# 14.4 Rl circuits  (Page 2/4)

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The time constant ${\tau }_{L}$ also tells us how quickly the induced voltage decays. At $t={\tau }_{L},$ the magnitude of the induced voltage is

$|{V}_{L}\left({\tau }_{L}\right)|=\epsilon {e}^{-1}=0.37\epsilon =0.37V\left(0\right).$

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 $I\left(0\right)=\epsilon \text{/}R.$ 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

$I\left(t\right)=\frac{\epsilon }{R}{e}^{\text{−}t\text{/}{\tau }_{L}}.$

The current starts at $I\left(0\right)=\epsilon \text{/}R$ and decreases with time as the energy stored in the inductor is depleted ( [link] ).

The time dependence of the voltage across the inductor can be determined from ${V}_{L}=\text{−}L\left(dI\text{/}dt\right)\text{:}$

${V}_{L}\left(t\right)=\epsilon {e}^{\text{−}t\text{/}\tau L}.$

This voltage is initially ${V}_{L}\left(0\right)=\epsilon$ , 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. Time variation of electric current in the RL circuit of [link] (c). The induced voltage across the coil also decays exponentially.

## An RL Circuit with a source of emf

In the circuit of [link] (a), let $\epsilon =2.0V,R=4.0\phantom{\rule{0.2em}{0ex}}\text{Ω},\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}L=4.0\phantom{\rule{0.2em}{0ex}}\text{H}\text{.}\phantom{\rule{0.2em}{0ex}}$ 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] .

## Solution

1. The inductive time constant is
${\tau }_{L}=\frac{L}{R}=\frac{4.0\phantom{\rule{0.2em}{0ex}}\text{H}}{4.0\phantom{\rule{0.2em}{0ex}}\text{Ω}}=1.0\phantom{\rule{0.2em}{0ex}}\text{s}.$
2. The current in the circuit of [link] (b) increases according to [link] :
$I\left(t\right)=\frac{\text{ε}}{R}\left(1-{e}^{\text{−}t\text{/}{\tau }_{L}}\right).$

At $t=0,$
$\left(1-{e}^{\text{−}t\text{/}{\tau }_{L}}\right)=\left(1-1\right)=0;\phantom{\rule{0.2em}{0ex}}\text{so}\phantom{\rule{0.2em}{0ex}}I\left(0\right)=0.$

At $t=2.0{\tau }_{L}$ and $t\to \infty ,$ we have, respectively,
$I\left(2.0{\tau }_{L}\right)=\frac{\text{ε}}{R}\left(1-{e}^{-2.0}\right)=\left(0.50\phantom{\rule{0.2em}{0ex}}\text{A}\right)\left(0.86\right)=0.43\phantom{\rule{0.2em}{0ex}}\text{A},$

and
$I\left(\infty \right)=\frac{\text{ε}}{R}=0.50\phantom{\rule{0.2em}{0ex}}\text{A}.$

From [link] , the magnitude of the induced emf decays as
$|{V}_{L}\left(t\right)|=\text{ε}{e}^{\text{−}t\text{/}{\tau }_{L}}.$

$\text{At}\phantom{\rule{0.2em}{0ex}}t=0,t=2.0{\tau }_{L},\phantom{\rule{0.2em}{0ex}}\text{and as}\phantom{\rule{0.2em}{0ex}}t\to \infty ,$ we obtain
$\begin{array}{ccc}\hfill |{V}_{L}\left(0\right)|& =\hfill & \epsilon =2.0\phantom{\rule{0.2em}{0ex}}\text{V},\hfill \\ \hfill |{V}_{L}\left(2.0{\tau }_{L}\right)|& =\hfill & \left(2.0\phantom{\rule{0.2em}{0ex}}\text{V}\right)\phantom{\rule{0.2em}{0ex}}{e}^{-2.0}=0.27\phantom{\rule{0.2em}{0ex}}\text{V}\hfill \\ & \text{and}\hfill & \\ \hfill |{V}_{L}\left(\infty \right)|& =\hfill & 0.\hfill \end{array}$

## Significance

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] . A generator in an RL circuit produces a square-pulse output in which the voltage oscillates between zero and some set value. These oscilloscope traces show (a) the voltage across the source; (b) the voltage across the inductor; (c) the voltage across the resistor.

## An RL Circuit without a source of emf

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

1. With the switches reversed, the current decreases according to
$I\left(t\right)=\frac{\epsilon }{R}{e}^{\text{−}t\text{/}{\tau }_{L}}=I\left(0\right){e}^{\text{−}t\text{/}{\tau }_{L}}.$

At a time t when the current is one-half its initial value, we have
$I\left(t\right)=0.50I\left(0\right)\phantom{\rule{0.2em}{0ex}}\text{so}\phantom{\rule{0.2em}{0ex}}{e}^{\text{−}t\text{/}{\tau }_{L}}=0.50,$

and
$t=\text{−}\left[\text{ln}\left(0.50\right)\right]{\tau }_{L}=0.69\left(1.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=0.69\phantom{\rule{0.2em}{0ex}}\text{s},$

where we have used the inductive time constant found in [link] .
2. The energy stored in the inductor is given by
${U}_{L}\left(t\right)=\frac{1}{2}L{\left[I\left(t\right)\right]}^{2}=\frac{1}{2}L{\left(\frac{\epsilon }{R}{e}^{\text{−}t\text{/}{\tau }_{L}}\right)}^{2}=\frac{L{\epsilon }^{2}}{2{R}^{2}}{e}^{-2t\text{/}{\tau }_{L}}.$

If the energy drops to $1.0\text{%}$ of its initial value at a time t , we have
${U}_{L}\left(t\right)=\left(0.010\right){U}_{L}\left(0\right)\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}\frac{L{\epsilon }^{2}}{2{R}^{2}}{e}^{-2t\text{/}{\tau }_{L}}=\left(0.010\right)\frac{L{\epsilon }^{2}}{2{R}^{2}}.$

Upon canceling terms and taking the natural logarithm of both sides, we obtain
$-\frac{2t}{{\tau }_{L}}=\text{ln}\left(0.010\right),$

so
$t=-\frac{1}{2}{\tau }_{L}\text{ln}\left(0.010\right).$

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
$t=-\frac{1}{2}\left(1.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)\text{ln}\left(0.010\right)=2.3\phantom{\rule{0.2em}{0ex}}\text{s}.$

## Significance

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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