AC Voltage Applied to a Capacitor

Figure 7.7 shows a simple rectangular loop. On the left side is an ac source (drawn as a circle with a ~ symbol) labelled $\varepsilon$. On the right side is a capacitor, drawn as two short parallel plates, labelled $C$. The top-left corner of the loop is labelled A, and the top-right corner is labelled B. Conductors complete the loop along the bottom.

The figure teaches the purely capacitive AC circuit — a circuit where only a capacitor is connected to an alternating voltage source. The key physical idea is that the capacitor does not block current completely (as it would in a DC circuit after charging). Instead, it alternately charges and discharges as the AC voltage reverses each half-cycle, allowing a continuous oscillating current to flow.

The textbook uses this figure to derive the relationship between voltage and current. From Kirchhoff’s loop rule, the source voltage equals the capacitor voltage at every instant:

where:

• $v$ = instantaneous voltage across the capacitor (and source)
• $v_m$ = peak voltage (amplitude)
• $\omega$ = angular frequency ($2\pi f$)
• $t$ = time
• $q$ = instantaneous charge on the capacitor
• $C$ = capacitance

The current is the rate of change of charge: $i = \frac{dq}{dt}$. Differentiating gives:

where the current amplitude is:

and capacitive reactance is defined as: …

Figure 7.8: Phasor Diagram and Waveform for a Purely Capacitive AC Circuit

This figure has two panels that together show the phase relationship between voltage and current in a circuit containing only a capacitor connected to an AC source.

Panel (a): Phasor Diagram at an instant $t_1$

• The diagram shows two rotating arrows (phasors) representing the voltage ($V$) and the current ($I$).
• The voltage phasor $V$ is drawn pointing up-and-right at an angle $\omega t_1$ from the horizontal axis. Its vertical projection is labelled $v_m \sin \omega t_1$, which is the instantaneous voltage at that instant.
• The current phasor $I$ is drawn a quarter-turn counter-clockwise ahead of $V$ — it is more vertical than $V$. This represents a phase difference of $\pi/2$ radians (90°). Its vertical projection is labelled $i_m \sin(\omega t_1 + \pi/2)$, showing that the current's instantaneous value is ahead of the voltage's by $\pi/2$.
• Key idea: The current phasor leads the voltage phasor by $\pi/2$ as they both rotate counter-clockwise.

Panel (b): Graph of $v$ and $i$ versus $\omega t$

• The horizontal axis is $\omega t$ (in radians), with ticks at $\omega t_1$, $\pi$, and $2\pi$.
• The solid curve represents the voltage $v = v_m \sin \omega t$.
• The dashed curve represents the current $i = i_m \sin(\omega t + \pi/2)$.
• The current reaches its maximum value one-quarter of a period earlier than the voltage does. For example, at $\omega t = 0$, the current is already at its peak $i_m$, while the voltage is zero. This visually confirms that current leads voltage by $\pi/2$.

Physical Idea Taught

In a purely capacitive AC circuit, the capacitor does not allow a steady current (as in DC), but it alternately charges and discharges as the AC voltage reverses. Because the current is the rate of change of charge ($i = dq/dt$), and the charge on the capacitor is proportional to the voltage ($q = Cv$), the current is proportional to the derivative of the voltage. The derivative of $\sin \omega t$ is $\omega \cos \omega t = \omega \sin(\omega t + \pi/2)$, which is why the current is $\pi/2$ ahead of the voltage.

Key Formulas Developed with This Figure

From the textbook derivation using Kirchhoff's loop rule and the relation $i = dq/dt$:

where the current amplitude is:

$i_m = \omega C v_m$ …