<Dd> V L (t) = V e − t R L V R (t) = V (1 − e − t R L). (\ displaystyle (\ begin (aligned) V_ (L) (t) & = Ve ^ (- t (\ frac (R) (L))) \ \ V_ (R) (t) & = V \ left (1 - e ^ (- t (\ frac (R) (L))) \ right) \,. \ end (aligned))) </Dd> <P> Thus, the voltage across the inductor tends towards 0 as time passes, while the voltage across the resistor tends towards 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--as the circuit reaches its steady - state, there is no further current change and ultimately no inductor voltage . </P> <P> These equations show that a series RL circuit has a time constant, usually denoted τ = 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 1 / e of its final value . That is, τ is the time it takes V to reach V (1 / e) and V to reach V (1 − 1 / e). </P> <P> The rate of change is a fractional 1 − 1 / e per τ . Thus, in going from t = Nτ to t = (N + 1) τ, the voltage will have moved about 63% of the way from its level at t = Nτ toward its final value . So the voltage across the inductor will have dropped to about 37% after τ, and essentially to zero (0.7%) after about 5τ . 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 t from V towards 0 . The resistor will be discharged to about 37% after τ, and essentially fully discharged (0.7%) after about 5τ . Note that the current, I, in the circuit behaves as the voltage across the resistor does, via Ohm's Law . </P>

What is the time constant of rl circuit