14.11: RLC Series Transient

A battery of constant $\text{EMF} V$ is connected to a switch, and an $R$, $L$ and $C$ in series. The switch is closed at time $t=0$. We'll first solve this problem by "conventional" methods; then by Laplace transforms. The reader who is familiar with the mechanics of damped oscillatory motion, such as is dealt with in Chapter 11 of the Classical Mechanics notes of this series, may have an advantage over the reader for whom this topic is new – though not necessarily so!

"Ohm's law" is

$$V=Q/C +RI + L\dot I, \label{14.11.1}$$

or

$$LC \ddot Q + RC \dot Q + Q = CV. \label{14.11.2}$$

Those who are familiar with this type of equation will recognize that the general solution (complementary function plus particular integral) is

$$Q=Ae^{\lambda _1t}+Be^{\lambda_2t}+CV, \label{14.11.3}$$

where

$$\lambda _1 = -\frac{R}{2L}+\sqrt{\frac{R^2}{4L^2}-\frac{1}{LC}}$$

and

$$\lambda _2=-\frac{R}{2L} - \sqrt{\frac{R^2}{4L^2}-\frac{1}{LC}}.\label{14.11.4}$$

(Those who are not familiar with the solution of differential equations of this type should not give up here. Just go on to the part where we do this by Laplace transforms. You'll soon be streaking ahead of your more learned colleagues, who will be struggling for a while.)

Case I

$\frac{R^2}{4L^2}-\frac{1}{LC}$ is positive. For short I'm going to write Equations $\ref{14.11.4}$ as

$$\lambda_1 = -a + k \ \text{and} \ \lambda_2 = -a - k. \label{14.11.5}$$

Then

$$Q=Ae^{-(a-k)t}+Be^{-(a+k)t}+CV \label{14.11.6}$$

and, by differentiation with respect to time,

$$I=-A(a-k)e^{-(a-k)t}-B(a+k)e^{-(a+k)t}.\label{14.11.7}$$

At $t=0$, $Q$ and $I$ are both zero, from which we find that

$$A= -\frac{(a+k)CV}{2k} \ \text{and} \ B = \frac{(a-k)CV}{2k}.\label{14.11.8}$$

Thus

$$Q=\left[ -\left( \frac{a+k}{2k}\right)e^{-(a-k)t}+\left(\frac{a-k}{2k}\right) e^{-(a+k)t}+1\right] CV \label{14.11.9}$$

and

$$I=\left[ \left( \frac{a^2-k^2}{2k} \right) \left(e^{-(a-k)t} - e^{-(a+k)t} \right) \right] CV.\label{14.11.10}$$

On recalling the meanings of $a$ and $k$ and the sinh function, and a little algebra, we obtain

$$I= \frac{V}{Lk}e^{-at} \sinh kt. \label{14.11.11}$$

Exercise $\PageIndex{1}$

Verify that Equation $\ref{14.11.11}$ is dimensionally correct. Draw a graph of $I$ : $t$. The current is, of course, zero at $t=0$ and $\infty$. What is the maximum current, and when does it occur?

Case II

$\frac{R^2}{4L^2}-\frac{1}{LC}$ is zero. In this case, those who are in practice with differential equations will obtain for the general solution

$$Q=e^{\lambda t}(A+Bt) + CV, \label{14.11.12}$$

where $$\lambda = -R/(2L), \label{14.11.13}$$

from which $$I= \lambda (A + Bt) e^{\lambda t} + Be^{\lambda t}. \label{14.11.14}$$

After applying the initial conditions that Q and I are initially zero, we obtain

$$Q = CV \left[ 1- \left( 1- \frac{Rt}{2L} \right) e^{-Rt/(2L)} \right] \label{14.11.15}$$

and

$$I=\frac{V}{L}t e^{-Rt/(2L)}. \label{14.11.16}$$

As in case II, this starts and ends at zero and goes through a maximum, and you may wish to calculate what the maximum current is and when it occurs.

Case III

$\frac{R^2}{4L^2}- \frac{1}{LC}$ is negative. In this case, I am going to write equations 14.11.4 as

$$\lambda_1 = -a + j \omega \ \text{and} \ \lambda_2 = -a - j\omega , \label{14.11.17}$$

where $$a = \frac{R}{2L} \ \text{and} \ \omega^2 = \frac{1}{LC} - \frac{R^2}{4L^2}. \label{14.11.18}$$

All that is necessary, then, is to repeat the analysis for Case I, but to substitute $-\omega^2$ for $k^2$ and $j\omega$ for $k$, and, provided that you know that $\sinh j\omega t = j \sin \omega t$, you finish with

$$I= \frac{V}{L\omega}e^{-at} \sin \omega t . \label{14.11.19}$$

This is lightly damped oscillatory motion.

Now let us try the same problem using Laplace transforms. Recall that we have a $V$ in series with an $R$, $L$ and $C$, and that initially $Q, \ I \ \text{and} \ \dot I$ are all zero. (The circuit contains capacitance, so $Q$ cannot change instantaneously; it contains inductance, so $I$ cannot change instantaneously.)

Immediately, automatically and with scarcely a thought, our first line is the generalized Ohm's law, with the Laplace transforms of $V$ and $I$ and the generalized impedance:

$$\bar{V} = [R + Ls + 1/(Cs)]\bar{I}. \label{14.11.20}$$

Since $V$ is constant, reference to the very first entry in your table of transforms shows that $\bar{V}= V/s$, and so

$$\bar{I} = \frac{V}{s[R + Ls + 1/(Cs)]} = \frac{V}{L(s^2 + bs + c)}, \label{14.11.21}$$

where $$b=R/L \ \text{and} \ c=1/(LC). \label{14.11.22}$$

Case I. $b^2 > 4c.$

$$\bar{I}= \frac{V}{L}\left( \frac{1}{(s-\alpha)(s-\beta)} \right)= \frac{V}{L} \left(\frac{1}{\alpha - \beta} \right) \left( \frac{1}{s-\alpha} - \frac{1}{s-\beta}\right). \label{14.11.23}$$

Here, of course, $$2 \alpha = -b + \sqrt{b^2-4c} \ \text{and} \ 2\beta = -b - \sqrt{b^2 - 4c} \label{14.11.24}$$

On taking the inverse transforms, we find that

$$I = \frac{V}{L} \left( \frac{1}{\alpha - \beta}\right) (e^{\alpha t} - e^{\beta t}). \label{14.11.25}$$

From there it is a matter of routine algebra (do it!) to show that this is exactly the same as Equation $\ref{14.11.11}$.

In order to arrive at this result, it wasn't at all necessary to know how to solve differential equations. All that was necessary was to understand generalized impedance and to look up a table of Laplace transforms.

Case II. $b^2 = 4c$.

In this case, Equation $\ref{14.11.21}$ is of the form

$$\bar{I} = \frac{V}{L} \cdot \frac{1}{(s-\alpha)^2}, \label{14.11.26}$$

where $\alpha = -\frac{1}{2}b$. If you have dutifully expanded your original table of Laplace transforms, as suggested, you will probably already have an entry for the inverse transform of the right hand side. If not, you know that the Laplace transform of $t$ is $1/s^2$, so you can just apply the shifting theorem to see that the Laplace transform of $te^{\alpha t}$ is $1/(s-\alpha)^2$. Thus

$$I= \frac{V}{L}t e^{\alpha t} \label{14.11.27}$$

which is the same as Equation $\ref{14.11.16}$.

[Gosh – what could be quicker and easier than that!?]

Case III. $b^2 < 4c$.

This time, we'll complete the square in the denominator of Equation $\ref{14.11.21}$:

$$\bar{I}= \frac{V}{L} \cdot \frac{1}{(s+\frac{1}{2}b)^2 + (c-\frac{1}{4}b^2)}=\frac{V}{L\omega}\frac{\omega}{(s+\frac{1}{2}b)^2+\omega^2}, \label{14.11.28}$$

where I have introduced $\omega$ with obvious notation.

On taking the inverse transform (from our table, with a little help from the shifting theorem) we obtain

$$I = \frac{V}{L\omega} \cdot e^{-\frac{1}{2}bt} \sin \omega t, \label{14.11.29}$$

which is the same as Equation $\ref{14.11.19}$.

With this brief introductory chapter to the application of Laplace transforms to electrical circuitry, we have just opened a door by a tiny crack to glimpse the potential great power of this method. With practice, it can be used to solve complicated problems of many sorts with great rapidity. All we have so far is a tiny glimpse. I shall end this chapter with just one more example, in the hope that this short introduction will whet the reader's appetite to learn more about this technique.