Editing RL circuit (section)

===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.