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In a current growing RL circuit for example, we know that when the voltage source is initially connected to the circuit current will start to flow with $di/dt>0$ and at its maximum, however this change in current causes the inductor to produce a back emf in accordance with Faraday's Law, this back emf causes the change in current to slow down i.e $d^2i/dt^2<0$, and the process goes on until the change in current is zero i.e $di/dt=0$.

This current as a function of time in a current growing RL circuit is modelled by this equation $$i(t)=i_{max}(1-e^{-t/τ})$$ where $$τ=L/R$$

and is the time constant for the circuit

From this expression we could see that as the resistance increases the current will grow faster, i.e it will take less time for the current to grow in the circuit, and if the resistance decreases the current will grow slower i.e it will take much more time for the current to grow in the circuit, but doesn't this imply that a resistor plays a role/ affects $d^2i/dt^2$.

So what is the connection between the resistance and $d^2i/dt^2$ or is it that my assumption is wrong in the first place?

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    $\begingroup$ RE: "as the resistance increases the current will grow faster"; are you sure? Remember $i_{max}$ also depends on resistance. $\endgroup$
    – The Photon
    Apr 17 '17 at 22:25
  • $\begingroup$ Oh thanks for the clarification. So is it safe to say that a resistor's role in an RL circuit is solely to control the steady state current point ( along side with the voltage source) ,with R's presence in the time constant τ=L/R just serving to compensate for that influence on the maximum current point? $\endgroup$
    – Honda
    Apr 18 '17 at 0:29
  • $\begingroup$ It does affect the time constant, and the steady state current. It just doesn't affect di/dt at t=0. $\endgroup$
    – The Photon
    Apr 18 '17 at 0:41
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what is the connection between the resistance and $d^2i/dt^2$

For the ideal RL circuit with a voltage step stimulus, $v_s(t) = V_{DC}\,u(t)$, the steady state current is

$$i(\infty) = \frac{V_{DC}}{R}$$

and then

$$i(t) = i(\infty)(1 - e^{-t/\tau}) = \frac{V_{DC}}{R}(1 - e^{-tR/L}),\qquad t \ge 0$$

$$\frac{di}{dt}=-\frac{V_{DC}}{R}(-\frac{R}{L})e^{-tR/L}=\frac{V_{DC}}{L}e^{-tR/L}$$

So the maximum rate of change of current does not depend on $R$. However, the maximum 'acceleration' of the current does.

$$\frac{d^2i}{dt^2}=\frac{V_{DC}}{L}(-\frac{R}{L})e^{-tR/L}=-V_{DC}\frac{R}{L^2}e^{-tR/L}$$

This makes sense for the reason that, if $R$ is zero, the rate of change is current is constant and thus, the 'acceleration' of current is zero.

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