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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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):

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I(t) \,} must reach a steady state current of:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 $t=0\,$, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I(0) = 0 \,} and the inductance Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I(0) = 0 \,} to the steady-state current Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I(t_{s}) = I_{s} \,} .

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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I(0) = 0 \,} , so therefore re-arranging the equation above we get:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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) \,}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = \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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{(s + \alpha)(s^{2} + \omega^{2})} } can be simplified as follows:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \cos{\phi_{s}} = \frac{R}{\sqrt{R^{2} + \omega^{2} L^{2}}} \,}

This can be re-arranged as follows:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \cos{\phi_{s}} = \frac{R}{L} \frac{1}{\sqrt{(\frac{R}{L})^{2} + \omega^{2}}} \,}

Likewise, the sine of the power angle is:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta} = 0^o. Here we see the classic transient current waveform for an RL switching (closing) circuit with the time constant R/L.

Related Topics

• Short Circuit
• Transformer Inrush