# AC Voltage Applied to a Capacitor

A pair of conductors separated by some medium can be used as a capacitor. Here we will try and see how a Capacitor behaves when ac is passed through it. We will study an ac circuit with a capacitor and try and get a measure of the current and other parameters.

## The Capacitor in an AC Circuit

Let us suppose I have a capacitor which is connected to a DC source and I find that no current flows through it, so if I connect a lamp to that circuit, then the lamp does not glow which mean no current flows through the capacitor. This seems to make sense because we know that there is an insulating medium present between the plates of a capacitor so current canâ€™t flow through it.

### Browse more Topics under Alternating Current

Now if I connect an AC source with the capacitor, I find that the current is flowing through it, and now if I put a lamp in the circuit the lamp glows, which means current flows in the AC circuit. So a capacitor is an insulator in a DC circuit and a conductor in the AC circuit!

## What Happens on AC Biasing?

Now how is it that current flows in the AC source and not where the DC source is connected? In the case of the DC source, the plates of the capacitor acquire a positive and negative charge respectively. Now during that very short while when the capacitor is getting charged to a voltage which is given by :

$$V=\frac{Q}{C}$$

During this very short interval the lamp will glow, but then after the same interval of time, the current stops flowing. In the case of an AC source, we have an alternating voltage which continuously charges and then discharges the capacitor. While charging the capacitor the voltage across the plates of the capacitor rises and the charge also builds up, and when the voltage across the plates decreases the charge will also decrease.

Then when the voltage increases in opposite direction (i.e when voltage reverses) the capacitor gets charged in the reverse order. Hence the plate which was positively charged will become negative and vice versa. Therefore once again the voltage decreases and comes back to zero and the charge on the plates of the capacitor also becomes zero. Hence, we have :

$$Q=C\times V$$

## Mathematical Expressions

The current flowing in the AC source is given by:

$$i= \frac{\mathrm{d} Q}{\mathrm{d} t}$$

Now the Alternating Voltage applied to a Capacitor is given by:

$$V=V_{m}sin\left ( \omega \times t \right )$$

Where Vm = amplitude of voltage and Ï‰ = angular frequency. Charge Q on capacitor is:

$$Q=C\times V$$

Which can be written as:

$$Q= C\times V_{m}\times sin\left ( \omega \times t \right )$$

Now we can write :

$$i= \frac{\mathrm{d} Q}{\mathrm{d} t}$$

Â $$i=\frac{\mathrm{d} \left ( C\times V_{m}\times sin\left ( \omega \times t \right ) \right )}{\mathrm{d} t}$$

$$i=C\times V_{m}\times \omega \times cos\left ( \omega \times t \right )$$

$$i=\frac{V_{m}\times sin\left ( \omega \times t +\frac{\pi }{2}\right )}{\frac{1}{\omega \times c}}$$

We can see that the current also varies sinusoidally having a phase difference ofÂ  90 degree.

$$i=i_{m}\times sin\left ( \omega \times t +\frac{\pi }{2} \right )$$

### Maximum Current

The maximum current will flow when sin(Î¸) i.e. $$sin\left ( \omega \times t +\frac{\pi }{2} \right )$$ will be unity.

### Capacitive Reactance

Take a look above, the term multiplied by the sin term denotes the value of current im. The term in the denominator can then be treated as some form of a resistance offered to the AC. Therefore, we have:

$$i_{m}=\frac{V_{m}}{X_{c}}$$

Capacitive reactance $${X_{c}}=\frac{1}{\omega \times c}$$

Therefore, we may also say that Capacitive reactance $$X_{c}=\frac{1}{2\pi fc}$$

The SI unit of Capacitive Reactance is Ohm same as that as resistance. The Capacitive reactance is inversely proportional to the frequency of the alternating voltage which is applied, thus for low frequency, the reactance is extremely high and for high frequencies, the reactance decreases.

This also explains that when the frequency is zero then capacitive reactance is infinite. So a capacitor does not allow a DC current to flow through it because the capacitive reactance is infinite.

## Solved Examples For You

Q. A 30 F capacitor is connected to a 240 V,60 Hz circuit. What is the current flow in this circuit?

Solution:Â $$X_{ c }=\frac { 1 }{ 2\pi fc } \\ X_{ c }=\frac { 1 }{ 2\pi \times \left( 60 \right) \times \left( 30\times 10^{ -6 } \right) } \\ X_{ c }=88.42\Omega$$

$$I=\frac{240 V}{88.42\Omega }$$

$$I=2.71\quad Ampere$$

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### 2 responses to “Representation of AC Current and Voltage by Rotating Vectors – Phasors”

1. Tanishk says:

Awesome

2. sridhar Dasari says:

Awesome post…