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Question
Derive an expression for e.m.f. and current in terms of turns ratio
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Solution
Expression for e.m.f and current :
- Let,`phi` be the magnetic flux linked per turn with both the coils at certain instant ‘t’
- Let ‘NP and ‘NS’ be the number of turns of primary and secondary coil,
NP`phi` = magnetic flux linked with the primary coil at certain instant ‘t’
NS`phi` = magnetic flux linked with the secondary coil at certain instant ‘t’ - Induced e.m.f produced in the primary and secondary coil is given by,
`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 turns ratio (transformer ratio) of the transformer - For an ideal transformer,
Input power = Output power - `e_pI_p=e_sI_s`
`e_s/e_p=I_p/I_s`.................(4) - From equation (3) and (4),
`e_s/p_p=N_s/N_p=I_p/I_s`
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