# RL Circuit Switching

## Introduction

The RL switching / closing transient is one of the most common electrical transients that is encountered in practice, and is also the basis for the computation of short circuit currents.

## Derivation

Consider the basic switching circuit in the figure to the right, consisting of an AC voltage source V, a switch S, a resistance R and an inductance L (all ideal circuit elements). At time $t=0\,$ , the switch S will close and complete the circuit. Suppose the voltage source can be characterised as a sinusoid (as a function of time):

$V(t)=V_{m}\sin(\omega t+\theta )\,$ Where $\theta \,$ is an arbitrary phase angle to capture the time of switching.

At the point of switching, the voltage is given by Kirchhoff's voltage law:

$V_{m}\sin(\omega t+\theta )=RI(t)+L{\frac {dI(t)}{dt}}\,$ The current $I(t)\,$ must reach a steady state current of:

$I_{s}={\frac {V}{Z}}={\frac {V}{R+j\omega L}}\,$ ... Equ. (1)

And have a steady state power factor:

$\cos {\phi _{s}}={\frac {R}{|Z|}}={\frac {R}{\sqrt {R^{2}+\omega ^{2}L^{2}}}}\,$ However, at $t=0\,$ , $I(0)=0\,$ and the inductance $L\,$ will prevent the circuit from reaching the steady-state current instantaneously. Therefore, there must be some transient that will provide a continuous transition path from $I(0)=0\,$ to the steady-state current $I(t_{s})=I_{s}\,$ .

Equation (1) can be re-written as follows:

$V_{m}\left[\sin(\omega t)\cos \theta +\cos(\omega t)\sin \theta \right]=RI(t)+L{\frac {dI(t)}{dt}}\,$ Taking the Laplace transform of both sides, we get:

$V_{m}\left({\frac {\omega \cos \theta }{s^{2}+\omega ^{2}}}+{\frac {s\sin \theta }{s^{2}+\omega ^{2}}}\right)=Ri(s)+sLi(s)-LI(0)\,$ We assume that the initial current $I(0)=0\,$ , so therefore re-arranging the equation above we get:

$i(s)={\frac {V_{m}}{L}}\left({\frac {1}{{\frac {R}{L}}+s}}\right)\left({\frac {\omega \cos \theta }{s^{2}+\omega ^{2}}}+{\frac {s\sin \theta }{s^{2}+\omega ^{2}}}\right)\,$ $={\frac {V_{m}}{L}}\left[{\frac {\omega \cos \theta }{(s^{2}+\omega ^{2})({\frac {R}{L}}+s)}}+{\frac {s\sin \theta }{(s^{2}+\omega ^{2})({\frac {R}{L}}+s)}}\right]\,$ ... Equ. (2)

It can be shown that the expression ${\frac {1}{(s+\alpha )(s^{2}+\omega ^{2})}}$ can be simplified as follows:

${\frac {1}{(s+\alpha )(s^{2}+\omega ^{2})}}={\frac {1}{\alpha ^{2}+\omega ^{2}}}\left({\frac {1}{s+\alpha }}-{\frac {s}{s^{2}+\omega ^{2}}}+{\frac {\alpha }{s^{2}+\omega ^{2}}}\right)\,$ Therefore, the inverse Laplace transforms of the terms in Equation (2) can be evaluated in a fairly straightforward manner:

1st term: ${\mathcal {L}}^{-1}\left[{\frac {\omega \cos \theta }{(s^{2}+\omega ^{2})({\frac {R}{L}}+s)}}\right]={\frac {\omega \cos \theta }{\left({\frac {R}{L}}\right)^{2}+\omega ^{2}}}\left[e^{-{\frac {R}{L}}t}-\cos(\omega t)+{\frac {R}{\omega L}}\sin(\omega t)\right]\,$ 2nd term: ${\mathcal {L}}^{-1}\left[{\frac {s\sin \theta }{(s^{2}+\omega ^{2})({\frac {R}{L}}+s)}}\right]={\frac {\sin \theta }{\left({\frac {R}{L}}\right)^{2}+\omega ^{2}}}\left[-{\frac {R}{L}}e^{-{\frac {R}{L}}t}+\omega \sin(\omega t)+{\frac {R}{L}}\cos(\omega t)\right]\,$ Combining the two terms together (and including the constants), we get the transient current:

$I(t)={\frac {V_{m}}{L\left({\frac {R}{L}}\right)^{2}+\omega ^{2}}}\left[(\omega \cos \theta -{\frac {R}{L}}\sin \theta )e^{-{\frac {R}{L}}t}+({\frac {R}{L}}\cos \theta +\omega \sin \theta )\sin(\omega t)-(\omega \cos \theta -{\frac {R}{L}}\sin \theta )\cos(\omega t)\right]\,$ ... Equ. (3)

Earlier, we found that the steady state power factor is:

$\cos {\phi _{s}}={\frac {R}{\sqrt {R^{2}+\omega ^{2}L^{2}}}}\,$ This can be re-arranged as follows:

$\cos {\phi _{s}}={\frac {R}{L}}{\frac {1}{\sqrt {({\frac {R}{L}})^{2}+\omega ^{2}}}}\,$ Likewise, the sine of the power angle is:

$\sin {\phi _{s}}={\frac {\omega }{\sqrt {({\frac {R}{L}})^{2}+\omega ^{2}}}}\,$ Using these two equations above, we can simplify Equation (3) even further:

$I(t)={\frac {V_{m}{\sqrt {({\frac {R}{L}})^{2}+\omega ^{2}}}}{L\left[\left({\frac {R}{L}}\right)^{2}+\omega ^{2}\right]}}\left[(\cos \theta \sin \phi _{s}-\sin \theta \cos \phi _{s})e^{-{\frac {R}{L}}t}+(\cos \theta \cos \phi _{s}+\sin \theta \sin \phi _{s})\sin(\omega t)-(\cos \theta \sin \phi _{s}-\sin \theta \cos \phi _{s})\cos(\omega t)\right]\,$ Using some angle sum and difference trigonometric identities, we get:

$I(t)={\frac {V_{m}}{\sqrt {(R^{2}+\omega ^{2}L^{2}}}}\left[(-\sin(\theta -\phi _{s})e^{-{\frac {R}{L}}t}+(\cos \theta \cos \phi _{s}+\sin(\theta -\phi _{s})\sin(\omega t)+(\cos(\theta -\phi _{s})\cos(\omega t)\right]\,$ Simplifying again with the same trig identities, we get the final equation:

$I(t)={\frac {V_{m}}{\sqrt {(R^{2}+\omega ^{2}L^{2}}}}\left[(\sin(\omega t+\theta -\phi _{s})-\sin(\theta -\phi _{s})e^{-{\frac {R}{L}}t}\right]\,$ ... Equ. (4)

## Interpretation

The figure right depicts a plot of the transient current in Equation (4) for the parameters R/L = 40 and switching angle $\theta$ = 0o. Here we see the classic transient current waveform for an RL switching (closing) circuit with the time constant R/L.