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प्रश्न
A transformer lowers e.m.f. from 220 V to 15 V. If 400 W power is given in primary, calculate (i) the current in primary coil and (ii) the current in secondary coil.
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उत्तर
(i) We know that Power = Ip × Ep
or 400 = Ip × 220
∴ Ip = 1.8 A
(ii) Using Ip × Ep = Is × Es
∴ `"I"_"s" = ("I"_"p" xx "E"_"p")/"E"_"s"`
`"I"_"s" = (1.8 xx 220)/15`
= 26.4 A.
संबंधित प्रश्न
Draw a labeled diagram of a step-down transformer.
For what purpose are the transformers used? Can they be used with a direct current source?
Describe, with the help of a suitable diagram, the working principle of a step-up transformer. Obtain the relation between input and output voltages in terms of the number of turns of primary and secondary windings and the currents in the input and output circuits.
Describe briefly, with the help of labelled diagram, working of a step-up transformer.
A step-up transformer converts a low voltage into high voltage. Does it not violate the principle of conservation of energy? Explain.
Name three losses of energy in a transformer. How are they minimized?
Distinguish between Step up and Step Down Transformer.
The primary of a transformer has 40 turns and works on 100 V and 100 W. Find a number of turns in the secondary to step up the voltage to 400 V. Also calculate the current in the secondary and primary.
Electrical energy is transmitted over large distances at high alternating voltages. Which of the following statements is (are) correct?
- For a given power level, there is a lower current.
- Lower current implies less power loss.
- Transmission lines can be made thinner.
- It is easy to reduce the voltage at the receiving end using step-down transformers.
A 60 W load is connected to the secondary of a transformer whose primary draws line voltage. If a current of 0.54 A flows in the load, what is the current in the primary coil? Comment on the type of transformer being used.
