# 14.4 Rl circuits  (Page 2/4)

 Page 2 / 4

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.

## 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] .

## 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.

define electric image.obtain expression for electric intensity at any point on earthed conducting infinite plane due to a point charge Q placed at a distance D from it.
explain the lack of symmetry in the field of the parallel capacitor
pls. explain the lack of symmetry in the field of the parallel capacitor
Phoebe
does your app come with video lessons?
What is vector
Vector is a quantity having a direction as well as magnitude
Damilare
tell me about charging and discharging of capacitors
a big and a small metal spheres are connected by a wire, which of this has the maximum electric potential on the surface.
3 capacitors 2nf,3nf,4nf are connected in parallel... what is the equivalent capacitance...and what is the potential difference across each capacitor if the EMF is 500v
equivalent capacitance is 9nf nd pd across each capacitor is 500v
santanu
four effect of heat on substances
why we can find a electric mirror image only in a infinite conducting....why not in finite conducting plate..?
because you can't fit the boundary conditions.
Jorge
what is the dimensions for VISCOUNSITY (U)
Branda
what is thermodynamics
the study of heat an other form of energy.
John
heat is internal kinetic energy of a body but it doesnt mean heat is energy contained in a body because heat means transfer of energy due to difference in temperature...and in thermo-dynamics we study cause, effect, application, laws, hypothesis and so on about above mentioned phenomenon in detail.
ing
It is abranch of physical chemistry which deals with the interconversion of all form of energy
Vishal
what is colamb,s law.?
it is a low studied the force between 2 charges F=q.q`\r.r
Mostafa
what is the formula of del in cylindrical, polar media
prove that the formula for the unknown resistor is Rx=R2 x R3 divided by R3,when Ig=0.
what is flux
Total number of field lines crossing the surface area
Kamru
Basically flux in general is amount of anything...In Electricity and Magnetism it is the total no..of electric field lines or Magnetic field lines passing normally through the suface
prince
what is temperature change
Celine
a bottle of soft drink was removed from refrigerator and after some time, it was observed that its temperature has increased by 15 degree Celsius, what is the temperature change in degree Fahrenheit and degree Celsius
Celine
process whereby the degree of hotness of a body (or medium) changes
Salim
Q=mcΔT
Salim
where The letter "Q" is the heat transferred in an exchange in calories, "m" is the mass of the substance being heated in grams, "c" is its specific heat capacity and the static value, and "ΔT" is its change in temperature in degrees Celsius to reflect the change in temperature.
Salim
what was the temperature of the soft drink when it was removed ?
Salim
15 degree Celsius
Celine
15 degree
Celine
ok I think is just conversion
Salim
15 degree Celsius to Fahrenheit
Salim
0 degree Celsius = 32 Fahrenheit
Salim
15 degree Celsius = (15×1.8)+32 =59 Fahrenheit
Salim
I dont understand
Celine
the question said you should convert 15 degree Celsius to Fahrenheit
Salim
To convert temperatures in degrees Celsius to Fahrenheit, multiply by 1.8 (or 9/5) and add 32.
Salim
what is d final ans for Fahrenheit and Celsius
Celine
it said what is temperature change in Fahrenheit and Celsius
Celine
the 15 is already in Celsius
Salim
So the final answer for Fahrenheit is 59
Salim
what is d final ans for Fahrenheit and Celsius
Celine
what are the effects of placing a dielectric between the plates of a capacitor
increase the capacitance.
Jorge
besides increasing the capacitance, is there any?
Bundi
mechanical stiffness and small size
Jorge
so as to increase the capacitance of a capacitor
Rahma
also to avoid diffusion of charges between the two plate since they are positive and negative.
Prince