Imagine you have a capacitor, which is like a little electric storage tank. Now, instead of giving it steady electrical power (like from a battery), we connect it to AC voltage. AC voltage is like the electricity that switches back and forth, like a swinging pendulum.
When you connect AC voltage to a capacitor, something cool happens. As the AC voltage changes direction, the capacitor starts filling up with electrical energy when the voltage is high. Then, when the voltage drops, the capacitor releases that stored energy. So, with AC voltage, the capacitor acts like a little energy sponge, soaking up electricity when it’s available and giving it back when needed.
AC Voltage and capacitor
A capacitor is a passive electronic device with two terminals that can store electrical energy in an electric field. When an alternating current (AC) voltage is applied across a capacitor, the behavior is different from that in a direct current (DC) circuit.
In a DC circuit, when a capacitor is connected to a voltage source, the current will flow for a short time required to charge the capacitor. After this short interval, the current stops flowing.
However, in an AC circuit, the voltage across the capacitor continuously changes in a sinusoidal manner. This leads to the capacitor continuously charging and discharging. As a result, there is a continuous flow of current in the circuit.
Let’s consider an AC voltage source applied across a capacitor. The potential difference or the AC voltage can be given as:
\(\displaystyle\boldsymbol{ v = v_m \sin(\omega t)}\)
where vm is the amplitude of the oscillating potential difference and ω is the angular frequency.
Also Read: AC Voltage Applied To An Inductor
The charge q on the capacitor is given by q = Cv, where C is the capacitance.
According to Kirchhoff’s rule, we can write from the above circuit:
\(\displaystyle v_m \sin(\omega t) = \frac{q}{C}\)
The current (i) through the circuit can be calculated using the relation \(\displaystyle i = \frac{dq}{dt}\). This gives us:
\(\displaystyle i = \frac{d(v_m C \sin(\omega t))}{dt} = \omega C v_m \cos(\omega t)\)
Using the relation \(\displaystyle\cos(\omega t) = \sin(\omega t + \frac{\pi}{2})\), we get:
\(\displaystyle i = i_m \sin(\omega t + \frac{\pi}{2})\)
where the amplitude of the current im is given by \(\displaystyle i_m = \omega C v_m\).
Here, we can see that the term \(\displaystyle\boldsymbol{\frac{1}{\omega C}}\) is equivalent to the resistance of this device and is termed as the capacitive reactance, denoted by XC. Thus, we can say that the amplitude of the current in this circuit is given as:
\(\displaystyle i_m = \frac{v_m}{X_C}\)
where, \(\displaystyle\boldsymbol{ X_C = \frac{1}{\omega C}}\).
The capacitive reactance restricts the passage of current in a purely capacitive circuit in the same way as resistance hinders the passage of current in a purely resistive circuit. It’s important to note that the capacitive reactance is inversely proportional to the frequency and the capacitance.
Graph of v and i versus ωt.
A comparison of Eq. i={{i}_{m}}\sin \left( {\omega t+\frac{\pi }{2}} \right)$ with the equation of source voltage, Eq. $$v = v_m \sin(\omega t)$$ shows that the current is π/2 ahead of voltage. The figure shows the phasor diagram at an instant t_1. Here the current phasor I is π/2 ahead of the voltage phasor V as they rotate counterclockwise. The figure shows the variation of voltage and current with time. We see that the current reaches its maximum value earlier than the voltage by one-fourth of a period.
instantaneous power
In an AC circuit with a capacitor, the current (i) and voltage (v) are related as follows:
\(\displaystyle i = i_m \sin(\omega t + \frac{\pi}{2})\)
\(\displaystyle v = v_m \sin(\omega t)\)
Here, im is the peak current, vm is the peak voltage, ω is the angular frequency, and t is time. The term \(\displaystyle\frac{\pi}{2}\) indicates that the current leads the voltage by a phase difference of \(\displaystyle\frac{\pi}{2}\) radians, or 90 degrees.
The instantaneous power (P) in an AC circuit is given by the product of the instantaneous voltage and current:
\(\displaystyle P = vi = v_m i_m \sin(\omega t) \sin(\omega t + \frac{\pi}{2})\)
However, the product of two sinusoidal functions can be rewritten using the trigonometric identity for the product of sines:
\(\displaystyle\sin(A)\sin(B) = \frac{1}{2}[\cos(A-B) – \cos(A+B)]\)
Applying this identity to the power equation gives:
\(\displaystyle P = \frac{1}{2} v_m i_m [\cos(\frac{\pi}{2}) – \cos(2\omega t + \frac{\pi}{2})]\)
Since \(\displaystyle\cos(\frac{\pi}{2}) = 0\), the equation simplifies to:
\(\displaystyle P = -\frac{1}{2} v_m i_m \cos(2\omega t + \frac{\pi}{2})\)
This equation shows that the instantaneous power oscillates at twice the frequency of the voltage and current, and its average value over a cycle is zero. This is because the capacitor stores energy in its electric field during one half of the cycle and returns it to the circuit during the other half. As a result, the net energy supplied to the capacitor over a complete cycle is zero.
However, this is not the standard form of the equation. We want to convert the above equation to the standard form of the equation. To do this, we can use the trigonometric identity that relates cosine and sine:
\(\displaystyle\cos(\theta + \frac{\pi}{2}) = -\sin(\theta)\)
Applying this identity to our power equation gives:
\(\displaystyle p_C = -\frac{1}{2} v_m i_m (-\sin(2\omega t))\)
Simplifying the equation gives the standard form of the equation for instantaneous power across a capacitor:
\(\displaystyle\boldsymbol{ p_C = \frac{1}{2} v_m i_m \sin(2\omega t)}\)
This equation shows that the instantaneous power also oscillates with time and has a frequency double that of the source frequency. The negative sign in the previous equation was due to the phase difference of \(\displaystyle\frac{\pi}{2}\) between the current and voltage. When we take that into account, we get the standard form of the equation.
The average power over a complete cycle is given by:
\(\displaystyle\langle p_C \rangle = \frac{1}{2} v_m i_m \langle \sin(2\omega t) \rangle\)
Since the average value of \(\displaystyle\sin(2\omega t)\) over a complete cycle is zero, the average power supplied to the capacitor over a complete cycle is zero. This is because the capacitor stores energy in its electric field during one half of the cycle and returns it to the circuit during the other half. As a result, the net energy supplied to the capacitor over a complete cycle is zero.
AC Biasing ?
When an alternating current (AC) voltage is applied to a capacitor, the capacitor experiences a cycle of charging and discharging. This is because a capacitor has the ability to store electrical energy in an electric field.
During the positive half cycle of the AC voltage, the capacitor charges up, storing energy. As the AC voltage decreases and becomes negative, the capacitor begins to discharge, releasing the stored energy.
This continuous process of charging and discharging creates a current flow in and out of the capacitor at a regular rate, which is dependent on the frequency of the AC supply. This is why capacitors are often used in AC circuits to filter or regulate the flow of current.
AC biasing in a capacitor results in a continuous cycle of charging and discharging, creating a current flow that corresponds to the frequency of the AC voltage.
FAQs
What happens when an AC voltage is applied to a capacitor?
- When an AC voltage is applied to a capacitor, the capacitor charges and discharges in accordance with the alternating voltage, leading to an alternating current flow through the capacitor.
How does the current through a capacitor behave in an AC circuit compared to a DC circuit?
- In an AC circuit, the current through a capacitor continuously varies with time, alternating in direction as the voltage changes polarity. In contrast, in a DC circuit, the current through a capacitor initially flows to charge the capacitor and then ceases once the capacitor is fully charged.
What is capacitive reactance, and how does it affect the flow of current in an AC circuit?
- Capacitive reactance XC is the opposition offered by a capacitor to the flow of alternating current. It is inversely proportional to the frequency of the AC voltage and the capacitance of the capacitor. Higher capacitive reactance leads to lower current flow for a given AC voltage.
Can a capacitor store energy in an AC circuit? If so, how?
- Yes, a capacitor can store energy in an AC circuit. When an AC voltage is applied to a capacitor, the capacitor charges and stores energy in its electric field during the positive half-cycle of the voltage and releases the stored energy during the negative half-cycle.
What is the phase relationship between voltage and current in a capacitor in an AC circuit?
- In a capacitor in an AC circuit, the current lags behind the voltage by \(\displaystyle 90^\circ\). This means that the current reaches its maximum value \(\displaystyle 90^\circ\) after the voltage has reached its maximum value.
How does the frequency of the applied AC voltage affect the behavior of a capacitor in a circuit?
- The frequency of the applied AC voltage affects the capacitive reactance XC of the capacitor. Higher frequencies result in lower capacitive reactance, leading to higher current flow through the capacitor for a given voltage.
What are the factors affecting the capacitance of a capacitor in an AC circuit?
- The capacitance C of a capacitor in an AC circuit depends on the physical properties of the capacitor, such as the area of the plates, the distance between the plates (dielectric thickness), and the permittivity of the dielectric material. Increasing the plate area or reducing the distance between the plates increases capacitance. Similarly, using a material with higher permittivity also increases capacitance.