## What is an Alternating Current?

*An electric current in which the magnitude of current and polarity both changes periodically with time is called an Alternating Current (AC)*

- Most of electric appliances run on alternating current.
- Electric supply for domestic, commercial and industrial use are mainly AC in nature.

### Advantages of AC over DC supply

AC supply has advantages over DC supply as –

- Generation, transmission and distribution of AC is more economical due to less expensive as compared to that of DC.
- The alternating voltage can be step down or step up easily using a
*transformer.* - AC travels on the surface of a
*conductor*, so costly conducting materials can be saved for long transmission by using inner core of cheaper material say steel e.g.(i.e. aluminium conductor steel reinforced) is widely used for electric supply lines.*ACSR* - AC can reach distant places with reduced loss of
*electric power*(by using a transformer). - Alternating voltage can be better controlled without any loss of electric power (say by using a choke coil).
- AC is easily converted into DC by using rectifiers.
- AC machines are smaller in size, having longer life and easy to use.

### Disadvantages of AC over DC supply

AC supply has some disadvantages over DC supply as –

- AC can be more harmful than DC of same intensity. This is because maximum value or peak value of AC is ( \sqrt {2} ) times that of the effective value. For example – well known domestic supply of ( 220 \ V \ AC ) has a peak value given by [( \sqrt {2} \times 220 \ V ) = 311 \ V ] and harm more as compared to ( 220 \ V \ DC ) .
- AC cannot be used in electrolysis processes such as electroplating, electrotyping and electrorefining, where only DC is used.
- AC in a wire is not uniformly distributed throughout its cross-section. The current density is much greater on the surface of the wire than that inside the wire. This concentration of AC near the surface of the wire is called
. Therefore, inner conducting material of the transmission wire remains unused.*skin effect*

## Peak value of an AC Current

If an AC supply is viewed on an oscilloscope, it will look like a sine or cosine wave as shown in figure.

- A sinusoidal alternating current is expressed as –

I = I_0 \sin ( \omega t ) ……… (1)

Here, ( I_0 ) is called as –

- The Maximum value or Peak value of alternating current or
*Amplitude*of alternating current.

- The
*angular velocity*( \omega ) of alternating current is given by –

\omega = \left ( \frac {2 \pi}{T} \right ) = 2 \pi \nu .

Here ( T ) is the *time period* and ( \nu ) is the *frequency.*

- The variation of alternating current with time is shown in figure.

Similarly, sinusoidal voltage of an alternating current source is given by –

V = V_0 \sin ( \omega t ) …….. (2)

Here, ( V_0 ) has the similar nomenclature as that of current. it may be – (1) maximum value or Peak value of alternating voltage or (2) *amplitude* of alternating voltage.

## Average value of an AC Current?

*Mean or average value of alternating current is defined as that value of steady current which sends the same amount of charge through a circuit in certain time interval as is sent by an alternating current through the same circuit in half cycle.*

- Average value or mean value of an AC over half cycle is found to be ( 63.7 \ \% ) of its peak value ( I_0 ) .

Therefore, \quad I_{av} = 0.637 ( I_0 )

Let, an alternating current is represented by ( I = I_0 \sin ( \omega t ) )

From definition of electric current, we have –

I = \left ( \frac {dq}{dt} \right )

Or, \quad dq = I dt

### Charge sent by AC in half cycle

The charge sent by the alternating current ( I ) in time ( dt ) will be –

dq = I_0 \sin ( \omega t ) dt

Therefore, the charge sent by AC in the first half cycle i.e. from ( t = 0 ) to \left ( t = \frac {T}{2} \right ) will be –

\int dq = \int\limits_{0}^{\frac {T}{2}} I_0 \sin ( \omega t ) dt

So, \quad q = I_0 \int\limits_{0}^{\frac {T}{2}} \sin ( \omega t ) dt

= I_0 \left [ \frac {- \cos ( \omega t )}{\omega} \right ]_{0}^{\frac {T}{2}}

= - \frac {I_0}{\omega} \left [ \cos ( \omega t ) \right ]_{0}^{\frac {T}{2}}

= - \frac {I_0}{2 \pi / T} \left [ \cos \left ( \frac {2 \pi}{T} \right ) t \right ]_{0}^{\frac {T}{2}}

Since, \quad \omega = \left ( \frac {2 \pi}{T} \right )

So, \quad q = - \frac {I_0 T}{2 \pi} \left [ \cos \left ( \frac {2 \pi}{T} \times \frac {T}{2} \right ) - \cos 0 \right ]

= - \frac {I_0 T}{2 \pi} \left [ \cos \pi - \cos 0 \right ]

= - \frac {I_0 T}{2 \pi} \left [ - 1 - 1 \right ]

= \left ( \frac {I_0 T}{\pi} \right ) …… (1)

### Charge sent by Average value of AC in half cycle

Let, ( I_{av} ) is the mean or average value of AC over positive half cycle, then the charge sent by it in time \left ( \frac {T}{2} \right ) will be –

q = I_{av} \times \left ( \frac {T}{2} \right ) …… (2)

Therefore, as per definition of mean or average value of current, we have –

\text {Charge sent by AC in half cycle} = \text {Charge sent by average value of AC in half cycle}

- Therefore, equation (1) and equation (2) must be equal.

Therefore, \quad I_{av} \times \left ( \frac {T}{2} \right ) = \left ( \frac {I_0 T}{\pi} \right )

So, \quad I_{av} = \left ( \frac {2 I_0}{\pi} \right ) = 0.637 \ I_0

Also, \quad V_{av} = \left ( \frac {2 V_0}{\pi} \right ) = 0.637 \ V_0

**TO BE NOTED –**

- The mean or average value of alternating current or voltage over a complete cycle becomes zero.
- Hence, ordinary DC instruments like
*Ammeter*and*Voltmeter*cannot measure AC current or voltage. In AC supply they will give zero reading.

## RMS value of an AC Current

*Root Mean Square value of alternating current is defined as that steady current which produces the same amount of heat in a conductor in a certain time interval as is produced by the AC in the same conductor during the time period of full cycle i.e. * ( T ) .

- Root Mean Square value of alternating current is also known as
or**effective value**.**virtual value**

Let, an alternating current flows through a conductor of *resistance* ( R ) for time ( dt ) .

- Then, heat produced in the conductor will be –

dH = I^2 R dt

= [ I_0 \sin ( \omega t )]^2 R dt

= I_0^2 R \sin^2 ( \omega t ) dt

### Heat produced by an AC in one cycle

- When AC current flows for time period from ( t = 0 ) to ( t = T ) , the heat produced will be –

\int dH = \int\limits_{0}^{T} I_0^2 R \sin^2 ( \omega t ) dt

So, \quad H = I_0^2 R \int\limits_{0}^{T} \sin^2 ( \omega t ) dt

From trigonometric relations we get –

\sin^2 \theta = \left [ \frac {1 - \cos ( 2\theta )}{2} \right ]

Therefore, \quad H = I_0^2 R \int\limits_{0}^{T} \left [ \frac { 1 - \cos ( 2 \omega t )}{2} \right ] dt

= \left ( \frac {I_0^2 R}{2} \right ) \int\limits_{0}^{T} \left [ 1 - \cos ( 2 \omega t ) \right ] dt

= \left ( \frac {I_0^2 R}{2} \right ) \left [ \int\limits_{0}^{T} dt - \int\limits_{0}^{T} \cos ( 2 \omega t ) dt \right ]

= \left ( \frac {I_0^2 R}{2} \right ) \left [ \left [ t \right ]_{0}^{T} - \left [ \frac {\sin ( 2 \omega t )}{2 \omega} \right ]_{0}^{T} \right ]

= \left ( \frac {I_0^2 R}{2} \right ) \left [ \left ( T - 0 \right ) - \frac {1}{2 \omega} \left [ \sin \left ( 2 \times \frac {2 \pi}{T} t \right ) \right ]_{0}^{T} \right ]

= \left ( \frac {I_0^2 R}{2} \right ) \left [ T - \frac {1}{2 \omega} \left [ \sin \left ( 2 \times \frac {2 \pi}{T} \times T \right ) - \sin 0 \right ) \right ]

= \left ( \frac {I_0^2 R}{2} \right ) \left [ T - \frac {1}{2 \omega} \left ( \sin 4 \pi - \sin 0 \right ) \right ]

= \frac {I_0^2 R}{2} \left [ T - \frac {1}{2 \omega} \left ( 0 - 0 \right ) \right ]

Therefore, \quad H = \left ( \frac {I_0^2 R}{2} \right ) T ……. (1)

### Heat produced by RMS value of AC in one cycle

Let, RMS value of an AC current is ( I_{rms} ) which flows through the conductor. Then –

H = ( I_{rms} )^2 R T ……. (2)

Therefore, as per definition of RMS value of current, we have –

\text {Heat produced by AC in one cycle} = \text {Heat produced by RMS value of AC in one cycle}

Therefore, equation (1) and equation (2) must be equal.

Hence, \quad ( I_{rms} )^2 RT = \left ( \frac {I_0^2 RT}{2} \right )

So, \quad I_{rms} = \left ( \frac {I_0}{\sqrt {2}} \right ) = 0.707 \ I_0

Also, \quad V_{rms} = \left ( \frac {V_0}{\sqrt {2}} \right ) = 0.707 \ V_0

## AC phasor diagram

*Phasor is a rotating vector which represents a quantity which is varying sinusoidal with time.*

An AC phasor is considered as rotating in anticlockwise direction with an angular speed ( \omega ) .

Instantaneous voltage of AC source is given by –

V = V_0 \sin ( \omega t )

So, an alternating current or voltage can be represented in phasor diagram as shown in figure.

In phasor diagram, length of phasor i.e. ( V ) is proportional to ( V_0 ) and it describes an angle ( \theta ) in time ( t ) , where ( \theta = \omega t ) and ( \omega ) is the *angular frequency* of the voltage of A.C source.

The vertical component of phasor i.e. [ V = V_0 \sin ( \omega t ) ] represents the instantaneous value of the quantity varying sinusoidal.