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प्रश्न
Describe the construction and working of a transformer with a neat labelled diagram.
Explain the working of a transformer with a neat labelled diagram.
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उत्तर
Principle: It is based on the principle of mutual induction, i.e., whenever the magnetic flux linked with a coil changes, an e.m.f is induced in the neighbouring coil.
Construction: A transformer consists of two sets of coils, primary P and secondary S, insulated from each other. Coil P is called the input coil, and coil S is called the output coil. The two coils are wound separately on a laminated soft iron core.
Working: When an alternating voltage is applied to the primary coil, the current through the coil goes on changing. Hence, the magnetic flux through the core also changes. As this changing magnetic flux is linked with both the coils, an e.m.f. is induced in each coil. The amount of the magnetic flux linked with the coil depends upon the number of turns of the coil.
Let `phi` be the magnetic flux linked per turn with both the coils at a certain instant ‘t’. NP and NS be the number of turns of the primary and secondary coils, respectively.
Then at the instant t, the magnetic flux linked with primary coil `phi_P = N_P phi` and with secondary coil `phi_S = N_S phi`.
The induced emf in primary and secondary coil will be
`e_p = -(dphi_p)/(dt) = -N_p(dphi)/(dt)` ....(1)
`e_s = -(dphi_s)/(dt) = -N_s(dphi)/(dt)` ....(2)
Dividing equation (2) by (1),
∴ `e_s/e_p = N_s/N_p` ....(3)
Equation (3) represents equation of transformer.
The ratio `N_s/N_p` is called the turns ratio (transformer ratio) of the transformer.
For an ideal transformer,
Input power = Output power
∴ ePIP = eSIS
∴ `e_s/e_p = I_p/I_s` ....(4)
From equation (3) and (4),
`e_s/e_p = N_s/N_p = I_p/I_s`
Notes
Students should refer to the answer according to their questions.
APPEARS IN
संबंधित प्रश्न
A group of students while coming from the school noticed a box marked "Danger H.T. 2200 V" at a substation in the main street. They did not understand the utility of a such a high voltage, while they argued, the supply was only 220 V. They asked their teacher this question the next day. The teacher thought it to be an important question and therefore explained to the whole class.
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(ii) Is it possible to use this device for bringing down the high dc voltage to the low voltage? Explain
(iii) Write the values displayed by the students and the teacher.
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