KAHIBARO
Table of Contents
Capacitive Response to Alternating Voltage
A capacitor behaves very differently in an alternating current circuit than it does in a direct current circuit. In a direct current situation, a capacitor eventually becomes fully charged and then stops allowing steady current through the circuit. In an alternating current circuit, the voltage keeps changing with time, so the capacitor is continually charging and discharging. Because of this repeated charge motion, current exists in the circuit even though no charge crosses the insulating gap between the capacitor plates.
The key idea is that a capacitor responds not to the amount of voltage alone, but to how quickly that voltage changes. A rapidly changing voltage causes a larger current than a slowly changing voltage.
Current Voltage Relationship
For any capacitor, the charge and voltage are related by
$$
q = Cv
$$
where $q$ is the charge on the capacitor, $C$ is the capacitance, and $v$ is the voltage across it.
Current is the rate of change of charge, so
$$
i = \frac{dq}{dt}
$$
Combining these gives the basic capacitor law in time dependent circuits,
$$
i = C\frac{dv}{dt}
$$
This equation is the foundation for understanding capacitors in AC circuits. It shows that current depends on the time derivative of voltage.
For a capacitor in any circuit,
$$
i = C\frac{dv}{dt}
$$
This means a capacitor allows more current when the voltage changes more rapidly.
If the voltage across the capacitor is constant, then $\frac{dv}{dt} = 0$, so the current is zero. If the voltage changes with time, then current flows.
Sinusoidal Voltage and Current
In AC circuits, the voltage is often sinusoidal. Suppose the voltage across a capacitor is
$$
v(t) = V_0 \sin(\omega t)
$$
where $V_0$ is the maximum voltage and $\omega$ is the angular frequency.
Using the capacitor law,
$$
i(t) = C\frac{dv}{dt}
$$
we get
$$
i(t) = C\frac{d}{dt}\left(V_0 \sin(\omega t)\right)
$$
so
$$
i(t) = \omega C V_0 \cos(\omega t)
$$
Since $\cos(\omega t) = \sin(\omega t + \pi/2)$, this can also be written as
$$
i(t) = I_0 \sin\left(\omega t + \frac{\pi}{2}\right)
$$
where
$$
I_0 = \omega C V_0
$$
This result shows that the current reaches its maximum one quarter of a cycle before the voltage does.
In a pure capacitive AC circuit, current leads voltage by $90^\circ$.
Phase Difference
The phase difference is one of the most important features of a capacitor in AC. The current does not rise and fall at the same time as the voltage. Instead, the current is ahead.
This happens because current depends on how fast the voltage is changing. When the voltage is crossing through zero, it is changing most rapidly, so the current is largest. When the voltage is at a maximum or minimum, it is momentarily not changing, so the current is zero.
The behavior can be summarized in the following table.
| Quantity | Expression |
|---|---|
| Voltage | $v(t) = V_0\sin(\omega t)$ |
| Current | $i(t) = I_0\sin(\omega t + \pi/2)$ |
| Current amplitude | $I_0 = \omega C V_0$ |
| Phase relation | Current leads voltage by $90^\circ$ |
Capacitive Reactance
In direct current circuits, resistance measures opposition to current. In AC circuits with capacitors, the similar quantity is called capacitive reactance. It tells us how strongly the capacitor opposes alternating current.
It is defined as
$$
X_C = \frac{1}{\omega C}
$$
and its unit is the ohm, $\Omega$.
Using $I_0 = \omega C V_0$, we can rewrite the amplitude relation as
$$
I_0 = \frac{V_0}{X_C}
$$
or for rms values,
$$
I_{\text{rms}} = \frac{V_{\text{rms}}}{X_C}
$$
So a capacitor acts somewhat like a frequency dependent resistance, except that it also shifts phase.
Capacitive reactance is
$$
X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}
$$
A larger frequency or larger capacitance gives a smaller reactance.
This means that high frequency signals pass more easily through a capacitor, while low frequency signals are opposed more strongly.
Frequency Dependence
The frequency dependence of a capacitor is central in AC circuits. Since
$$
X_C = \frac{1}{2\pi f C}
$$
we see that if the frequency $f$ increases, then $X_C$ decreases. That means the current becomes larger for the same applied voltage.
At very low frequency, the capacitor has a very large reactance. It behaves almost like an open circuit. At very high frequency, the reactance becomes small, and the capacitor allows alternating current much more easily.
This is why capacitors are useful in signal circuits, filters, and timing circuits. The detailed study of those applications belongs elsewhere, but the reason behind them is this frequency dependent reactance.
Pure Capacitive Circuit
Consider a circuit containing only an AC source and a capacitor. The source voltage is
$$
v(t) = V_0\sin(\omega t)
$$
Then the current is
$$
i(t) = \omega C V_0 \cos(\omega t)
$$
and the rms relation is
$$
I_{\text{rms}} = \frac{V_{\text{rms}}}{X_C}
$$
The capacitor does not dissipate energy in the way a resistor does. Instead, energy is stored in the electric field when the capacitor charges, and returned to the circuit when it discharges.
Instantaneous Power
The instantaneous power delivered to the capacitor is
$$
p(t) = v(t)i(t)
$$
For a sinusoidal voltage and current, this power changes sign during the cycle. At some times energy flows from the source into the capacitor, and at other times energy flows back from the capacitor to the source.
Because of this back and forth energy exchange, the average power over a complete cycle in an ideal capacitor is zero.
An ideal capacitor in a pure AC circuit has zero average power consumption over one full cycle.
This does not mean nothing happens. Current still flows, and energy is continuously stored and released. It only means the capacitor does not consume net energy over time if it is ideal.
Physical Picture
A good mental picture is to imagine the capacitor as an energy storage device that is repeatedly filled and emptied. When the applied voltage increases, charge builds on the plates and energy is stored in the electric field. When the voltage decreases, the stored energy is released back into the circuit.
Even though electrons do not cross the dielectric between the plates, the changing electric field allows the circuit to respond as if current were passing through the capacitor.
Comparison with a Resistor
A resistor and a capacitor respond differently to AC.
| Feature | Resistor | Capacitor |
|---|---|---|
| Basic relation | $v = iR$ | $i = C\frac{dv}{dt}$ |
| Phase difference | Voltage and current in phase | Current leads voltage by $90^\circ$ |
| Opposition to AC | Resistance $R$ | Reactance $X_C = 1/(\omega C)$ |
| Average power | Positive, energy dissipated | Zero for ideal capacitor |
This comparison helps show that a capacitor is not simply another kind of resistor. Its behavior depends on time variation and phase.
Graphical Interpretation
If voltage is a sine wave, current is also a sine wave, but shifted forward by one quarter cycle. When voltage is at zero and rising, current is maximum positive. When voltage is at maximum, current is zero. When voltage is at zero and falling, current is maximum negative.
Simple Circuit Representation
A pure capacitive AC circuit is very simple in structure.
Key Results
For a capacitor in an AC circuit, the essential equations are the time relation
$$
i = C\frac{dv}{dt}
$$
the reactance
$$
X_C = \frac{1}{\omega C}
$$
and the rms current voltage relation
$$
I_{\text{rms}} = \frac{V_{\text{rms}}}{X_C}
$$
These formulas describe how capacitors respond to alternating voltage, how phase shift appears, and why frequency matters so much. Understanding these ideas is the basis for studying more complicated AC circuits that include resistors, inductors, or combinations of components.