance is, all the available voltage is across the inductor at first.
- I think.

renzoms:
So, following your reply, the voltage across the inductor in the circuit with a 50 V voltage supply is available to the inductor at the beginning. The voltage rises from 0 V to 50 V.

If the voltage rises imperfectly, I imagine it would rise to 50 V linearly, like a portion of an AC triangle wave. This would happen very quickly. The very steep triangle/triangular/linear increase to 50 V when the voltage source powers up would create a unique response across the components of the RL circuit.
For me, where I am in my studies, I think it's best to consider the voltage source to work ideally.

So the voltage rises to 50 V, creates a current 25 mA. The current 25 mA creates the back emf. The back emf, the self-induced voltage, creates the voltage drop across the inductor. Next, the voltage across the inductor decreases and the current passing through the inductor increases as the back emf decreases.

What causes the back emf to decrease more quickly.

I don't understand how the resistance causes the back emf to decrease more quickly.

I have the equations where the time constant L/R becomes smaller, while I would like to understand it in another way than the formula.

jmw:
Solving for the inductor voltage, \$v_L(t) = V_S e^{-t/(L/R)}\$. So \${v_L}'(t) = -V_S (R/L) e^{-t/(L/R)}\$, and immediately \$|{v_L}'(t)| \propto R\$.

One thing that might be useful is to apply Norton's theorem and replace the voltage source + series resistor with a current source in parallel with a resistor. So your series R-L becomes a parallel R-L driven by a current source. A higher resistance means the current will preferentially flow through the inductor, "charging" it faster so it reaches the steady state of a short circuit faster. Hence dV/dt must be higher with higher resistance.

StillTrying:
"I have the equations where the time constant L/R becomes smaller, while I would like to understand it in another way than the formula."

I can't understand anything from a formula, I have to know exactly how it works first before the formulas start to make sense, I probably wouldn't even believe Ohms law unless I already knew that 1V/1R = 1A. :)
www.eevblog.com/forum/beginners/the-art-of-electronics-book-is-awesome!/msg2851596/#msg2851596

All of your the rises and falls are exponential curves.

The start is easy where the current is close to zero so there's no volts across the resistor, with 50V across the inductor the back emf  has to be 10X higher so the current has to be rising 10X faster, in spite of the higher resistance.

But I can't think where the crossover points are in your 5V 200R and 50V 2k cases where the current rises to a level where the volt drop across the resistance changes things. I'd have to simulate and study it.

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