The voltage across the inductor therefore drops to about
$37\text{\%}$ of its initial value after one time constant. The shorter the time constant
${\tau}_{L},$ the more rapidly the voltage decreases.
After enough time has elapsed so that the current has essentially reached its final value, the positions of the switches in
[link] (a) are reversed, giving us the circuit in part (c). At
$t=0,$ the current in the circuit is
$I(0)=\epsilon \text{/}R.$ With Kirchhoff’s loop rule, we obtain
$IR+L\frac{dI}{dt}=0.$
The solution to this equation is similar to the solution of the equation for a discharging capacitor, with similar substitutions. The current at time
t is then
This voltage is initially
${V}_{L}(0)=\epsilon $ , and it decays to zero like the current. The energy stored in the magnetic field of the inductor,
$L{I}^{2}\text{/}2,$ also decreases exponentially with time, as it is dissipated by Joule heating in the resistance of the circuit.
An
RL Circuit with a source of emf
In the circuit of
[link] (a), let
$\epsilon =2.0V,R=4.0\phantom{\rule{0.2em}{0ex}}\text{\Omega},\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}L=4.0\phantom{\rule{0.2em}{0ex}}\text{H}\text{.}\phantom{\rule{0.2em}{0ex}}$ With
${\text{S}}_{1}$ closed and
${\text{S}}_{2}$ open (
[link] (b)), (a) what is the time constant of the circuit? (b) What are the current in the circuit and the magnitude of the induced emf across the inductor at
$t=0,\phantom{\rule{0.2em}{0ex}}\text{at}\phantom{\rule{0.2em}{0ex}}t=2.0{\tau}_{L}$ , and as
$t\to \infty $ ?
Strategy
The time constant for an inductor and resistor in a series circuit is calculated using
[link] . The current through and voltage across the inductor are calculated by the scenarios detailed from
[link] and
[link] .
If the time of the measurement were much larger than the time constant, we would not see the decay or growth of the voltage across the inductor or resistor. The circuit would quickly reach the asymptotic values for both of these. See
[link] .
After the current in the
RL circuit of
[link] has reached its final value, the positions of the switches are reversed so that the circuit becomes the one shown in
[link] (c). (a) How long does it take the current to drop to half its initial value? (b) How long does it take before the energy stored in the inductor is reduced to
$1.0\text{\%}$ of its maximum value?
Strategy
The current in the inductor will now decrease as the resistor dissipates this energy. Therefore, the current falls as an exponential decay. We can also use that same relationship as a substitution for the energy in an inductor formula to find how the energy decreases at different time intervals.
Solution
With the switches reversed, the current decreases according to
Upon canceling terms and taking the natural logarithm of both sides, we obtain
$-\frac{2t}{{\tau}_{L}}=\text{ln}(0.010),$
so
$t=-\frac{1}{2}{\tau}_{L}\text{ln}(0.010).$
Since
${\tau}_{L}=1.0\phantom{\rule{0.2em}{0ex}}\text{s}$ , the time it takes for the energy stored in the inductor to decrease to
$1.0\text{\%}$ of its initial value is
This calculation only works if the circuit is at maximum current in situation (b) prior to this new situation. Otherwise, we start with a lower initial current, which will decay by the same relationship.
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