# 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

Figure 1. Basic RL switching circuit

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:

$V_{m} \sin(\omega t + \theta) = R I(t) + L \frac{dI(t)}{dt} \,$

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:

$V_{m} \left[ \sin(\omega t)\cos \theta + \cos(\omega t)\sin \theta \right] = R I(t) + L \frac{dI(t)}{dt} \,$

Taking the Laplace transform of both sides, we get:

$V_{m} \left( \frac{\omega \cos \theta}{s^{2} + \omega^{2}} + \frac{s \sin \theta}{s^{2} + \omega^{2}} \right) = R i(s) + s L i(s) - L I(0) \,$

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

Figure 2. RL switching transient current

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.