Our goal was to see how the peak current and voltages vary as functions of the frequency $\omega$. Equations 32.13 and 32.14 are rather complex and best interpreted by looking at graphs. FIGURE 32.11 is a graph of $V_{\text{R}}$ and $V_{\text{C}}$ versus $\omega$.

FIGURE 32.11 Graph of the resistor and capacitor peak voltages as functions of the emf angular frequency $\omega$.

You can see that our qualitative predictions have been borne out. That is, $V_{\text{R}}$ increases from 0 to ${\epsilon }_0$ as $\omega$ is increased, while $V_{\text{C}}$ decreases from ${\epsilon }_0$ to 0. The explanation for this behavior is that the capacitive reactance $X_{\text{C}}$ decreases as $\omega$ increases. For low frequencies, where $X_{\text{C}}\gg R$, the circuit is primarily capacitive. For high frequencies, where $X_{\text{C}}\ll R$, the circuit is primarily resistive.

The frequency at which $V_{\text{R}}=V_{\text{C}}$ is called the crossover frequency ${\omega }_{\text{c}}$. The crossover frequency is easily found by setting the two expressions in Equations 32.14 equal to each other. The denominators are the same and cancel, as does ${\epsilon }_0$, leading to

In practice, $f_{\text{c}}={\omega }_c\text{/}2\pi$ is also called the crossover frequency.

We’ll leave it as a homework problem to show that $V_{\text{R}}=V_{\text{C}}={\epsilon }_0\text{/}\sqrt{2}$ when $\omega ={\omega }_{\text{c}}$. This may seem surprising. After all, shouldn’t $V_{\text{R}}$ and $V_{\text{C}}$ add up to ${\epsilon }_0?$

No! $V_{\text{R}}$ and $V_{\text{C}}$ are the peak values of oscillating voltages, not the instantaneous values. The instantaneous values do, indeed, satisfy $v_{\text{R}}+v_{\text{C}}=\epsilon$ at all instants of time. But the resistor and capacitor voltages are out of phase with each other, as the phasor diagram shows, so the two circuit elements don’t reach their peak values at the same time. The peak values are related by ${\epsilon }_0{\text{}}^2=V_{\text{R}}{\text{}}^2+V_{\text{C}}{\text{}}^2$, and you can see that $V_{\text{R}}=V_{\text{C}}={\epsilon }_0\text{/}\sqrt{2}$ satisfies this equation.