Advertisements
Advertisements
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.
Advertisements
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
Name two factors on which the magnitude of an induced e.m.f. in the secondary coil depends.
A transformer lowers e.m.f. 220 V to 12 volts. If the number of turns in primary are 8800, how many turns are in secondary coil?
An ideal transformer has 100 turns in the primary and 250 turns in the secondary. The peak value of the AC is 28 V. The rms secondary voltage is nearest to ______
Explain step up and step down transformer?
In a transformer, the number of turns in the primary and the secondary are 410 and 1230 respectively. If the current in primary is 6A, then that in the secondary coil is
A 200V/120V step-down transformer of 90% efficiency is connected to an induction stove of resistance 40 Ω. Find the current drawn by the primary of the transformer.
For an ideal step-down transformer, the quantity which is constant for both the coils is ______.
A transformer consisting of 300 turns in the primary and 150 turns in the secondary gives output power of 2.2 kW. If the current in the secondary coil is 10 A, then the input voltage and current in the primary coil are ______.
For what purpose are the transformers used?
