lecture 1 transformer

9
Single Phase Transformer

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Page 1: Lecture 1  transformer

Single Phase Transformer

Page 2: Lecture 1  transformer

• The efficiency of electrical power transmission has been improved by the use of higher voltages.

• This is one of the main reasons that alternating current (AC) has nearly entirely replaced direct current (DC) for power transmission and distribution.

• While it is true that the AC generator is better than the DC generator for producing higher voltages.

• The transformer is the device or machine primarily responsible for the wide use of AC today.

Page 3: Lecture 1  transformer

• Basically, the transformer is a device for transferring electrical energy from one circuit to another circuit without a change in frequency.

• The transformer accomplishes the change in voltage without use of moving parts, and therein lies its great advantage.

Page 4: Lecture 1  transformer

• The cost per kilowatt is comparatively low, and the efficiency is high.

• As a matter of fact, the transformer is the most efficient piece of electrical machinery, and efficiencies of 98 and 99% are not at all uncommon.

• Since there are no moving parts, maintenance is simpler and cheaper, and the required insulation for the extremely high voltages obtained can more easily be constructed.

Page 5: Lecture 1  transformer

• A transformer is a static (or stationary) piece of apparatus by means of which electrical power in one circuit is transformed into electric power of the same frequency in another circuit.

• It can raise (step up) or lower (step down) the voltage in a circuit but with a corresponding decrease or increase in current.

• Physically, a transformer is mutual induction between two circuits linked by a common magnetic flux.

Page 6: Lecture 1  transformer

Voltage Transformation Ratio (K)

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

transformer

(We will discuss about EMF in next class)

Page 7: Lecture 1  transformer

• For Ideal Transformer, input VA= output VAV1 x I1= V2 x I2= 1/k

Page 8: Lecture 1  transformer

Solve the problem:

The maximum flux density in the core of a 250/3000 volts,50 Hz single phase transformer is 1.2 Wb/m2. If the e.m.f. per turn is 8 volt, determine-

• Primary and secondary turns

Page 9: Lecture 1  transformer

Sol:

E1= N1 x e.m.f. induced/ turn• N1= 250/8= 32• N2= 3000/8= 375