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The Differential Equation

We begin by picturing a generic series-LCR circuit driven by a sinusoidal voltage ${\cal E}(t) = {\cal E}_0 \cos (\omega t) = \Re \; e^{i \omega t}$. It is convenient to use the complex form^21.2 for calculations; just remember that none of the actual physical quantities like current or voltage will actually have a measurable imaginary part.^21.3 The voltage amplitude ${\cal E}_0$ is taken to be pure real.

  

Figure:  An LCR circuit driven by an AC voltage.

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

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

When the AC power supply is first turned on, we generally have a very complicated behaviour involving the resonant (or "natural") frequency

$$\omega_0 \equiv { 1 \over \sqrt{LC} }$$ (21.2)

and the damping rate

$$\gamma \equiv {R \over 2L } = {1 \over 2 \tau_{_{RL}}} \; ,$$ (21.3)

just like the damped mass on a spring when an oscillatory driving mechanism is applied. If there were no resistor in the circuit, this "ringing" would go on indefinitely; but with the damping caused by R it eventually dies away and the circuit settles down to the only plausible "steady-state" motion, namely for Q to oscillate at the same frequency as the driving voltage:

$$Q(t) = Q_0 e^{i \omega t} \; .$$ (21.4)

Bearing in mind that the constant amplitude Q₀ may not be entirely real, let's see if this trial solution (4) "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

$${\cal E}_0 e^{i \omega t} + \omega^2 L Q_0 e^{i \omega t} - . . . . . . Q_0 e^{i \omega t} - i \omega R Q_0 e^{i \omega t} = 0 \; ,$$ (21.5)

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

$$Q_0 = { {\cal E}_0/L \over {1 \over LC} - \omega^2 + i {R \over L} \omega }$$ (21.6)

or [recalling the definitions (2) and (3)]

$$Q_0 = { {\cal E}_0/L \over \omega_0^2 - \omega^2 + 2i \gamma \omega } \;.$$ (21.7)

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

$$I(t) = i \omega Q(t) = I_0 e^{i \omega t}$$ (21.8)

where

$$I_0 = i \omega Q_0 = { i \omega {\cal E}_0/L \over \omega_0^2 - \omega^2 + 2i \gamma \omega } \;.$$ (21.9)

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 Q₀ and I₀, we can think in terms of an effective resistance $R_{\rm eff}$ such that

$${\cal E} = I R_{\rm eff} \qquad \hbox{\rm or} \qquad R_{\rm eff} = { {\cal E}_0 \over I_0 } \; .$$ (21.10)

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

$$R_{\rm eff} = R - i X_C + i X_L$$ (21.11)

where

$$X_C \equiv {1 \over \omega C} \qquad \hbox{\rm and} \qquad X_L \equiv \omega L$$ (21.12)

are respectively the capacitive reactance and the inductive reactance of the circuit. These are quantities that "act like" (and have the units of) resistances - 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 is always in phase with the driving voltage. This is not necessarily the case for the voltage drops across R, C and L:

 

$$- \Delta {\cal E}_R I R \; , \cr - \Delta {\cal E}_C$$ = (21.13)

From Eqs. (11) and (13) 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 inductance will be in the positive imaginary direction - i.e. its real part will be zero at that instant, as will that of the voltage drop across the capacitor, which is then in the negative imaginary direction.

  

Figure:  The "Phase Circle".

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. All three 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 $\pi/2$ and the voltage across the inductance leads that across the resistor by an angle of $\pi/2$.^21.4 At any other instant the real, measurable value of any of these voltages is just its real part, i.e. the projection of its complex vector onto the real axis.