AMPERE

The ampere is constant current.
Electric current is the same as electric quantity in movement, or quantity per unit time.

I = dq / dt

I = electric current (ampere)
dq = electric quantity (coulomb)
dt = time (s)

CURRENT

A flow of electricity through a conductor is called Current.

Unit : ampere (A)


http://members.tripod.com/ep_381/EI2_LAB4.pdf

RESISTANCE:

It is defined as the property of a substance due to which it opposes the flow of electricity (electrons) through it.

Unit : ohm (Ω)

R ∞ l / A

R = ρ l/A

ρ - resistivity (Ωm) or Specific resistance



CONDUCTANCE:

It is reciprocal of resistance.

G = 1 / R = A / (ρ l) = σ A/l

σ = Conductivity (mho/m) or Specific Conductance

Unit : mho


Ohm’s Law:

The ratio of potential difference (V) between any two points on a conductor to the current (l) flowing between them, is constant, provide the temperature of conductor does not change.

V / I = constant

V / I = R

Why Transformer Rating in kVA?

Cu loss of a transformer depends on current and iron loss on voltage. Hence, total transformer loss depends on volt-ampere (VA) and not on phase angle between voltage and current i.e. it is dependent of load power factor. That is why rating of transformers is in kVA and not in Kw.

LOSSES IN A TRANSFORMER:

1. Core or iron Loss: It includes both hysteresis loss and eddy current loss. Because the core flux in a transformer remains practically constant for all loads (its variation being 1 to 3% from no-load to full-load), the core loss is practically the same at all loads. These losses are minimized by using steel of high silicon content for the core and by using very thin laminations. Iron or core loss is found from the O.C. test. The input of the transformer when on no-load measures the core loss.

2. Copper Loss: This loss is due to the ohmic resistance of the transformer windings. Total Cu loss=I12R1+I22R2=I12R01=I22R02. It is clear that Cu loss is proportional to (current)2 or kVA2. In other words, Cu loss at half the full-load is one-fourth of that at full-load. The value of Cu loss is found from the short-circuit test.

EFFICIENCY OF A TRANSFORMER:

Efficiency = Output/input

But a transformer being a highly efficient piece of equipment, has very small loss, hence it is impractical to try to measure transformer efficiency by measuring input and output. These quantities are nearly of the same size. A better method is to determine the losses and then to calculate the efficiency from;

Efficiency = Output / (Output + Losses) = Output / (Output + Cu loss + iron loss)

= (Input-losses) / Input = 1 – (Losses / input)

Efficiency can be computed by determining core loss from no-load or open-circuit test and Cu loss from the short-circuit test.

The ordinary or commercial efficiency of a transformer is given by the ratio

Output in watts / Input in watts

AUTO-TRANSFORMER:

It is a transformer with one winding only, part of this being common to both primary and secondary. Obviously, in this transformer the primary and secondary are not electrically isolated from each other as is the case in a 2-winding transformer. But its theory and operation similar to that of a two-winding transformer. Because of one winding, it uses less copper and hence is cheaper. It is used where transformation ratio differs little from unity.


E.M.F. EQUATION OF A TRANSFORMER







Let N1 =No. of turns in primary

N2 =No. of turns in secondary

φm =Maximum flux in core in Webbers = Bm × A

f =frequency of a.c. input in HZ

flux increases from its zero value to maximum value φm in one quarter of the cycle i.e. in 1/4f second.

Average rate of change of flux = φm / (1 / 4f) = 4fφm Wb/s or volt

Now, rate of change of flux per turn means induced e.m.f. in volts.

Average e.m.f. / turn = 4fφm volt


In an ideal transformer on no load,

V1 = E1 and E2 = V2

Where V2 is the terminal voltage.

VOLTAGE TRANSFORMATION RATIO (K)

E2/E1=N2/N1=K

This constant K is known as voltage transformation ratio.

If N2>N1 i.e. K>1, then transformer is called step-up transformer.

If N21 i.e. K<1, style="mso-bidi-font-weight:normal">step-down transformer.

Again for an ideal transformer,

Input VA=Output VA

V1I1=V2I2 or I2/I1=V1/V2=1/K

Hence, currents are in the inverse ratio of the voltage transformation ratio.




TRANSFORMER:


WORKING PRINICIPLE OF TRANSFORMER:

A transformer is a device that

1. Transfers electric power from one circuit to another

2. It does so without a change of frequency

3. It accomplishes this by electromagnetic induction

4. Where the two electric circuits are in mutual inductive influence of each other.

Constructionally, the transformers are of two general types,

1. Core-Type

2. Shell-Type. Another recent development is spiral-core or wound-core types the trade name being spirakore transformer. The spiral-core transformer employs the newest development in core construction. The advantages of such construction are

1. A relatively more rigid core

2. Lesser weight and size per kVA rating

3. Lower iron losses at higher operating flux densities

4. Lower cost of manufacture.

Transformers are generally housed in tightly-fitted sheet-metal tanks filled with special insulating oil*. This oil has been highly developed and its function is two-fold. By circulation, it not only keeps the coils reasonably cool, but also provides the transformer with additional insulation not obtainable when the transformer is left in the air.

Classifying the transformers is according to the type of cooling employed. The following types are in common use:-

1. Oil-filled self-cooled

2. Oil-filled water-cooled

3. Air-blast type.


IDEAL TRANSFORMER:

An ideal Transformer is one which has no losses i.e. its windings have no ohmic resistance, there is no magnetic leakage and hence which has no I2R and core losses. In other words, an ideal transformer consists of two purely inductive coils wound on a loss-free core.