# ALTERNATING CURRENT

## How do produced the alternating current. Values of the current and voltage in any time. The maximum values of AC current and voltage. Subject expression and solved solved questions.

**Definition Of The Alternating Current**

The current that its magnitude and direction changes in time is called the alternating current.

**Generation Of the Alternating Current**

If it is moved a bobbin or conductive things in a magnetic area (example: between polars of a magnet) , a change of magnetic flux occurs. The occured magnetic flux change occur a voltage across the conductive. This voltage is called the electromotor force.

This occurred voltage is generate electrons that prepared to flow through the conductive object. If the electrons is find a way for flow, these electrons are flowing along this way. In this way occured current is called the inductive current. The greater the electromotive force generated, the greater the current flowing through the wire.

The voltage that generationed using magnetic flux changed is called alternating voltage. The current that occurred by this voltage is called the alternating current.

Let we take a magnet in circular form and fix to a place. We will wrap a fine Copper wire to on a reel. Next let we placed this bobbin in the magnet ring. Next we will place the end of this reel in shaft bearings in a way that the reel can be turn freely. Let we link up the ends of this spiral to a collector. Next we will make a brush assembly in a way do not erode the collector. We will place all these in a protective casing. So we created a simple alternating current generator. If we turned the reel that the fine wire wrapped, we occur the alternating current.

Even the basic structure of biggest generators are in this way.

The graph above is shown change over time of voltage that taken from alternating current source.

The current get out from the polar A between 0 – t_{1} interval, therefore the polarity B is negative (-) polarity. In this interval the current is out of from polar A and flowing through a resistance, and comes in the polarity B.

The current is ouput from the "B" pole in range interval t_{1} - t_{2}. In this interval the polar "B" is positive (+) , and the polar "A" is negative (-) polar. The current emerges from polar "B" and flows through a resistance, and come to the polar "A".

In the range t_{2} - t_{3}, beings the same situation as in the range 0 - t_{1}.

The current not get out two direction in the DC current circuit, it is get out only from one polar. Its graph only a straight line form, no ripple.

The batteries, accumulators and solar cell are example to the direct current. The alternating current can be convert to the direct current. For this, used the rectifier circuits.

**Period And Angular Frequency In Alternating Current Circuits**

We can think the system in the figure as a shaft connected to the reel that wrapped with wire.

If the shaft that connected to the winding wire turns "ϑ" in "t" time, the angular velocity of the shaft is

ω = | ϑ | |

t |

The time elapsed for the arrowed rod starting to move from point 0 and come back again to the point 0 is called one period. The number of revolutions of the wire winding per unit time is called frequency. The frequency is reverse of the period.

f = | 1 | |

T |

T= | 1 | |

f |

According to the SI unit system, the unit of the period is seconds (represented by s), and the unit of the frequency is 1/second (s^{-1}=Hertz).

When the system of reel do one full cycle, the system is rotates up to angle of 2π.

If we write 2π instead of ϑ in the expression of the ω = ϑ/t

It will be ω = | 2•π | |

T |

1 | = f due to | |

T |

It will be ω = 2•π•f

Φ = B•A•cosϑ

The reel rotate up to ω.t at the "t" time, this rotate is equal to the angle of ω•t. It will be ϑ = ω•t.

Φ = B•A•cos ω•t

Let we write these formulas in its places in the Electromotive force equations.

ɛ = | ∆Φ | |

∆t |

ɛ = | - ∆ (B•A•cos ωt) | |

∆t |

The equality above is equal to the below.

ɛ = B•A•ω•sinω•t

If the reel has number of N winding, the Electromotive force equation is

ɛ =N• B•A• ω•sinωt

Here, the ɛ is the value of the alternating voltage.

**The Maximum And Instantaneous Values Of The Alternating Current**

Let we study the expression of ɛ = B.A. ω.sinω.t

ω•t = 90° →sinωt =1,

ω•t = 270° →sinωt =-1,

According to these knowledge, the maximum electromotive force will be

ɛ_{max} = N•B•A•ω .

ɛ_{max} = Maximum value of the alternating voltage.

When the reel system perform a full rotate, the angle of the system is being once 90 degrees and once at 270 degrees . The these times are hill point of the sinusoidal wave. In these times is being the inductive voltage is maximum.

We can write the at any moment value of alternating voltage , using the expression of the ɛmax = N•B•A• ω.

ɛ = ɛ_{max}•sinωt

The equality above gives the value of the alternating voltage at any moment.

If the value of ωt being 180° or 0°, in this case sinωt = 0. At these moments beings

ɛ = 0. These moments are points that the sinusoidal wave cut the x-axis.

**The Maximum Value Of The Alternating Current**

I = | ɛ | due to |

R |

I_{max} = | ɛ_{max} | |

R |

The maximum value of the alternating current at any moment is

I = I_{max}. sinωt

**Example:**

The maximum value of the voltage of a alternating voltage source is 300 V.

This generator starting operated at t = 0 moment.

How many volts the value of the voltage of this source after 5 miliseconds?

(sin18° = 0.3 )

**Solution:**

ɛ = ɛ_{max}•Sinωt

ɛ = 300•sin(2•π•f•t)

ɛ = 300•sin(2•π•60•5/1000)

ɛ = 300•sin(π•600/1000)

ɛ = 300•sin(6π/10)

π | = 18° | |

10 |

sin(π/10) = sin18°

Sin18° = 0.3

ɛ = 300•sin18° = 300•0.3

ɛ = 90 V

**Example:**

An alternating voltage source is occurs from number of 400 winding that its each have area of 0.2 m2. The wire winding are in magnetic area of 0.6 T. The shaft to which these windings are connected rotates 50 times per second.

How many Volts the value of maksimum electromotive force of this generator?

ɛ_{max} = N•B•A• ω

N = 400

B = 0.6 T

A = 0.2 m^{2}

f = 50 Hz

ɛ_{max} = 400•0.6•0.2• (2•π•50)

ɛ_{max} = 48•(100•3.14)

ɛ_{max} = 15072 V

RISE KNOWLEDGE

28 January 2019

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