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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.