Transient States in RL Circuits: The Journey to Steady-State

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Transient States in RL Circuits: The Journey to Steady-State 

In this article, we attempt to obtain the complete solution of an RL circuit, including the transient state. We first consider the case of a DC voltage, where the current gradually builds up and after that, we take up the special case of AC Voltage which is often skipped in elementary courses.



Fig: Growth of current in cases of DC and AC voltages


DC Voltage:

A DC Voltage refers to a steady voltage that does not vary with time. The below diagram shows a simple DC circuit involving a resistor `R` and an inductor `L`. A cell acts as the source of a DC emf E.



Fig: A simple RL circuit with DC emf

From Kirchhoff's Voltage Law, we know that the magnitude of the net Voltage drop across a loop is equal to the magnitude of the net emf in that loop. The voltage drop due to the resistor is `iR` and due to the inductor is `L \frac{di}{dt}`. Thus we may write,

`L\frac{di}{dt} + IR = \mathcal{E}`

`\frac{di}{dt} + \frac{R}{L}i =\frac{ \mathcal{E}}{L}`


This is a linear differential equation of first order and can be simply solved using an interesting factor. The integrating factor here is `e^{\int_{ }^{ } \frac{R}{L} dt} = e^{\frac{Rt}{L}}`. The above then transforms into,


`ie^{\frac{Rt}{L}} = \int_{\text{}}^{\text{}} (\frac{ \mathcal{E}}{L} e^{\frac{Rt}{L}}) dt`

` ie^{\frac{Rt}{L}} = \frac{ \mathcal{E}}{L} e^{\frac{Rt}{L}} \cdot \frac{L}{R}+ c`

`ie^{\frac{Rt}{L}} = \frac{ \mathcal{E}}{R} e^{\frac{Rt}{L}} + c`


If we put the the initial condition of `i = 0` when `t= 0`, we get ` c = \frac{- \mathcal{E}}{R}`. Whence, putting the value of `c` and taking the term `e^{\frac{Rt}{L}}` to the right we get the equation for current as,

$$ \boxed{i = \frac{\mathcal{E}}{R}(1-e^{\frac{-Rt}{L}})} $$

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This is the complete solution for an RL Circuit connected to a DC voltage.




Fig: Variation of current in the circuit.

See how the current gradually grows? We thus make the following inferences.

Inferences:

  • At t = 0: At `t = 0`, the factor `e^{\frac{-Rt}{L}}` is equal to `1`, so that the factor `(1-e^{\frac{-Rt}{L}})= 0`. Thus, the current at the beginning is zero as we desired it to be. 
  • As t increases from 0: As `t \uparrow 0`, the factor `e^{\frac{-Rt}{L}}` becomes a little and little lesser than 1, thus making `(1-e^{\frac{-Rt}{L}})` a little and little more than 0. i.e. `i` increases a little and little more from 0. This increase continues, however with the rate of increase declining slowly.
  • As t →∞:  As `t \rightarrow ∞`,  `e^{\frac{-Rt}{L}} \rightarrow 0`, implying `(1 - e^{\frac{Rt}{L}}) \rightarrow 1`. Consequently , `i \rightarrow \frac{\mathcal{E}}{R}`.


AC Voltage:

We saw the variation of current in a DC RL circuit and explored its transient state which is already quite popular. Now, we turn to the special case of RL circuits connected to a AC voltage. This will give us many new insights into the actual behavior of alternating current in the transient state. It may also break some of your previously held notions!

The following is a simple AC circuit involving a resistor `R` and an inductor `L` connected in series. A voltage source creates a sinusoidally varying voltage `V_{0} sin(wt)` in the circuit.



Fig: A simple RL circuit with Alternating emf


Again by Kirchhoff's law, 

` L\frac{di}{dt} + iR = V_{0} sin(wt) `

` \frac{di}{dt} + \frac{R}{L} i = \frac{V_{0}}{L} sin(wt) `

This is again a ordinary linear differential equation, and thus we use the integrating factor `e^{\frac{Rt}{L}`. The above then transforms into,

` d(ie^{\frac{Rt}{L}}) = \frac{V_{0}}{L} e^{\frac{Rt}{L}) sin(wt) dt `

Integrating both sides, we have, 

` \int` ` d(ie^{\frac{Rt}{L}}) = \frac{V_{0}}{L}` ` \int ` ` e^{\frac{Rt}{L}) sin(wt) dt `

` i = \frac{V_{0}e^{\frac{-Rt}{L}}}{L}`  `\int ` `e^{\frac{Rt}{L}}  sin(wt) dt `


To integrate the right hand side, we can write `sin(wt) = Im(e^{jwt})`. Then we have, 

` i = \frac{V_{0}e^{\frac{-Rt}{L}}}{L} ` `\int` ` e^{\frac{Rt}{L}} Im(e^{jwt}) dt `

` i = \frac{V_{0}e^{\frac{-Rt}{L}}}{L} ` `\int` `Im(e^{\frac{Rt}{L} + jwt}) dt`

It only remains for us to integrate the expression. Alternatively, we could have also used the Laplace Transform. The complete solution can be seen here (I have come across a few minor mistakes in the pdf, it will be corrected soon). In either case, we arrive at the final equation, 

$$ \boxed{ i = \frac{V_{0}}{\sqrt{R^{2} + (wL)^{2}}} [sin(wt - φ)+ sin(φ) e^{\frac{-Rt}{L}}]} $$

where `φ = tan^{-1}(\frac {wL}{R}) `

This is the complete solution of an RL Circuit connected to an AC Voltage.




Fig: Variation of current in the circuit 

Inferences: 

  • At t = 0: At `t = 0` , the factor `e^{\frac{-Rt}{L}} = 1` , so that the factor `sin(φ)e^{\frac{-Rt}{L}} = sin(φ)`. Consequently, the whole term inside the bracket is equal to 0. Thus, the current at the beginning is zero as set by us.
  • As `t` increases from 0: As `t` increases, the factor `e^{\frac{-Rt}{L}}` becomes smaller & smaller, this reducing the effect of the term involving `sin(φ)`, that produced the variation in the sine wave. Our curve thus approaches a sinusoid.
  • As ` t \rightarrow ∞ `: As  `t \rightarrow  ∞`,  `sin(φ) \rightarrow 0`. Thus, `sin(ωt - φ) + sin(φ)e^{\frac{Rt}{L}} \rightarrow sin(ωt - φ)`. Thus, we achieve the popular steady state equation, 

`i(t) \approx \frac{V}{\sqrt{R^{2} + (ωL)^2}} sin(ωt - φ) `


Analysis:

The deviation factor of `sin(φ)` is quite interesting. It's particularly peculiar because it seems rather arbitrary. One step ahead would be to open this, to see what exactly it is composed of. Since we had, 

`tan(φ) = \frac{ωL}{R} `

` \implies sin(φ) = \frac{wL}{\sqrt{R^{2} + (wL)^{2}}} `

But this still is not particularly helpful, because it just doesn't look anything familiar. So this is just not enough. Then, let's try some other changes. What if our sine voltage, instead of starting from zero, started at some initial phase, say, `Φ`? I.e. what if `v = V_{0} sin(wt + Φ)` ?. In this case the complete current equation would be of the form[1], 


`  i = \frac{V_{0}}{Z} [sin(wt + Φ - φ)+ sin(φ - Φ) e^{\frac{-Rt}{L}}] `


The deviation factor is now `sin(φ - Φ)`. So for example if we were to start to from the peak value of voltage (as is standard in some regions), we would require a cosine wave. For this we may set `Φ = + \frac{π}{2}`. Then, the deviation factor would become,

 `sin(φ - \frac{π}{2}) = - cos(φ) `





Do It Yourself!

Coming soon..,

Power Consumption:

Below is a graph of power consumption and its variation with time. See that every time, the power consumed by the passive circuit elements is more than power supplied by them. A proper do-it-yourself will be coming soon.



Fig: Graph representing variation of different parameters in an AC RL Circuit

Red: Voltage
Blue: Current
Purple shaded: Power
Green: Total Energy Consumed 




_______________________________________________



Notes:

[1] For notational simplicity, we use the impedance of the series circuit,

`Z = \sqrt{R^2 + (wL)^2} `






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