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प्रश्न
Explain the working of the transformer.
उत्तर
Principle:
The principle of a transformer is the mutual induction between two coils. That is, when an electric current passing through a coil changes with time, an emf is induced in the neighbouring coil.
Working:
- If the primary coil is connected to a source of alternating voltage, an alternating magnetic flux is set up in the laminated core.
- If there is no magnetic flux leakage, then the whole magnetic flux linked with primary coil is also linked with the secondary coif This means that rate at which magnetic flux changes through each turn is same for both primary and secondary coils.
- As a result of flux change, emf is induced in both primary and secondary coils. The emf induced in the primary coil εp is almost equal and opposite to the applied voltage υp and is given by
υp = εp = -Np `("d" Φ_"B")/"dt"` .....(1) - The frequency of alternating magnetic flux in the core is the same as the frequency of the applied voltage. Therefore, induced emf in secondary will also have the same frequency as that of applied voltage.
- The emf induced in the secondary coil eg is given by
εs = -Ns `("d" Φ_"B")/"dt"`
where Np and Ns are the number of turns in the primary and secondary coil, respectively. If the secondary circuit is open, then εs = υs where υs is the voltage across secondary coil.
υs εs = -Ns `("d" Φ_"B")/"dt"` .....(2)
From equation (1) and (2),
`"v"_"s"/epsilon_"s" = "N"_"s"/"N"_"p"` = K ....(3)
This constant K is known as voltage transformation ratio. For an ideal transformer,
Input power υp ip = Output power υsis
where iυp and is are the currents in the primary and secondary coil respectively. Therefore,
`"v"_"s"/"v"_"p" = "N"_"s"/"N"_"p" = "i"_"p"/"i"_"s"`
Equation (4) is written in terms of amplitude of corresponding quantities,
`"V"_"s"/"V"_"p" = "N"_"s"/"N"_"p" = "I"_"p"/"I"_"s"` = K ....(4)
(i) If Ns > Np ( or K > 1)
∴ Vs > Vp and Is < Ip.
This is the case of a step-up transformer in which voltage is decreased and the corresponding current is decreased.(ii) If Ns < Np (or K < 1)
∴ Vs < Vp and Is > Ip
This is a step-down transformer where voltage is decreased and the current is increased.
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संबंधित प्रश्न
The teachers of Geeta’s school took the students on a study trip to a power generating station, located nearly 200 km away from the city. The teacher explained that electrical energy is transmitted over such a long distance to their city, in the form of alternating current (ac) raised to a high voltage. At the receiving end in the city, the voltage is reduced to operate the devices. As a result, the power loss is reduced. Geeta listened to the teacher and asked questions about how the ac is converted to a higher or lower voltage.
1) Name the device used to change the alternating voltage to a higher or lower value. State one cause for power dissipation in this device.
2) Explain with an example, how power loss is reduced if the energy is transmitted over long distances as an alternating current rather than a direct current.
3) Write two values each shown by the teachers and Geeta.
Can a transformer be used with direct current source? Give reason.
The primary coil of a transformer has 200 turns while the secondary coil has 1000 turns. What type of transformer is this? if the input voltage is 10V, what will be the output voltage?
The primary and secondary coils of a transformer each have an inductance of 200 x 10-6 H. The mutual inductance between the windings is 4 x 10-6 H. What percentage of the flux from one coil reaches the other?
For an ideal step-down transformer, the quantity which is constant for both the coils is ______.
Read the following paragraph and answer the question:
Long distance power transmissions
The large-scale transmission and distribution of electrical energy over long distances is done with the use of transformers. The voltage output of the generator is stepped up. It is then transmitted over long distances to an area sub-station near the consumers. There the voltage is stepped down. It is further stepped down at distributing sub-stations and utility poles before a power supply of 240 V reaches our homes.
A power transmission line feeds input power at 2300 V to a step down transformer with its primary windings having 4000 turns. The number of turns in the secondary in order to get output power at 230 V are ______.
A transformer is essentially an a.c. device. It cannot work on d.c. It changes alternating voltages or currents. It does not affect the frequency of a.c. It is based on the phenomenon of mutual induction. A transformer essentially consists of two coils of insulated copper wire having different numbers of turns and wound on the same soft iron core.
The number of turns in the primary and secondary coils of an ideal transformer is 2000 and 50 respectively. The primary coil is connected to a main supply of 120 V and secondary coil is connected to a bulb of resistance 0.6 Ω.
The value of voltage across the secondary coil is ______.
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.
1 MW power is to be delivered from a power station to a town 10 km away. One uses a pair of Cu wires of radius 0.5 cm for this purpose. Calculate the fraction of ohmic losses to power transmitted if
- power is transmitted at 220 V. Comment on the feasibility of doing this.
- a step-up transformer is used to boost the voltage to 11000 V, power transmitted, then a step-down transfomer is used to bring voltage to 220 V. (ρCu = 1.7 × 10–8 SI unit)
