Editing RL circuit (section)

==Series circuit==
[[image:series-RL.png|thumb|right|250px|[[series and parallel circuits#Series circuits|Series]] RL circuit]]

By viewing the circuit as a [[voltage divider]], we see that the [[voltage]] across the inductor is:
:<math>V_L(s) = \frac{Ls}{R + Ls}V_\mathrm{in}(s)\,,</math>

and the voltage across the resistor is:
:<math>V_R(s) = \frac{R}{R + Ls}V_\mathrm{in}(s)\,.</math>

===Current===
The current in the circuit is the same everywhere since the circuit is in series:
:<math>I(s) = \frac{V_\mathrm{in}(s)}{R + Ls}\,.</math>

===Transfer functions===
The [[transfer function]] to the inductor voltage is

:<math> H_L(s) = \frac{ V_L(s) }{ V_\mathrm{in}(s) } = \frac{ Ls }{ R + Ls } = G_L e^{j \phi_L} \,.</math>

Similarly, the transfer function to the resistor voltage is

:<math> H_R(s) = \frac{ V_R(s) }{ V_\mathrm{in}(s) } = \frac{ R }{ R + Ls } = G_R e^{j \phi_R} \,.</math>

The transfer function, to the current, is

:<math> H_I(s) = \frac{ I(s) }{ V_\mathrm{in}(s) } = \frac{ 1 }{ R + Ls }  \,.</math>

====Poles and zeros====
The transfer functions have a single [[pole (complex analysis)|pole]] located at

:<math> s = -\frac{R}{L} \,.</math>

In addition, the transfer function for the inductor has a [[zero (complex analysis)|zero]] located at the [[origin (mathematics)|origin]].

===Gain and phase angle===
The gains across the two components are found by taking the magnitudes of the above expressions:

:<math>G_L = \big| H_L(\omega) \big| = \left|\frac{V_L(\omega)}{V_\mathrm{in}(\omega)}\right| = \frac{\omega L}{\sqrt{R^2 + \left(\omega L\right)^2}}</math>

and

:<math>G_R = \big| H_R(\omega) \big| = \left|\frac{V_R(\omega)}{V_\mathrm{in}(\omega)}\right| = \frac{R}{\sqrt{R^2 + \left(\omega L\right)^2}}\,,</math>

and the [[phase (waves)|phase angles]] are:

:<math>\phi_L = \angle H_L(s) = \tan^{-1}\left(\frac{R}{\omega L}\right)</math>

and

:<math>\phi_R = \angle H_R(s) = \tan^{-1}\left(-\frac{\omega L}{R}\right)\,.</math>

===Phasor notation===
These expressions together may be substituted into the usual expression for the [[phasor]] representing the output:<ref name=":0">{{Cite web |date=2021-04-06 |title=RL Circuit : Working, Phasor Diagram, Impedance & Its Uses |url=https://www.elprocus.com/rl-circuit-working-uses/ |access-date=2022-03-16 |website=ElProCus - Electronic Projects for Engineering Students |language=en-US}}</ref>
:<math>\begin{align}
 V_L &= G_{L}V_\mathrm{in} e^{j \phi_L}\\
 V_R &= G_{R}V_\mathrm{in}e^{j \phi_R}
\end{align}</math>

===Impulse response===
The [[impulse response]] for each voltage is the inverse [[Laplace transform]] of the corresponding transfer function. It represents the response of the circuit to an input voltage consisting of an impulse or [[Dirac delta function]].

The impulse response for the inductor voltage is

:<math> h_L(t) = \delta(t) -\frac{R}{L} e^{-t\frac{R}{L}} u(t) = \delta(t) -\frac{1}{\tau} e^{-\frac{t}{\tau}} u(t) \,,</math>

where {{math|''u''(''t'')}} is the [[Heaviside step function]] and {{math|''τ'' {{=}} ''{{sfrac|L|R}}''}} is the [[time constant]].

Similarly, the impulse response for the resistor voltage is

:<math> h_R(t) = \frac{R}{L} e^{-t \frac{R}{L}} u(t) = \frac{1}{\tau} e^{-\frac{t}{\tau}} u(t) \,.</math>

=== Zero-input response ===
The '''zero-input response''' (ZIR), also called the '''natural response''', of an RL circuit describes the behavior of the circuit after it has reached constant voltages and currents and is disconnected from any power source. It is called the zero-input response because it requires no input.

The ZIR of an RL circuit is:

:<math>I(t) = I(0)e^{-\frac{R}{L} t} = I(0)e^{-\frac{t}{\tau}}\,.</math>

===Frequency domain considerations===
These are [[frequency domain]] expressions. Analysis of them will show which frequencies the circuits (or filters) pass and reject. This analysis rests on a consideration of what happens to these gains as the frequency becomes very large and very small.

As {{math|''ω'' → ∞}}:
:<math>G_L \to 1 \quad \mbox{and} \quad G_R \to 0\,.</math>

As {{math|''ω'' → 0}}:
:<math>G_L \to 0 \quad \mbox{and} \quad G_R \to 1\,.</math>

This shows that, if the output is taken across the inductor, high frequencies are passed and low frequencies are attenuated (rejected). Thus, the circuit behaves as a ''[[high-pass filter]]''. If, though, the output is taken across the resistor, high frequencies are rejected and low frequencies are passed. In this configuration, the circuit behaves as a ''[[low-pass filter]]''. Compare this with the behaviour of the resistor output in an [[RC circuit]], where the reverse is the case.

The range of frequencies that the filter passes is called its [[Bandwidth (signal processing)|bandwidth]]. The point at which the filter attenuates the signal to half its unfiltered power is termed its [[cutoff frequency]]. This requires that the gain of the circuit be reduced to 
:<math>G_L = G_R = \frac{1}{\sqrt 2}\,.</math>

Solving the above equation yields
:<math>\omega_\mathrm{c} = \frac{R}{L} \mbox{ rad/s} \quad \mbox{or} \quad f_\mathrm{c} = \frac{R}{2\pi L} \mbox{ Hz}\,,</math>
which is the frequency that the filter will attenuate to half its original power.

Clearly, the phases also depend on frequency, although this effect is less interesting generally than the gain variations.

As {{math|''ω'' → 0}}:
:<math>\phi_L \to 90^{\circ} = \frac{\pi}{2} \mbox{ radians} \quad \mbox{and} \quad \phi_R \to 0\,.</math>

As {{math|''ω'' → ∞}}:
:<math>\phi_L \to 0 \quad \mbox{and} \quad \phi_R \to -90^{\circ} = -\frac{\pi}{2} \mbox{ radians}\,.</math>

So at [[Direct current|DC]] (0&nbsp;[[Hertz|Hz]]), the resistor voltage is in phase with the signal voltage while the inductor voltage leads it by 90°. As frequency increases, the resistor voltage comes to have a 90° lag relative to the signal and the inductor voltage comes to be in-phase with the signal.

===Time domain considerations===
:''This section relies on knowledge of {{mvar|e}}, the [[E (number)|natural logarithmic constant]]''.

The most straightforward way to derive the time domain behaviour is to use the [[Laplace transform]]s of the expressions for {{mvar|V<sub>L</sub>}} and {{mvar|V<sub>R</sub>}} given above. This effectively transforms {{math|''jω'' → ''s''}}. Assuming a [[Heaviside step function|step input]] (i.e., {{math|''V''<sub>in</sub> {{=}} 0}} before {{math|''t'' {{=}} 0}} and then {{math|''V''<sub>in</sub> {{=}} ''V''}} afterwards):

:<math>\begin{align}
 V_\mathrm{in}(s) &= V\cdot\frac{1}{s} \\
 V_L(s) &= V\cdot\frac{sL}{R + sL}\cdot\frac{1}{s} \\
 V_R(s) &= V\cdot\frac{R}{R + sL}\cdot\frac{1}{s}\,.
\end{align}</math>
<!-- Despite the titles of these images, they do apply in an RL circuit too, as long as they are captioned correctly! -->
[[Image:Series RC resistor voltage.svg|thumb|right|230px|Inductor voltage step-response.]]
[[Image:Series RC capacitor voltage.svg|thumb|right|230px|Resistor voltage step-response.]]
<!-- Despite the titles of these images, they do apply in an RL circuit too, as long as they are captioned correctly! -->
[[Partial fraction]]s expansions and the inverse [[Laplace transform]] yield:
:<math>\begin{align}
 V_L(t) &= Ve^{-t\frac{R}{L}} \\
 V_R(t) &= V\left(1 - e^{-t\frac{R}{L}}\right)\,.
\end{align}</math>

Thus, the voltage across the inductor tends towards 0 as time passes, while the voltage across the resistor tends towards {{mvar|V}}, as shown in the figures. This is in keeping with the intuitive point that the inductor will only have a voltage across as long as the current in the circuit is changing &mdash; as the circuit reaches its steady-state, there is no further current change and ultimately no inductor voltage.

These equations show that a series RL circuit has a time constant, usually denoted {{math|''τ'' {{=}} ''{{sfrac|L|R}}''}} being the time it takes the voltage across the component to either fall (across the inductor) or rise (across the resistor) to within {{math|{{sfrac|1|''e''}}}} of its final value. That is, {{mvar|τ}} is the time it takes {{mvar|V<sub>L</sub>}} to reach {{math|''V''({{sfrac|1|''e''}})}} and {{mvar|V<sub>R</sub>}} to reach {{math|''V''(1 − {{sfrac|1|''e''}})}}.

The rate of change is a ''fractional'' {{math|1 − {{sfrac|1|''e''}}}} per {{mvar|τ}}. Thus, in going from {{math|''t'' {{=}} ''Nτ''}} to {{math|''t'' {{=}} (''N'' + 1)''τ''}}, the voltage will have moved about 63% of the way from its level at {{math|''t'' {{=}} ''Nτ''}} toward its final value. So the voltage across the inductor will have dropped to about 37% after {{mvar|τ}}, and essentially to zero (0.7%) after about {{math|5''τ''}}. [[Kirchhoff's circuit laws#Kirchhoff's voltage law|Kirchhoff's voltage law]] implies that the voltage across the resistor will ''rise'' at the same rate. When the voltage source is then replaced with a [[short circuit]], the voltage across the resistor drops exponentially with {{mvar|t}} from {{mvar|V}} towards 0. The resistor will be discharged to about 37% after {{mvar|τ}}, and essentially fully discharged (0.7%) after about {{math|5''τ''}}. Note that the current, {{mvar|I}}, in the circuit behaves as the voltage across the resistor does, via [[Ohm's law|Ohm's Law]].

The delay in the rise or fall time of the circuit is in this case caused by the [[back emf|back-EMF]] from the inductor which, as the current flowing through it tries to change, prevents the current (and hence the voltage across the resistor) from rising or falling much faster than the [[Time constant|time-constant]] of the circuit. Since all wires have some [[inductance|self-inductance]] and resistance, all circuits have a time constant. As a result, when the power supply is switched on, the current does not instantaneously reach its steady-state value, {{mvar|{{sfrac|V|R}}}}. The rise instead takes several time-constants to complete. If this were not the case, and the current were to reach steady-state immediately, extremely strong inductive electric fields would be generated by the sharp change in the magnetic field &mdash; this would lead to breakdown of the air in the circuit and [[electric arc]]ing, probably damaging components (and users).

These results may also be derived by solving the [[differential equation]] describing the circuit:
:<math>\begin{align}
 V_\mathrm{in} &= IR + L\frac{dI}{dt} \\
 V_R &= V_\mathrm{in} - V_L \,.
\end{align}</math>

The first equation is solved by using an [[integrating factor]] and yields the current which must be differentiated to give {{mvar|V<sub>L</sub>}}; the second equation is straightforward. The solutions are exactly the same as those obtained via Laplace transforms.

===Short circuit equation===

For [[short circuit]] evaluation, RL circuit is considered. The more general equation is:
:<math> v_{in} (t)=v_L (t)+ v_R (t)=L\frac{di}{dt} + Ri </math>

With initial condition:
:<math> i(0) = i_0 </math>

Which can be solved by [[Laplace transform]]:
:<math> V_{in}(s)=sLI-Li_0+RI</math>
Thus:
:<math> I(s)=\frac{Li_o+V_{in}}{sL+R}</math>

Then antitransform returns:
:<math> i(t)=i_0 e^{-\frac{R}{L}t}+\mathcal{L}^{-1}\left[\frac{V_{in}}{sL+R}\right]</math>

In case the source voltage is a [[Heaviside step function]] (DC):
:<math> v_{in}(t)=Eu(t)</math>

Returns:
:<math> i(t)=i_0 e^{-\frac{R}{L}t}+\mathcal{L}^{-1}\left[\frac{E}{s(sL+R)}\right] = i_0 e^{-\frac{R}{L}t}+\frac{E}{R}\left( 1 - e^{-\frac{R}{L}t} \right) </math>

In case the source voltage is a sinusoidal function (AC):
:<math> v_{in}(t)=E\sin(\omega t) \Rightarrow V_{in}(s)= \frac{E\omega}{s^2+\omega^2} </math>

Returns:
:<math> i(t)=i_0 e^{-\frac{R}{L}t}+\mathcal{L}^{-1}\left[\frac{E\omega}{(s^2+\omega^2)(sL+R)}\right] = i_0 e^{-\frac{R}{L}t}+  \mathcal{L}^{-1}\left[\frac{E\omega}{2j\omega}
\left(\frac{1}{s-j\omega} - \frac{1}{s+j\omega}\right)\frac{1}{(sL+R)}\right]</math>
:<math> = i_0 e^{-\frac{R}{L}t}+ \frac{E}{2jL} \mathcal{L}^{-1}
\left[ \frac{1}{s+\frac{R}{L}} \left( \frac{1}{\frac{R}{L}-j\omega} - \frac{1}{\frac{R}{L}+j\omega} \right)
+\frac{1}{s-j\omega}\frac{1}{\frac{R}{L}+j\omega} - \frac{1}{s+j\omega}\frac{1}{\frac{R}{L}-j\omega}
 \right] </math>
:<math> = i_0 e^{-\frac{R}{L}t}+ \frac{E}{2jL} e^{-\frac{R}{L}t} 2j \text{Im}\left( \frac{1}{\frac{R}{L}-j\omega} \right)
 + \frac{E}{2jL} 2j \text{Im}\left( e^{j\omega t} \frac{1}{\frac{R}{L}+j\omega} \right)
</math>
:<math> = i_0 e^{-\frac{R}{L}t}
 + \frac{E\omega}{L \left[  \left(\frac{R}{L}\right)^2  + \omega^2 \right] } e^{-\frac{R}{L}t}
 +  \frac{E}{L \left[  \left(\frac{R}{L}\right)^2  + \omega^2 \right] } \left[ \frac{R}{L}\sin(\omega t) -\omega\cos(\omega t) \right]</math>
:<math> i(t) = i_0 e^{-\frac{R}{L}t}
 + \frac{E\omega}{L \left[  \left(\frac{R}{L}\right)^2  + \omega^2 \right] } e^{-\frac{R}{L}t}
 +  \frac{E}{L \sqrt{ \left(\frac{R}{L}\right)^2  + \omega^2 } } \sin\left[\omega t-\tan^{-1}\left(\frac{\omega L}{R}\right)\right]
</math>