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Question
Derive an expression for ratio of e.m.f.s and currents in terms of number of turns in primary and secondary coil.
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Solution
Derivation:
Let ‘Φ’ be the magnetic flux linked per turn with both coils at a certain instant of time ‘t’.
Let the number of turns of the primary and secondary coils be ‘Np’ and ‘Ns’, respectively.
Therefore, the total magnetic flux linked with the primary coil at certain instant of time ‘t’ is NpΦ. Similarly, the total magnetic flux linked with the secondary coil at certain instant of time ‘t’ is NsΦ.
Now, the induced emf in a coil is
`e=(d phi)/dt`
Therefore, the induced emf in the primary coil is
`e_p=(d phi_p)/dt=(dN_p phi)/dt=-N_p(d phi)/dt` .......(1)
Similarly, the induced emf in the secondary coil is
`e_s=(d phi_s)/dt=(dN_s phi)/dt=-N(d phi)/dt` ........(2)
Dividing equations (1) and (2), we get
`e_s/e_p=(-N_s(d phi)/dt)/(-N_p(d phi)/dt)=N_s/N_p` ........(3)
The above equation is called the equation of the transformer and the ratio `N_s/N_p`is known as the turns ratio of the transformer.
Now, for an ideal transformer, we know that the input power is equal to the output
power.
∴`P_p=P_s`
∴`e_p i_p=e_s i_s`
∴`e_s/e_p=i_p/i_s`
From equation (3), we have
`e_s/e_p=N_s/N_P`
∴`e_s/e_p=N_s/N_p=i_p/i_s`
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