**This post answers the question “What is a series RL circuit?”. It shows how to make an analysis of series $RL$ circuit with the step input signal.**

The circuit and the input voltage signal are depicted below. Let’s find voltage and current wave-forms of the inductor and it’s time constant.

Let’s consider the voltage source signal as ${V}_{S}\left(t\right)=\left\{\begin{array}{l}0,t0\\ V,t\ge 0\end{array}\right.$.

As in the example with the series RC circuit, we must apply KVL ${i}_{L}R+L\frac{d{I}_{L}}{dt}={v}_{S}$. The total solution of this differential equation is the sum of homogenous and particular solution.

In order to find homogenous solution we must resolve the equation ${i}_{L}R+L\frac{d{i}_{L}}{dt}=0$. Let’s assume that homogenous solution is in the form ${i}_{L}=A{e}^{st}$. So $RA{e}^{st}+sL{e}^{st}=0$, then $s=\u2013\frac{R}{L}$ and the homogenous solution is ${i}_{L}=A{e}^{\u2013\frac{R}{L}t}$.

In order to find particular solution, we must resolve the equation ${v}_{S}={i}_{Lp}R+L\frac{d{i}_{Lp}}{dt}$. Particular solution of this equation can be ${i}_{Lp}=const$. In this case $const=\frac{V}{R}$.

So total solution is ${i}_{L}=\frac{V}{R}+A{e}^{\u2013\frac{R}{L}t}$. Knowing that for $t=0$, ${i}_{L}=0$ then $A=\u2013\frac{V}{R}$. Then ${i}_{L}=\frac{V}{R}(1\u2013{e}^{\u2013\frac{R}{L}})$.

Here we can obtain ${v}_{L}=L\frac{d{i}_{L}}{dt}=V{e}^{\u2013\frac{R}{L}t}$.

The current and voltage waveforms for inductor are depicted below.

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