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I have a simple circuit of a coil parallel to a resistance and then in series with another resistance. I wanted to understand the behavior of this circuit so I ran a simulation, and what I noticed is that the current $I_1$ through $R_1$ is changing like the voltage $U_1$. I also noticed that $U_1+U_2=U_0$, that is logical, because we can't make voltage from nothing... What I am interested is, why does the current and voltage change and how can I calculate them? Circuit

$R_1=1.2\,\Omega, R_2=806.7\,\Omega,L=400H,U_0=240V$

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  • $\begingroup$ what are U1 and U2. I think when you switch on the circuit the current increases and you will have inductive load which results in slower increase in current and same is true when current is decreased the coil will cause slow decrease in current. $\endgroup$
    – hsinghal
    Commented Jun 15, 2016 at 21:22
  • $\begingroup$ $U_1$ is the voltage on the resistor $R_1$ and $U_2$ is the voltage on $R_2$ and $L$ $\endgroup$
    – CryoDrakon
    Commented Jun 15, 2016 at 21:25
  • $\begingroup$ What voltage have you measured in $U_2$? I suppose it is small.. $\endgroup$ Commented Jun 15, 2016 at 23:19
  • $\begingroup$ I think it is because $U_1 = I_1R_1$, with $R_1$ constant, $U_1$ is proportional to $I_1$ $\endgroup$ Commented Jun 15, 2016 at 23:23
  • $\begingroup$ Have you looked up coil (inductor) in your textbook or on the internet? $\endgroup$ Commented Jun 15, 2016 at 23:24

1 Answer 1

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The inductor works the same as it does in AC circuits : the PD across it is related to the current through it by $V = L\frac{dI}{dt}$.

You are correct : the current I through L increases in proportion to $1-e^{-kt}$ towards a maximum value. Meanwhile $\frac{dI}{dt}$ decreases from a maximum value to zero. When the maximum current flows it is constant, and there is no longer any impedance from L because then $\frac{dI}{dt}=0$. L then acts as a short circuit across $R_2$.

http://tuttle.merc.iastate.edu/ee201/topics/capacitors_inductors/RL_transients.pdf

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