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The Differential Equation

Applying Kirchhoff's rule of single-valued potential around this loop, we have

$${\cal E} - {1 \over C} Q - R \dot{Q} = 0 \; .$$ (21.1)

When the AC power supply is first turned on, we might expect to see some complicated behaviour that eventually fades away, so that the circuit can settle down to the only plausible "steady-state" motion, namely for $Q$ to oscillate at the same frequency as the driving voltage. We express this expectation as a trial solution:

$$Q(t) = Q_0 e^{i \omega t} \; .$$ (21.2)

Bearing in mind that the constant amplitude $Q_0$ may not be entirely real, let's see if this trial solution (2) "works" - i.e. satisfies the differential equation.

One motive for using the complex exponential form is that it is so easy to take derivatives: each time derivative of $Q(t)$ just "pulls down" another factor of $i \omega$. Thus

$${\cal E}_0 e^{i \omega t} - {1 \over C} Q_0 e^{i \omega t} - i \omega R Q_0 e^{i \omega t} = 0 \; ,$$ (21.3)

from which we can remove the common factor $e^{i \omega t}$ and do a little algebra to obtain

$$Q_0 = { {\cal E}_0/R \over {1/RC} + i \omega } = {{\cal E}_0/R \over \lambda + i \omega}$$ (21.4)

where

$$\lambda \equiv {1 \over RC } \equiv {1 \over \tau} \; .$$ (21.5)

Now, the charge on a capacitor cannot be measured directly; what we usually want to know is the current $I \equiv \dot{Q}$. Since the entire time dependence of $Q$ is in the factor $e^{i \omega t}$, we have trivially

$$I(t) = i \omega Q(t) = I_0 e^{i \omega t}$$ (21.6)

where

$$I_0 = i \omega Q_0 = { i \omega {\cal E}_0/R \over \lambd . . . . . . \omega } = { {\cal E}_0/R \over 1 - i \lambda/\omega } \;.$$ (21.7)

Since everything we might want to know (${\cal E}$, $Q$ and $I$) has the same time dependence except for differences of phase encoded in the complex amplitudes $Q_0$ and $I_0$, we can think in terms of an effective resistance $R_{\rm eff}$ such that

$${\cal E} = I R_{\rm eff} \qquad \hbox{\rm or} \qquad R_{\rm eff} = { {\cal E}_0 \over I_0 } \; .$$ (21.8)

With a little more algebra we can write the effective resistance in the form

$$R_{\rm eff} = R - i X_C$$ (21.9)

where

$$X_C \equiv {1 \over \omega C}$$ (21.10)

is the capacitive reactance of the circuit. This is a quantity that "act like" (and has the units of) a resistance - just like $R$, the first term in $R_{\rm eff}$.

The current through the circuit cannot be different in different places (due to charge conservation) and follows the time dependence of the driving voltage but (because $R_{\rm eff}$ is generally complex) is not generally in phase with it, nor with the voltage drop across $C$:

$$- \Delta {\cal E}_R = I R \; , \qquad \hbox{\rm but} \cr - \Delta {\cal E}_C$$ (21.11)

From Eqs. (9) and (11) one can easily deduce the phase differences between these voltages at any time (for example, $t=\pi/\omega$) when ${\cal E}$ has its maximum negative real value: the voltage drop across $R$ will be real and positive (it is always exactly out of phase with the driving voltage) but the voltage drop across the capacitor will be in the negative imaginary direction - i.e. its real part will be zero at that instant.

A convenient way of looking at this is with the "Phase Circle" shown in Fig. 21.2, where the "directions" of the voltage drops in "complex phase space" are shown as vectors. Both voltage drops "rotate" in this "phase space" at a constant frequency $\omega$ but their phase relationship is always preserved: namely, the voltage across the capacitor lags that across the resistor by an angle of $\pi/2$.^21.4At any instant the actual, measurable value of any voltage is just its real part, i.e. the projection of its complex vector onto the real axis.

Footnotes

. . . . ^21.4

There are many ways of remembering this phase relationship; I prefer to think of it this way: when the current starts flowing there is immediately a voltage drop across the resistor, but it takes a while to charge up the capacitor, so it lags behind. Use whatever works for you.