14.10: Generalized Impedance

We have dealt in Chapter 13 with a sinusoidally varying voltage applied to an inductance, a resistance and a capacitance in series. The equation that governs the relation between voltage and current is

$$V=L\dot I + RI + Q/C.\label{14.10.1}$$

If we multiply by $C$, differentiate with respect to time, and write $I$ for $\dot Q$, this becomes just

$$C \dot V = LC \ddot I + RC \dot I + I.\label{14.10.2}$$

If we suppose that the applied voltage $V$ is varying sinusoidally (that is, $V=\hat{V}e^{j\omega t}$, or, if you prefer, $V=\hat{V}\sin \omega t$), then the operator $d^2/dt^2$, or "double dot", is equivalent to multiplying by $-\omega ^2$, and the operator $d/dt$, or "dot", is equivalent to multiplying by $j\omega$. Thus Equation $\ref{14.10.2}$ is equivalent to

$$j\omega CV = -LC\omega^2I+ jRC\omega I + I.\label{14.10.3}$$

That is, $$V=[R+jL\omega + 1/jC\omega]I.\label{14.10.4}$$

The complex expression inside the brackets is the now familiar impedance Z, and we can write

$$V=IZ.\label{14.10.5}$$

But what if $V$ is not varying sinusoidally? Suppose that $V$ is varying in some other manner, perhaps not even periodically? This might include, as one possible example, the situation where $V$ is constant and not varying with time at all. But whether or not $V$ varying with time, Equation $\ref{14.10.2}$ is still valid – except that, unless the time variation is sinusoidally, we cannot substitute $j\omega$ for $d/dt$. We are faced with having to solve the differential Equation $\ref{14.10.2}$.

But we have just learned a neat new way of solving differential equations of this type. We can take the Laplace transform of each side of the equation. Thus

$$C\bar{\dot V} = LC \bar{\ddot I} + RC \bar{\dot I} + \bar{I}.\label{14.10.6}$$

Now we are going to make use of the differentiation theorem, equations 14.7.2 and 14.7.3.

$$C(s\bar{V}-V_0) = LC(s^2\bar{I} - sI_0 - \dot I_0) + RC(s\bar{I} - I_0) + \bar{I}.\label{14.10.7}$$

Let us suppose that, at $t=0$, $V_0$ and $I_0$ are both zero – i.e. before $t=0$ a switch was open, and we close the switch at $t=0$. Furthermore, since the circuit contains inductance, the current cannot change instantaneously, and, since it contains capacitance, the voltage cannot change instantaneously, so the equation becomes

$$\bar{V} = (R+Ls+1/Cs)\bar{I}.\label{14.10.8}$$

This is so regardless of the form of the variation of $V$: it could be sinusoidal, it could be constant, or it could be something quite different. This is a generalized Ohm's law. The generalized impedance of the circuit is $R+Ls+\frac{1}{Cs}$. Recall that in the complex number treatment of a steady-state sinusoidal voltage, the complex impedance was $R+jL\omega+\frac{1}{jCw}$.

To find out how the current varies, all we have to do is to take the inverse Laplace transform of

$$\bar{I}=\frac{\bar{V}}{R+Ls+1/(Cs)}.\label{14.10.9}$$

We look at a couple of examples in the next sections.